# Mechanics - Dynamics

August 17, 2017 | Author: Sherif Yehia Al Maraghy | Category: Momentum, Acceleration, Force, Velocity, Collision

#### Short Description

Newton's 1st, 2nd and 3rd laws of motion, applications on pulleys, lifts, problems containing resistance and frictio...

#### Description

Dynamics

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Revision Introduction : Dynamics is a branch of mechanics which deals with moving bodies under the Action of some influences of forces .  Velocity Vector

 V : A velocity vector is a velocity with direction .

The velocity is measured by : Km / hr 1 km / hr  Or we can say : 1 km / hr   Displacement vector

, m / sec Or

cm / sec where :

1000 m 5  m / sec 60  60 sec 18

‫ي ح فظ‬

1000  100cm 250  cm / sec 60  60 sec 9

‫ي ح فظ‬

S :

The displacement of a particale during an interval of time t  t2  t1 is the change of the position of the particle from its initial position A

Dicplacement Distance

at t1 to its terminal position B at t2 . ----------------------------------------------------------------------------------------------------------------------

Equations of motion “uniform acceleration”

 1 V  u  a t

" we use it if displacement is not given "

1 S  u t  a t 2 " we use it if velocity is not given " 2 2 2  3  V  u  2 a s " we use it if time is not given " These equations are used only when a body is travelling in a straight line with

2

Constant  uniform  acceleration  If positive  Or deceleration  If negative  . Notes

 i  S  here equals to the whole displacement of the whole trip  ii  If the body started from rest : Initial velocity  u  vo  Zero  iii  If the body stopped after a time  t  : Final velocity  V  Zero  iv Uniform velocity means that : S  V T  as a  0 

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Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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The Vertical motion “Downwards” 1 V  U  gt "This equation to calculate the velocity of a body for a time period of t seconds" 1 2 gt 2 " This equation can be used to calculate the distance traveled by a body in a time t"

2 S  U T  3 V 2

 U 2  2 gs

" the above equation gives a relation between the final velocity v of the body and the distances traveled by the body"

Important Notes :

1 g , v , u , s are all positive in this case    .  2  The distance in this case  measure of the displacement.  3  The expression : Left to fall or fell means U  0 .  4  The expression : The body is projected by Veloctiy... " This is the Initial velocity : "U"

Second : Vertical Motion Upwards  1 V  u  g t 1 2 gt 2  3  V 2  u 2  2g s

2 S  u t 

Very Important Remarks : Cases of Displacements  S 

---------------------------------------------------------------------------------------------------------------------Dynamic – 3rd secondary

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Finding the Time and Distance of maximum height :  t Time of Max. height    Smax 

u g

S max

u2 2g

t max

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Expressions could help you  Left to fall  Fall from rest  Comes to rest

initial velocity = zero  u  0  initial velocity = zero  u  0  terminal velocity = zero . V  0 

Gravity will be negative  g  -9.8 m/sec 2 

 Projected upwards    

Uniform velocity acceleration = zero And S = V t Maximum velocity acceleration = zero Maximum height velocity = zero V  0  Mass of drop (each ) drop when it reaches the ground

Mass = original mass of the drop + (Rate of increase  time)  Retarded acceleration = -ve ----------------------------------------------------------------------------------------------------------------------

Differentiation vector Using the algebraic measure of vectors S  r  r0 dS dr  dt dt dS 2  V  dt dV 3  A  dt ----------------------------------------------------------------------------------------------------------------------

1

Dynamic – 3rd secondary

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Very important remarks Distance during a given interval of time If  S  is a distance covered during n th seconds for example : Ex  1 : The distance covered in the 5 th seconds :

45  4.5 s from the begining of time 2 Ex  2  : The distance covered in the 5 th and 6 th seconds :

Here 5th starts from 4  5 seconds  Then t 

4 6  5 s from the begining of time 2 Ex  3  : The distance covered in the 3 rd , 4 th and 5th seconds : Time starts from 4  5  6 seconds  Then t 

Time starts from 2  3  4  5 seconds  Then t 

25  3.5 seconds 2

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The Relative Velocity We know that the motion is something relative to another and it changes in description by the Change of an observer . For Example : If you look from the train window to the moving cars in the same direction of the motion of the train , They seen to move slowly , While we feel the contrary if the cars are moving in the opposite direction of the train . So , If the velocity vector of A is VA and the velocity vector of B is VB Then , The relative velocity vector of B with respect to A is denoted VBA Where : VB A  VB  VA

Or

we can say : VB  VBA  VA

Velocity of B  Velocity of B w.r. to A  Velocity of A

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Dynamic – 3rd secondary

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Example (1) A car moves on a straight road with velocity 82 km / hr . If it meets a motor - cyclist moves on

the same road with velocity 43 km / hr , Find the relative velocity of the cyclist with respect to the car in two cases :

 First  : The cyclist and the car move in the same diection .  Second  : The cyclist and the car move in the opposite direction . Answer 1st step : let the positive direction in the direction of the relative velocity  here VBA   First : both are in the same direction x

let VA  Vcar  82 x

VA  82

VB  Vcycle  43 x  VBA  VB  VA

VB  43

A VBA  43 x  82 x  -39 x

B

C

 The Magnitude of the relative velocity of the cyclist with respect to the car  39 km / hr  Second : both are in opposite direction

VBA  VB  - VA

x

VB VA

A B VBA  43 x  82 x  VBA  125 x   Its Magnitude is 125 km / hr . ---------------------------------------------------------------------------------------------------------------------Example (2) A body moves in a straight line with a uniform acceleration of magnitude 2.5 cm/sec 2 , and Initial velocity 10 cm/sec in the same direction of the acceleration , find :

 i  The distance covered by the body after 8 seconds  ii  After how many seconds the body is at 700 cm far from the begining point. Given : u  vo  10 cm/sec

Answer a  2.5 cm/sec 2

Note : The velocity is in the same direction of the acceleration means that the velocity of the body will increase as time increase " i.e: Imagine a bicycle is moving on a bridge"

i   ii 

1 1 2 S  u t  at 2   S  10  8    2.5  8   160 cm 2 2 1 5 700  10t   2.5  t 2  700  10t  t 2  Multiply by 4  2 4 2 2  2800  40t  5t   5t  40t  2800  0  divide by 5   t 2  8t  560  0 

 t  28 t  20   0

  t  20 sec

Thus the body is at 700 far from the initial point after 20 seconds Dynamic – 3rd secondary

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Example (3) A body fell vertically downwards from height of 78.4 m above the ground , Find :

1 The velocity of the body after 3 seconds .  2  The time taken to reach the ground .  3  The velocity when it reaches the ground . Answer

The word "fell" means U  0

u 0  ve

1 After 3 second here means t  3 sec .

g

V  U  gt  9.8  3  29.4 m / sec

2 S

 78.4 m

U 0

g  9.8 m / sec .

9.8 m / sec

1 2 1 g t  78.4   9.8 t 2  t 2  16  2 2 So , We need 4 sec to make the body reach the ground . S  ut 

S 78.4 2

t  4 sec.

 3  To get the velocity when the body reach the ground : We have to know either the time of the body to reach the ground , Or the total distance of the body to reach ground. So

V U  gt

V  9.8  4  39.2 m / sec .

Or V 2  U 2  2 g s  V  2  9.8 78.4   39.2 m / sec . ----------------------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Newton’s laws of motion

 Mass :

The mass of a body is a  ve scalar quantity which is proportional with the weight of this body . The mass of a body is denoted by m . Units of mass :

1 Ton  1000 Kgm & 1 Kgm  1000 gm & 1Gram  1000 milligram .

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1st : Momentum ‫قوة الدفع‬ Definition : The momentum vector of a particle , At a certain instant is defined as the product of the mass of the particle and its velocity vector at this instant , Momentum is denoted by H . Rule of Momentum :-

H mv

From this definition , It is clear that the momentum of a body at a certain instant is a vector in the same direction of the velocity vector . For example, a heavy truck moving fast has a large momentum—it takes a large and prolonged force to get the truck up to this speed, and it takes a large and prolonged force to bring it to a stop afterwards. If the truck was lighter, or moving slower, then it would have less momentum. Momentum can be defined as "mass in motion." All objects have mass; so if an object is moving, then it has momentum - it has its mass in motion

------------------------------------------------------------------------------------------------------Units of Momentum : Where m  Mass

H  mv

, And v

 Velocity .

 Its units may be for an example : gm . cm / sec

& kg . m / sec

& kg . m / hr .

------------------------------------------------------------------------------------------------------Note : I always prefer to use the unit kgm . m / sec .

------------------------------------------------------------------------------------------------------Example (1) A body of mass 7.5 gm , Moves from rest in a straight line with acceleration 9 cm / sec 2 in the 1 direction of its motion , Find its momentum after minutes from the begining of the motion . 2 Answer Since there is only one body in the problem , Then we don't need to show a direction . To find the momentum : H  mv 1 So we must find v : u  0 a  9 cm / sec 2 t   60  30 sec . 2 v  u  at  v  9  30   270 cm / sec .  H  mv  7.5  270  2025 gm . cm / sec . Dynamic – 3rd secondary

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Example (2) Find the momentum of a stone of mass 500 gm , When it is let to fall 4.9 meters vertically

downwards . Answer H  mv  So we have to get velocity : u  0 " fall" s  4.9 m . So ,

v2  u 2  2 g s

v 2  2  9.8  4.9   96.04

 v  9.8 m / sec .  H  500  9.8  4900 gm . m / sec .

------------------------------------------------------------------------------------------------------Example (3) Find the height from which a body of mass 500 gm , falls such that the magnitude of the momentum when it collides with the ground equals the magnitude of the momentum of a body of mass 70 gm moving with velocity of magnitude 432 km / hr . Answer 5 v2  432   120 m / sec .  So m1v1  m2v2 18  500 V1  70  120   V1  16.8 m / sec . u 0

v1  16.8 m / sec

 v 2  u 2  2g S

s  ??

16.8 

2

 2  9.8  S

S  14.4 m .

------------------------------------------------------------------------------------------------------Example (4) 1 kg and its velocity at the opening 5 of the gun is 200 m / sec , Find the momentum of the bullets fired per second in gm.cm / sec Answer In order to find the momentum of bullets fired per second . A gun fires 300 bullets per minute ,If the mass of each bullet is

300  5 bullets per second . 60 1 Each one of them has mass  kg . 5 Then the mass of the bullets in one seconds is :

 Number of bullets fired 

Momemtum

1 5   1000 gm 5  H  mv 

1000 gm .

1000  200  100  2  107 gm .cm / sec .

------------------------------------------------------------------------------------------------------Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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The change in momentum It is the change of the velocity of an object from T1 to T2  H 2  H 1  Rule :

The change in momentum :

H 2  H 1  m  v2  v1 

 It is also called Impulse

------------------------------------------------------------------------------------------------------Example (5) A rubber ball of mass 300 gm moves horizontally with uniform velocity 135 cm / sec . It collides 4 by a vertical wall , And rebounds in a perpendicular direction to the wall after loosing of the 5 magnitude of its velocity before collision, Find the magnitude of the ball change in momentum due to collision with the wall . Answer let the positive direction be in the direction of the 2 nd velocity v1  135 m / sec .   v1  135 x

v1

The velocity after , Collision is opposite to x 4 1 After loosing of its velocity  v2  135   27 x 5 5

ve v2

Change of momentum  H 2  H 1  m v2  v1 H 2  H 1  300 27  135 x  48600 x  Its magnitude  48600 gm . cm / sec

------------------------------------------------------------------------------------------------------Example (6) A ball is left to fall from a height of 16.9 m , And its momentum when it impinges with the ground is 5460 gm . m / sec , Find its mass . If the ball rebounds to a height of 4.9 m , Then Find the change of its momentum just before and after impact . Answer Let the positive direction be    . u 0  Before impact : u  0 s  16.9 m . ve v 0 2 2 2 v1  u  2 g s   v1  2  9.8 16.9  4.9 m v1 2 v1  331.24   v1  18.2 x 16.9 m And H 1 is the momentum vector of the ball before impact . u  ?? 5460 u  Not the same as V  5460  m  18.2  m   300 gm . 18.2 the direction changed  Momentum after impact : u  ?? v  0 " as the ball will become finally at rest" s  4.9 v2 2  u 2  2 g s  0  u 2  2  9.8  4.9   u2  9.8 m / sec .   u 2  -9.8 x The change of momentum before and after impact is : m v2  v1  300 9.8  18.2  8400 x .  Its magnitude is 8400 gm .m / sec .

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (7) A bullet of mass 120 gm is fired with a velocity of 390 m / sec towards a wooden body of mass 3 kgm .which is at rest . If the bullet is imbedded in it and the system moves after that with a certain velocity . Find their velocity , Given that the momentum of the system doesnot change due to impact . Answer In this problem, we want to find the velocity of the bullet and the wood after impact x v1 be its velocity just before impact . v 2

 v1  390 x v1

Let v2 be the velocity of the system after impact . And m1 be the mass of the bullet and m2 is the mass of the system . As the momentum of the system does not change due to impact .  H1  H 2

 m1 v1  m2 v2

 120  390 x   3000  120  v2

  v2 

m1  120 gm m 2  3120 gm

120  390   15 x

]

3120 Thus the system will move after impact with velocity 15 m / sec in the same direction of th bullet .

------------------------------------------------------------------------------------------------------Example (8) A fixed cannon fired a projectile of mass 5 kg with a velocity of 350 m / sec in a horizontal direction towards a tank moving with a velocity of 45 km / hr and it hit it , Find the absolute value of the momentum of the projectile , Then Calculate the magnitude of the momentum of the projectile relative to the tank if :1 The tank is moving away of the canon  2  The tank is moving towards the canon . Answer Let x be the unit vector in direction of the projectile :

vT

vP x

H  m v  5  350 x  1750 kg . m / sec .

1 The tank is moving away of the canon means , That are in the same direction

.

5 x  337.5 x 18 H  m vPT  5  337.5 x  1687.5 x

 VPT  VP  VT  350 x  45  Its momentum

 The magnitude of the moment of the projectile  1687.5 kg . m / sec .

2

vPT  vP  vT 5 x  362.5 x 18 H  m vPT  5  362.5 x  1812.5 x

 350 x  45  Its momentum

x

vP

vT

 The magnitude of its momentum is 1812.5 kg . m / sec . Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (9) Water vapour condinces on the surface of water drops , while it is falling at rate 10.5 milligram / sec , The mass of one falling water drop is 0.25 gm , Find the momentum in gm . m / sec of one water drop when it reaches the surface of the ground from a height 1000 meters .

Answer 10.5 Water condineces at rate  0.0105 gm / sec . 1000 So , To know the total mass of the water drop in the 1000 m : We must get the time then . 1 1 100 s  u t  g t2  1000   9.8  t 2  t sec . 2 2 7 100 Then the time water drop travel from 1000 m till it reaches the ground is sec . 7 100 Then the mass of the water vapour at this time  0.0105   0.15 gm . 7 The total mass of the drop  0.15  0.25  0.4 gm . 100 To get the velocity :v  u  g t  9.8   140 m / sec . 7 The momentum H  mv  0.4  140  56 gm . m / sec .

------------------------------------------------------------------------------------------------------Example (10) A swimmer of mass 40 kgm jumped vertically from rest to the water surface of a swimming

pool , He collided with the surface after 1.5 sec , Then he douse vertically into the water in a retarded motion with uniform acceleration of magnitude 2.7 m / sec 2 and covered a distance is 5.4 meter before he starts the ascending , Calculate magnitude of his change of momentum due to collision with water . Answer  First  Find magnitude of the swimmer velocity directly before collision with water : v  u  g t  0  9.8 1.5   14.7 m / sec

u 0

Let u be a unit vector directed vertically downwards .  The velocity directly before collision  v1  14.7 x

 Second 

t  1.5 sec

Magnitude of swimmer velocity directly after collision with water :

v2  u 2  2 a s  u   5.4  2

x

2

 

 0  u 2  2  2.7  5.4

u  ??

 u  5.4 meter / sec .

 The velocity directly after collision  v2  5.4 x

V  ??

a  -2.7

 Change of mometum due to collision  H 2  H 1  m v2  v1

5.4 m

V 0

 40 5.4 x  14.7 x  372 x  Its magnitude  372 kgm . m / sec . rd Dynamic – 3 secondary

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Chapter one – Newton’s laws of motion

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Example (11) A rocket is projected vertically upwards with velocity 180 km / hr , If its mass at any instant is m   25  0.001t  kgm , Find the rate of change in its momentum after 15 seconds . Answer Very important note : When the mass of the body comes variable in the problem , we must use the rule of derivative only . 180 u km / sec  u  0.05 km / sec . 3600 v u  g t  v  0.05  9.8 t

H  m v   25  0.001t 0.05  9.8 t  x dH  -9.8  25  0.001t   0.0010.05  9.8t  dt

At t  15 sec .

dH  -244  853  0.14695  -244.706 kg . m / sec 2 . dt

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Newton’s first law Every body remains in its state of rest or uniform motion unless it is compelleted to change that state by an external action called a force .

Discussion of the first law :

1The law assumes that bodies which are at rest state or have

uniform motion state is a

natural state of the body .

 2 The law assumes that every body can not change any of its natural state by itself , So this law is called the law of inertia .

 3  The law assumes that the existence of an external action called a force can only change the state of the body .

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Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Cases of uniform motion of a body A uniform motion on a horizontal plane under the

Case (1)

action of a horizontal force .

N

When a body of weight  w  moves horizontally by force F ,

Motion

You have to know that it meets an opposite resistance called " Resistance force " and it is denoted by R . In this case

F R

Or

Also , We can say that : N

F

R

F  R 0 "Normal reaction of the road "

W

" Weight "

W

------------------------------------------------------------------------------------------------------A uniform motion of a body on a horizontal plane under

Case (2)

F sin  N

The action of an inclined force .

It a force is inclinedby angle  to the horizontal . Then : N  F sin   W And

R

R  F cos 

Motion

F

F cos 

W

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Case (3)

Vertical uniform motion :

R

Direction of

If a body of weight W  moves uniformly vertically in a liquid , Then

motion

R W

W

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Case (4)

Uniform motion on an inclined plane : F

F Sin  N

 N  F sin   W Cos 

Motion

 F cos   R  W Sin 

R

W Sin

F cos 

 

W cos 

W

------------------------------------------------------------------------------------------------------Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Notes :

1 The resistance of the plane to the moving body is always parallel to the plane and in the opposite direction to the motion of the body .

 2  The force generated by the motor of a car or a train is always in the same direction of the motion .  3  When we say that the body is moving with maximum velocity , This means that it is moving with a uniform motion Then a  0   4  If the resultant of forces , Acting on a body , Vanishes at any moment during its motion , Then it moves from this moment with a uniform motion . " So , Sum of forces  0 when the body moves with a uniform motion " .

 5  In many times , The reaction R is variable as the velocity of the moving body .  If R  V , Then : R  A V where A is constant    If R  V 2 , Then : R  A V 2  

R1 V1  R2 V2

R1 V12  R2 V2 2

------------------------------------------------------------------------------------------------------Example (1) A train of mass 250 Ton moves with a uniform velocity along a horizontal plane . The force of the engine is 2000 kg .wt ,Find the magnitude of the resistance for each ton of the mass . Answer The train moves with a uniform velocity, then it will stop soon according to resistance F  2000 kg . wt R we have to use Newton's 1st law   F  R  2000  The resistance per ton  2000  250  8 kg .wt

Case 1 

------------------------------------------------------------------------------------------------------Example (2) A locomotive of mass 6 tons pulls a number of wagons the mass of each equals 3 tons along a horizontal straight road with uniform velocity. If the magnitude of the driving force of the locomotive equals 900 kgm.wt, and the resistance to the motion of the train equals 15 kgm.wt per each ton of its mass . Find the number of the pulled wagons . Answer The train moves with a uniform velocity, then it will stop soon according to resistance The mass of the whole system is : Mass of the locomotive  Mass of number of wagons  The mass  6  3n Motion 900 kg . wt It moves with a uniform velocity : R

So we have to use Newton's 1st law  F  R  900  6  3n  15   900  90  45 n  45 n  810   n  Dynamic – 3rd secondary

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810  18 wagons . 45

Chapter one – Newton’s laws of motion

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Example (3) A body is pulled along a horizontal straight road by a force of magnitude 1350 kgm.wt , and 3 inclined at an angle of Sin to the horizontal, so the body moved with a uniform motion 5 3 against the road resistance which is equal to of its weight . Calculate the weight of the body 4 and the normal reaction of the road . Answer 1350 Sin  F  1350 The body moves in a uniform velocity :  F  R N 3 4 3 1350 cos   W  1350   W   W  1440 kg . wt 3  4 5 4 w 1350 cos  4 3 Also , N  1350 sin   W  N  1350   1440 5 3 W  N  630 kg.wt Sin   5 3 5 4  4 Cos   5

------------------------------------------------------------------------------------------------------Example (4) A car of weight 3 ton.wt moves with a uniform velocity up a plane inclines to the horizontal 1 . If the resistance of the plane is 4 kg.wt per ton, find the 60 driving force of the car in kg . wt . Answer N R  4  3  12 kg . wt F The component of W in the plane direction downwards R 1 1 1 is W sin   3   Ton . wt   1000  50 kg . wt 60 20 20 W sin   W cos   The car moves uniformly With a uniform velocity  at an angle  where Sin  

 F  R  W sin   F  12  50  62 kg . wt W

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Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (5) The engine of a car of weight 5 ton.wt is stopped while it is moving downwards a road 1 inclined to the horizontal at an angle  where Sin   with a uniform velocity , 100 calculate the resistance for each ton of its mass in kg.wt. Answer The engine stopped while moving downwards means N R

that the car is moving by its weight only  No force exists  R  5 Sin   5 

1 1  ton.wt 100 20

Motion

1  R  1000  50 kg.wt 20

5 Sin 

So , the resistance per each ton is

 5Cos 

50  10 kg.wt 5

5

------------------------------------------------------------------------------------------------------Example (6) A car of weight 4.5 ton.wt moves along the line of the greatest slope of a plane inclined 1 at an angle of Sin , if the engine of this car is stopped when moving downwards with 50 uniform velocity . If the car engine is turned on, find the magnitude of the driving force of this car such that it ascends this plane with uniform velocity given that the resistance of the plane to the car is unchanged in the two cases of motion . Answer 4.5 ton.wt  4.5  1000  4500 kg.wt Case  1 : When the car engine is stopped and the car

R

N

moves downward

The car moves with its weight only   R  4500 Sin   4500 

Motion

1  90 kg.wt 50

4500 Sin 

4500

Case  2  : When the car engine is turned on and the car is moving upwards . F  90  4500 

1  180 kg . wt 50

R  90

- 16 -

F

N

Motion

4500 Sin 

Dynamic – 3rd secondary

4500 Cos 

 4500

4500 Cos 

Chapter one – Newton’s laws of motion

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Example (7) A car of weight 2.5 ton . wt moves on a straight horizontal road with a uniform velocity 1 , The driver 32 stopped the motor of the car . So , It moved down the ramp with a uniform velocity , Given when it reached a ramp inclined to the horizontal at an angle whose sin is

3 of the resistance of the horizontal road . 8 Calculate the driving force of the motor ofthe car along the horizontal road measured in kg .wt Answer 3 Given : R2  R1 8  When the car moves in the horizontal road with uniform velocity : Motion Then : F1  R1  1 N  When the car moved in the inclined plane : that the resistance of the ramp is equal to

First : You have to know that : no force downward .

F1

Then the car is moving with its weight only So , And

R2  2500 sin   2500  R2 

3 R1 8

1  78.125 kg . wt 32 N

R1

R

W  2500

Motion

 8 1 2500 sin  2500 cos   78.125  208 kg . wt 3 3 2500  1 Ramp From  1 F1  R1  208 kg . wt 3 ------------------------------------------------------------------------------------------------------ R1 

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (8) A body of mass 30 kg is placed on a plane inclined to the horizontal at an angle  and is pulled by a force of 20 kg.wt acting along the line of the greatest slope up the plane, it moves uniformly up the plane against resistance of  R  kg.wt. When the tension is reduced to 10 kg.wt, the body can move down the plane uniformly, find the measure of the angle of inclination of the plane, given that the resistance of the plane doesn't change in the two cases .

Answer  From the figure : R  30 Sin   20   

 1

N

20

 When the tension of the force reduced to 10 kg . wt , The body moved downwards and the 10 kg.wt

Motion

Became another resistance . So , R  10  30 sin 

R

30 sin

 R  30 Sin   10     2  Substitute  2  in  1 : 30 sin   10  3 sin   20  60 sin   30  Sin  

1 2

30 cos 

30 10

N

R

Motion

   30 o 30 sin

30

30 cos 

------------------------------------------------------------------------------------------------------Example (9) A body of weight 16 kg.wt is placed on a plane inclined to the horizontal at an angle of measure 30 ,the body is pulled by a rope of force F kg . wt upwards and this force inclined to the greatest slope at an angle 30 , If the body moves uniformly upwards the plane when the resistance of the plane to the body is 4 kg . wt , Find F and the pressure of the body on the plane . Answer o o   F cos 30  4  16 sin 30   F cos 30 o  12 F N 12 F sin 30  F  8 3 kg . wt 30 cos 30   N  F sin 30  16 Cos 60 o 4 30 1 3  N  8 3   16  o 2 2 16 sin30 30 16  N  8 3  4 3  4 3 kg . wt

F cos 30 

16 cos 30o

------------------------------------------------------------------------------------------------------Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (10 “important”) A body of mass 5 kg is placed on a horizontal plane and is attached to two horizontal ropes , The measure of the angle between them is 120 , When the two ropes are pulled by a force of 200 gm . wt each then the body moves on the plane uniformly , Find the magitude and the direction of the force of resistance of the plane against the motion of the body . Answer N Let F be the resultant of the two tensions

R 2  T 2  T 2  2 T T  Cos  " Static"  R 2  2T 2  2T 2 Cos   2  200   2  200  Cos120 o 2

2

T  2

R

 R 2  40000   R  200 gm . wt

 2

F

T

5

------------------------------------------------------------------------------------------------------Example (11) A mettalic ball of weight 30 gm.wt is left to fall down in long vertical tube full of a viscial liquid. If the resistance of the liquid to the motion of the ball varies directly as the magnitude of the velocity of its motion in the tube , And given that the magnitude of the resistance of the liquid to the ball equals 25 gm . wt , When the magnitude of the velocity of the ball equals 12 cm / sec , Calculate the magnitude of its velocity when it is uniform . R

Answer R  W   R1  30 gm.wt And

R V

R  25 gm.wt  from  1 :

R1 V1      1 R2 V2

when V  12cm / sec

30

30 V1 30  12    V1   14.4 cm/sec 25 12 25

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Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (12) A train of mass 45 tons moves on a straight horizontal road and the driving force of the engine equals 6.25 tons . wt , Find the uniform velocity with which it moves , Gives that the resistance to its motion is proportional to the square of its velocity and the resistance equals 25 kg . wt for each ton of its mass when its velocity equals 15 km / hr . Answer When the train moves with a uniform velocity : R  R1  6250 kg . wt R V

2

6250

R1 V12        1 R2 V2 2

45

R2  25  45  1125 kg .wt when V2  15 : 6250 V12 from  1 :    V  25 1125  15 2

2 km / hr .

------------------------------------------------------------------------------------------------------Example (13) An Aviator is tied to a parachute descends vertically downwards . Given that the air resistance 8 is directly proportional to the cube of its velocity at any time , This resistance is equal to the 125 weight of the man and the parachute when its velocity is 20 km / hr . Find the maximum velocity the man descends with .

Answer Let the weight of the man and the parachute is W kg . wt The man moves with uniform velocity  R  W and

R V3

 

R

R1 V13      1 R2 V2 3

8 W  when V  20 km / hr 125 W V13 from  1 :   V 3  125000  V  50 km / hr . 3 8 W  20  125

And

R1 

W

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Dynamic – 3rd secondary

- 21 -

Chapter one – Newton’s laws of motion

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Newton’s Second law

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Rate of change of momentum with respect to the time is proportional to the impressed force and takes place in the direction in which the force acts .

1

The sympolic form of the second law  Equation of motion  :

Suppose a force F acts on a body of mass m for a time t, and causes its velocity to change from u to v This changing in velocities lead the appearence of acceleration. The second law states that :

d dt

m v  

  ma  K F

F

And if the unit of the force magnitude is that which produces one unit of acceleration magnitude When Acts on a body with one unit vector of mass   1  1  K  1  K  1

Two rules are formed :

1 Vector form of the 2nd

law when the mass is Constant . ma  F

2

Where a is the acceleration vector .

Algebraic form of the 2 nd law when the mass is Constant . ma  F

Note: In our problems, Newton 2 nd law is the basic rule which will we use always

------------------------------------------------------------------------------------------------------Very important note : The force vector F Or its algebraic measure F and the acceleration a Must have the same direction a So from the opposite figure : F R We can say : F  R  ma Or F  R  ma

-------------------------------------------------------------------------------------------------------

Differences between Newton’s first law and Newton’s second law Newton’s first law (1) Uniform velocity (Motion) Or Maximum velocity exists without changing “Special case of Newton’s 2nd law” (2) This means that a = 0 (3) R F

Newton’s second law (1) Uniform Acceleration Or changing the uniform velocity due to force or resistance “The original rule which will be always used” (2) This means that a  0 (3) R F

F  R Or F  R  0

F  R  ma Or F  ma  R

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Fundamental units of forces a

Newton

  kg.m /sec 2 

It is the magnitude of the force which while acting on a body of mass one kilogram it produces an acceleration of magnitude 1 m / sec 2 .

b

Dyne

 gm.cm / sec  2

It is the magnitude of the force which while acting on a body of mass one gram it produces an acceleration of magnitude 1 cm / sec 2 . Note  1

Newton  10 5 Dyne . Dyne  10 -5 Newton .

So when applying our Rule : F  m a Dyne

Cm / sec 2

gm

F

 M

Newton

 

a  m / sec 2

Kg

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The relation between the weight of a body and its mass If a body of mass  m  is left to fall , It desends vertically by a uniform acceleration g because the earth attracts it by a force W  called weight . We replace F  ma

BY 

w  mg

M

G

W

------------------------------------------------------------------------------------------------------   

Example (1) If the mass of a body m  15Kg  Its weight : W  mg  15  9.8  147 N If the mass of a body m  3 Ton  Its weight : W  mg  3  1000  9.8  29400 N If the mass of a body m  200 gm  Its weight : W  mg  200  980  196000 Dyne. w 117.6 The body whose weight  117.6 N  Its mass : m    12 Kg g 9.8

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Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Summary of units 1  Kgm .w t  9.8  2  gm .w t  980 3  Kgm .w t  Ton . wt  1000 Kgm .

Newton 

Newton

1 gm . wt 980 1 Dyne  Kgm.wt 9.8  10 5

Dyne 

Dyne

9.8 10 5 Dyne 

1 Kgm . wt 9.8

wt  10 6 gm .

wt  1000  9.8 N  980  10 6 Dyne

 980  1000   Kg  gm   dyne 1000  980

Also , Note :

If the mass of a body  X gm

 Its weight  X gm.wt

If the mass of a body  X Kgm

 Its weight  X Kgm.wt

Proof : If m  7 Kgm

  W  7  9.8 Newton  W 

7  9.8  7 Kgm . wt 9.8

------------------------------------------------------------------------------------------------------Example (2) A body of mass 8.4 Kgm is moving in a straight line with uniform acceleration of magnitude 350 cm / sec 2 . Find the magnitude of the force acting on that body .

 First  In Newton m  8.4 Kgm F  ma

 Second  In Kgm.wt

Third  In dyne

Answer  a  350  100  3.5 m / sec 2  F  8.4  3.5  29.4 Newton

 F  29.4  10 5  2940000 Dyne   F  29.4  9.8  3 Kgm .wt

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Very important note From now on: 1 If the resistance is not mentioned in the problem, then F  m a only

 2  If the force is not mentioned in the problem, then -R  m a  3  If the engine is stopped, then F  0

only

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Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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We have three cases in this chapter Case I : Motion under a single force Example (1) A car of mass 245 kg moves with uniform velocity 27 km / hr , the brakes are used to stop the car after covering a distance 9 meters , then find the magnitude of the force of the brakes in kg.wt Answer the uniform velocity changed when the brakes are used, then F  0 There is no driving force  and resistance  R  , acceleration appear

5 u  27   7.5 m / s 18

s9 m

 retardation  and V final  0

R  ??

a

V 0

R

F

-25 m / sec 2 8 And F  R  m a  But there is no driving force mentioned  : -25 -6125 6125 1 -R  245  Newton   9.8    R   9.8  78 kg.wt 8 8 8 8 V 2  u 2  2 a s   0  7.5   2  9  a 2

 a

------------------------------------------------------------------------------------------------------Example (2) A force of magnitude 4.2  10 dyne acts on a body in the state of rest to move it in a straight 7

line with uniform acceleration of magnitude 6 m / sec 2 , find the magnitude of momentum of 1 a minute from the instant of start in kg .m / sec . 2 Answer from now on, if the resistance is not mentioned in the problem, then F  m a only

this body after

F  4.2  107  10 5  420 Newton

F  ma

 420  m  6

 m  70 kgm 1  60  30 sec 2  V  u  at  0  6  30  180 m / sec

And

u 0

And

t

Magnitude of momentum  mv  70  180  12600 kgm . m / sec

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Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (3) A car of mass 1.6 tons moves along a horizontal straight road , And when it is moving with velocity of magnitude 45 km / hr , The driving force of the car is stopped , And the brackes are used so the car stopped after a small interval of time , find the distance which the car travelled in this interval if the total resistance to the car is constant and its magnitude  5  10 4 N . Answer When the brakes are used , then F  0 There is no driving force  And resistance  R  appears

 retardation  and V final  0

acceleration appear

F  R  m a   - R  m a  -5  10 4  1.6  1000  a -125 a m / sec 2 R 4 5 25 And u  45   m / sec , V 0 18 2

So

a

x F

2

So v  u  2 a s 2

2

-125  25  0     2 S 4  2 

 S  2.5 m

------------------------------------------------------------------------------------------------------Example (4) A body is projected with velocity of 560 cm/sec on a rough horizontal plane so that it stopped after 2 seconds from the begining of the motion, calculate the coefficient of friction between the body and the plane. Answer When we are talking about rough planes  use either resistance force or FF R a the body is stopped after 2 seconds  F  0 FF   R

where R  m g   FF    m g 

u  560 So, So

V  u  at

t2

FF

F

V 0

 0  560  2 a   a  - 280 cm / sec

mg

F  FF  m a   - FF  m a  - m g  -280 m   

-280 2  -980 7

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Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (5) When a train was moving along a straight road with velocity 63 km / hr , its last wagon whose mass is 7 ton is separated so it stopped due to a constant resistance of magnitude 500 kg.wt , find the time taken until it came to rest Answer let x be a unit vector in the direction of F   F and a have the same direction 5 u  63   17.5 m / s V 0 R  500 kg.wt  9.8  4900 Newton 18 a And F  R  m a  But there is no driving force mentioned  : F R -7 2  -R  ma   - 4900  7000 a  a  m / sec 10 7 And V  u  at  0  17.5  t  t  25 sec 10

------------------------------------------------------------------------------------------------------Example (6) When a cannon of mass 1.2 tons fires a projectile , it retreats on a horizontal ground a distance of 90 cm , if the resistance of the ground to its motion is 300 kgm . wt , Then find the velocity with which the cannon starts retreating. Answer Retreats a distance here means that : R  300 kgm.wt  9.8  2940 Newton S  0.9 m V  0  Comes to rest  u  ?? a And F  R  ma   - R  ma F R 2  - 2940  1200 a  a  -2.45 m / sec And V 2  u 2  2a s  0  u 2  2  2.45 0.9    u  2.1m / sec

------------------------------------------------------------------------------------------------------Example (7) A body of mass 600 kg moves horizontally in the space with uniform velocity of magnitude 864 km / hr . It intered a dusty cloud which acted upon by a resistance , Its magnitude equals 1 kgm . wt per each kilogram of the body mass . Find the velocity of the body at the instant of 4 getting out of the cloud , If it remained 40 sec through it . Answer there is change in velocities  acceleration appears  use Newton's 2 nd law 1 R   600  150 kg.wt   R  150  9.8  1470 Newton 4 -49 And - R  m a  -1470  600  a  a  m / sec 2 20 5 u  864   240 m / sec t  40 sec 18  -49  V  u  at  V  240    V  142 m / sec.   40 20   rd Dynamic – 3 secondary

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Chapter one – Newton’s laws of motion

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Example (8) A bullet of mass 35 gm is fired horizontally with velocity of magnitude 40 meter / sec towards a 4 fixed vertical target of wood ,it penetrated the target , and lost of the magnitude of its velocity 5 1 after sec , calculate the magnitude of the target resistance assuming that it is constant , 100 And find the distance which the bullet moves through the target before it comes to rest . Answer we have two kinds of bullet problems: 1 If the bullet penetrated the body with given thickness , the we will solve the problem from the begining of the body .  2  If the bullet penetrated the body without given thickness , the we will solve the problem far from the body . v 8 v 0 1 In our problem , we will use the 2 nd option u  40 t  100 1 1 u  40 , V  40   8 m / sec , t sec 5 100 1 So V  u  at  8  40  a   a  -3200 m / sec 2 100 3 F  R  m a  - R  35  10  -3200  R  112N 2 V 2  u 2  2 a s  0   8   2  -3200  S  S  0.01m

------------------------------------------------------------------------------------------------------Example (9) A bullet of mass 32 gm is fired horizontally with velocity of magnitude 100 meter / sec towards a fixed target of wood to embed through it a distance 16 cm before coming to rest, find the magnitude of the wood resistance in kg.wt assuming that it is constant , and if the same bullet is fired with the same velocity at another fixed target made of the same wood of the first target and its thickness is 7 cm only, find the magnitude of the velocity with which the bullet gets out of this target. Answer In this problem, thickness of wood is given , then we will solve the problem from the begining of the wood  u  100 , V  0 , S  0.16 m 2 2 2 So V  u  2a S  0  100   2 0.16  a  a  -31250 m / sec 2 u  100 F  R  m a  - R  32  10 3  -31250 v 0 5000  R  1000 Newton   9.8   kg.wt S  0.16 m 49 When the target is of thickness 7 cm u  100 , V  ?? , S  0.07 m , a  -31250 m / sec 2 2 V 2  u 2  2 a s  V 2  100   2  -31250 0.07   V  75 m / s

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Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (10) A bullet of mass 49 gm is fired horizontally with velocity of magnitude 74 meter / sec at a body which is at rest and formed of two adjacent layers , one of which is fiberglass andthe other is wooden , if the bullet passes through the layer of fiberglass which is of thickness 7 cm , then it is embedded in the wooden layer at a distance of 10 cm before it stops , if the resistance of the wood is three times that of the fiberglass , then find the resistance of each in kgm.wt . Answer the problem mentioned the thickness which the bullet penetrated , then we will start our problem from the begining of the wood f .g wood a * Motion from A  B  Inside the fiberglass  : 0.07 0.1 u  74 m / s V  ?? S  0.07 m A

B

VB 2  u A2  2 a s   VB 2  74   2 0.07  a    1 2

49 -1000 - R1  m a  - R1  a  a R1     2  1000 49

A

U A  74

B

C

VB  ?? VC  0

20 2  -1000   VB 2  74   2  0.07   R1   5476  R1     3  7  49  * Motion from B  C  Inside the wood  : - R2  m a'  -3R1 

49 a'  1000

 a' 

-3000 R1     4  49

Also VB here is the initial velocity between B  C VB  u B  ??

VC  0

S  0.1 m

20 R1  0.2a' 7 20 20 600  -3000   0  5476  R1  0.2  R1    0  5476  R1  R1 7 7 49  49  VC 2  VB 2  2 a' s   0  VB 2  2 0.1 a'   0  5476 

740 5476  49 1813 R1  5476   R1   Newton   9.8   37 kg.wt 49 740 5  R2  3  37  111 kg.wt

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

- 28 -

Chapter one – Newton’s laws of motion

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Example (11) A body of mass 12 kg is at rest , a force of magnitude 3 kg.wt acts on the body for 12 seconds , and after this time , the driver force stopped and the body moved with uniform velocity , then find the distance the body covered after 30 seconds from the beginning . Answer If the resistance is not mentioned in the problem, then F  m a only * For the 1st 12 seconds " during the action force " u A  0 t  12 sec t  18 sec Motion from A  B

t  12 sec

u 0

F  3 kg.wt  9.8  29.4 Newton

F

A C B 29.4 2 And F  R  m a   F  m a  29.4  12 a  a   2.45 m/sec 12 1 1 2 So S  u t  at 2  S   2.45 12   176.4 m 2 2 So VB  u A  at  VB  2.45 12   29.4 m / sec Motion from B  C : When the driver force stopped , it means that F  0 and there is no

acceleration body moves with uniform velocity for the next 18 seconds  S V t  29.4  18  529.2 m   the total distance after 30 seconds  176.4  529.2  705.6 m

------------------------------------------------------------------------------------------------------Example (12) A force of 600 Dynes acts on two bodies (A) and (B) placed at rest on a horizontal smooth plane , the masses of them are 60 gm , 40 gm respectively and attached together by a string . After 5 seconds from the action of the force , the body (B) is separated moves with a uniform velocity and the body (A) only remains under the action of the force , Calculate the distance between the two bodies 5 seconds after the instant of braking the string . Answer * When the two bodies are connected : uniform motion F  ma  600   60  40  a No acceleration a u 0 600 100 a  600  a   6 cm / sec 2 F A A B F B 100 uB  ?? u A  ?? Thus the bodies move with acceleration 6 cm / sec 2 S At t  5 sec  V  u  a t  6  5   30 cm/sec After 5 seconds the string cut and the two bodies separated when their velocities u  30 cm / s : Body  B  move without force  F  a  0 Body  A  moves with F  600 dyne F  m a   600  60 a'  a'  10 cm/sec 2 Then B moves with uniform motion : u  30 cm / s a'  10 cm/sec 2 t  5 sec  S B  vt  30  5  150 cm 1 1 So S B  u t  a' t 2  S B  30  5 + 10  25  2 2  S B  275 cm

Then the distance between the two bodies after 5 seconds is 275  150  125 cm

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Case 2 : Motion under more than two forces Equations of motion of famous applications : Application Motion on a horizontal plane

Drawing

Equation F  R  m a

x

under a horizontal force F

R

F

and a and resistance R u

Motion on a horizontal plane under a force F inclined at an angle  With the horizontal and

F

F Sin N

R

a resistance R . Motion on a horizontal plane

 F Cos 

under a force F inclined at an

N F Sin

angle  with the vertical

F Cos  mg

Motion vertically downward

Ground surface R

through a Ground under

F

Inside

the weight  mg  and the earth

x

the ground

mg

R

F sin  ma N  F cos   m g  0

No driver force except Its weight m g R ma acceleration appears inside the ground

Motion vertically upwards

F R m g ma

F

under a driving force F

R

and weight

Motion vertically downwards

acceleration appears in the

x

air when there is another force act on the body

mg F R

under the weight mg and a

x

resistance R and a lifting

m g  R  F  ma

mg

force F

F 0 R 0 a 0

F sin  N  m g  0

mg x

resistance

F cos   R  ma

Very important Notes

If the engine force stopped  steam shutt off

.

If the resistance are neglected . If the body moved with uniform velocity or maximum V .

 R  Opposite to direction of motion

 m g  Always Downwards

 F  According the given direction Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (13) A car of mass 3tons starting from rest along a horizontal straight road with uniform

acceleration and against constant resistance of magnitude 50 kg . wt , If the magnitude of the car velocity is 39.2 m / sec after 2 minutes from the starting, calculate the magnitude of the driving force of the car in kg . wt Answer Let u be a unit vector in the direction of motion . u 0

V  39.2 m / sec

t  2  60  120 sec . a R 49 V  U  at  39.2  0  120a  a m / sec 2 150 Equation of motion of the car is resultant of the forces acting on the car  m a

x

49 150  F  490  980  1470 Newton  9.8  150 kgm . wt

F  R  ma

F  R  m a  F  50  9.8  3  1000 

------------------------------------------------------------------------------------------------------Example (14) A train of mass 5.6 ton moves from rest with a uniform acceleration along a horizontal straight 1 road, it travelled a distance 1260 meter during a minute from the starting, find the magnitude 2 of the resistance per each ton of its mass if the magnitude of the driving force  3000 kgm . wt .

Answer 1260 2520 0  30  2520 m / min  V   42 m / sec and t  15 sec 0.5 60 2 x 42 V  u  at  Then : 42  15a  a   2.8 m / sec 2 a R 15 And the equation of motion of train is : F  R  m a

V

R  F  ma

 F

R  3000  9.8  5.6  1000  2.8  13720 Newton  9.8   1400 kg.wt

Magnitude of the resistance per ton  1400  5.6  250 Kg .wt .

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (15) A piece of iron of mass 3.5 kg starts motion from rest on a rough horizontal plane when a horizontal force of magnitude 3.15 kg.wt acts on it, if the coefficient of friction between the body and the plane equals 0.3, find the velocity of the piece of iron after 5 seconds from the begining of motion. Answer When we are talking about rough planes  use either resistance force or FF FF   R

where R  3.5 g  3.5  9.8  34.3 Newton

F  FF  m a   3.15  9.8  0.3  34.3  3.5 a

And

 a  5.88 m / sec u 0 So ,

a

a  5.88

V  u  at

R F  3.15  9.8

FF

2

t 5

 V  5.88  5   29.4 m / s

3.5 g

------------------------------------------------------------------------------------------------------Example (16) A body of mass 360 gm is placed on a rough horizontal plane whose coefficient of friction with 3 . A force of magnitude 420 gm.wt acted on the body in a direction inclined to 3 the horizontal at an angle of measure 30 o , so that it moved from rest with a uniform acceleration. Find the velocity of the body after 9 seconds from the begining of motion, and if the action of the force is stopped after that, determine the time that the body makes until it comes to rest. Answer When we are talking about rough planes  use resistance force or FF R the body is stopped after 2 seconds  F  0 420 FF   R where R  360  980  420  980 Sin30 o  147000 dyne 420 Sin30 o the body equals

3  147000  360 a FF 3 u 0 a  754.4 t 9

30 o

F  FF  m a  420  980Cos 30 o   a  754.4 cm / sec

2

Then

420 Cos 30 o

360 g

So, V  u  at  V  754.4  9   6789.64 cm / sec When the action of the force stopped: Then the body is moving under the action of resistance  FF  only

R 420 Sin30 o

FF

3 R  360  980  352800 dyne and  3 360 g 3 So, - FF  m a'  -  R  m a'    352800  360 a'   a'  -565.8 cm/sec 2 3 Then u  6789.64 a'  -565.8 V 0 So,

V  u  at

 0  6789.64  565.8 t   t  12 sec

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (17) A train of mass 30 ton started from rest and moved with uniform acceleration along a straight horizontal road under the action of a driving force of magnitude 6750 kg.wt and a resistance its magnitude is 25 kg.wt per each ton of its mass, and after 5 minutes of the start, the driving force is stopped and the resistance still constant , find the distance which the train moved after that instant before it comes to rest . Answer The train moved from O to A under the driving force F and the resistance R and from A to B only under R . If U is a unit vector in the direction of motion . Equation of the train during OA is : F  R  m a

a

u

 6750  9.8  25  9.8  30  30  1000a  9.8  6750  750   3000a  9.8  6000  30000a

O

R

F

a' A

R

B

a  1.96 m / sec 2

Magnitude of the velocity at A : Va  U  at  0  1.96   5  60   588 m / sec. Equation of motion during AB :

 when engine stopped and

u  588 changed to V  0    F  0 and acceleration must change 

-R  m a' , where a' is the algebraic measure of the acceleration during AB  - 25  30  9.8  30  1000 a'  a'  -0.245 m / sec 2 And

VB 2  VA2  2 a s   0   588   2  0.245S 2

 0.49 S   588 

2

 S  705600 m .

------------------------------------------------------------------------------------------------------Example(18) 1 kg moves along a horizontal st. road with uniform velocity and the magnitude of 2 1 the resistance to its motion equals of its weight, find magnitude of the driving force of the car motor 5 in gm.wt. And if the force increases with magnitude 15 gm.wt , find magnitude of this acceleration . Answer 1 1 1 Uniform velocity means a  0   F  R  F  R  w  m g   0.5  9.8  0.98 N 5 5 5 0.98 1 F  R   kg.wt  1000  100 gm.wt, And when force increased with 15 gm.wt : 9.8 10 Then F  R  m a   115  980  100  980  0.5  1000 a A car of mass

 14700  500a  a  29.4 cm/sec 2 Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (19) A locomotive pulls a train moves horizontally in a straight road with a uniform velocity 45 km / h, the mass of the locomotive and the train together is 160 ton and the assistance force is 5 kg.wt for each ton of the mass ,calculate the driving force of the locomotive. If the last wagon is separated from the train, given that the mass of the wagon is 16 ton , find the distance between it and the rest of the train 2 minutes after the instant of separating . Answer Before separation : The train is moving with uniform velocity this means that : F  R  5  9.8  160  7840 Netwon driving force for all train . After separation : The wagon moves with no force  Resistance only  and the train . goes

 force  Resistance 

The initial veocity of the locomotive and the wagon at the instant of sparation was u  45 km / h So , After separation regard to wagon :

F

R

Its initial velocity directly after separation was: 5  12.5 m / s 18 offcourse its acceleration is decreasing u  45 

So

F  R  ma

a  -0.049 m/s

a  - 0.049 m/s 2

F

R

R

 -R  ma

 -5  9.8  16  16  1000a   u  12.5 m/sec

a

u  12.5 m/sec S2

u  12.5 m/sec

2

t  2  60  120 sec

S1

So , lets find the distance of the wagon after 2 minutes : 1 1 2 S1  u t  at 2  S1  12.5 120    -0.049 120   S1  1147.2 meters 2 2 After separation regard to the locomotive  train  : Hence there is a driving force still and resistance , the train has initial velocity u  12.5 m/s after separation directly , off course , Its acceleration will increase

 as it becomes lighter 

  F  R  m a'  7840  5  9.8  160  16    160  16   1000 a'

 784  144000 a'   a' 

784 49  m / sec 2 144000 9000

49 1  49  2 m / s2 t  120 sec  S 2  12.5 120      120   1539.2 m 9000 2  9000  Then the distance between them after 2 min. covered by each of them : u  12.5 m/s

S 2  S1

a' 

 1539.2  1147.2  392 meters

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (20) A train of mass 60 tons moves with a uniform velocity 44.1 km / h against a resistance of magnitude 10 kg.wt . for each ton of the mass of the train . If the last cab is separated from the train , given that the mass of the cab is one ton , find the time taken by the separated cab till it comes to rest and the velocity of the train at this moment . Answer The train is uniform   F R  The driving force F  10  9.8  60 F  F  5880 Newton is the driving force of the train R after sepration with respect to the cap  - R  m a   -10  9.8  1  1  1000a   a  -0.098 m/s 2 5  12.25 m/s a  -0.098 V 0 18 V  u  at  0  12.25  0.098t  t  125 seconds

u  44.1 

After seperation with respect to the train :  Its acceleration increased   F  R  m a'  5880  10  9.8  60  1  60  1  1000 a'  a' `  49 m/s 2 t  125 sec 29500 R 49 27  V  u  a' t  12.25  125   12 m/s 29500 59 When u  12.25 m/s

&

a' 

49 m/s 2 29500

a' R u  12.25 m/sec

F

u  12.25 m/sec

------------------------------------------------------------------------------------------------------Example (21) A helicopter of mass 2 tons rises vertically with acceleration of magnitude 4.9 m / s 2 against resistances of air of magnitude 750 kg.wt, find the driving force of the engine of the helicopter, also find the force of the engine if the helicopter rises with a uniform velocity assuming that the resistance does not change . Answer Note : the acceleration appears in the air beside the gravity as the helicopter changes its velocity  F  R  m g  ma F  F  750  9.8  2  1000  9.8  2  1000  4.9 F  driving force   36750 Newton   9.8   3750 kg.wt a If the helicopter rises with uniform velocity :  F  R mg 0  F  R mg R  F  750  9.8  2  1000  9.8  26950 Newton  2750 kg.wt mg Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (22) A body of mass 500 gm fell vertically downwards from a height of 22.5 meters above the surface, it embedded through the ground . Find the distance which the body covered through the ground before coming to rest given that magnitude of the ground resistance is constant and equals 372.4 Newton . Answer 372.4 u 0 S  22.5 m VB 2  u 2 +2 g s  VB 2  2  9.8  22.5   441

22.5

 VB  21 m/s From B  C "Inside sand  acceleration appears"

B

m g  R  m a   0.5  9.8   372.4  0.5 a  a  -735 m/s 2

mg vc 0

Vc 2  VB 2  2a s  0  441  2 735  S

sand a

C

 S  0.3m  30 cm

------------------------------------------------------------------------------------------------------Example (23) A body of mass 40 gm fell vertically downwards from a certain height above the surface of a sandy ground , Then penetrated through the ground 4 cm before coming to rest, given that magnitude of the ground resistance is constant and equals 6.44 kgm.wt . Find the height from which the body fell. Answer In sand , there is no driving force except body' s weight . u 0 6.44 So from B   C : "Inside ground  acceleration appears" m g  6.44  9.8  0.04 a   0.04  9.8  6.44  9.8  0.04a  a  -1568 m/s 2

S

Vc 2  VB 2  2a s  0  VB 2  2  -1568  0.04 

B VB  ?

 VB 2  125.44   VB  11.2 m/sec u 0

VB  11.2

S ?

0.04 m

V 2  u 2  2g s  11.2   2  9.8 S  S  6.4 m

mg sand

a

2

C

Vc  0

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

- 36 -

Chapter one – Newton’s laws of motion

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Example (24) A hummer of mass 500 kg fell down from a point 0.4 meters high above a piece of iron, It was embedded in it a distance one centimetre . Find the magnitude of the pressure of the hummer on the piece of iron. R Answer We will deal with this problem as if a body is fell vertically downwards u 0 From A  B : u 0 S  0.4 m A And V 2  u 2  2g s  V 2  2  9.8  0.4  7.84 0.4 m  V  2.8 m / s mg B From B  C : "Inside iron  acceleration appears" 0.01 m VB  2.8 m / s Vc  0 S  0.01 v 0 V 2  u 2  2a s

 0   2.8   2 0.01 a 2

C

 a  -392 m / sec 2

a

m g  R  m a  R  m  g  a   500  9.8   -392  

R  200900 Newton   9.8 

 R  20500 kg.wt

------------------------------------------------------------------------------------------------------Example (25) A body of mass 5 kgm is placed on a horizontal plane, and pulled by a string which is inclined at angle of measure 60 o to the horizontal so it moved on the plane along a straight line with uniform acceleration, where it covered a distance 18 meters during half a minute starting from 1 of its weight , find the magnitude 5 of the tension in the string and magnitude of the normal reaction of the plane on the body. Answer 18  S  18m during 30 sec.  V   0.6 m / s 30 0  30 And t   15 sec T 2 T sin 60 o 0.6 1  V  u  at  0.6  15a   a   m / sec a 15 25 1 o T cos60 o  R  ma  T cos60 o  w  ma R 60 o T cos60 5 1 1 1 1 1  T  mg  ma  T   9.8  5  5  2 5 2 5 25 1 1  T   9.8  T  20 Newton 2 5 mg To get the normal reaction: T sin60 o  N  m g  20 sin60 o  N  9.8  5  N  31.68 32 Newton rest. Given that magnitude of the resistance to its motion equals

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Case 3 : Vector form of the equation of Motion

 F  acted upon a particle of mass  m  , And it acquired an acceleration  a  , And the displacement vector of the particle during an interval " t" from instant of start of motion , and  V  If a force

is the velocity vector at the end of this interval , then : The equation of motion of the particle is :

 1

d H d mv   F dt dt

2 m 

a  F

 we use this rule if the mass of the body is variable

 we use this rule if the mass of the body is scalar where v 

ds dt

a 

and

 m  f t 

 m  any number 

dv dt

------------------------------------------------------------------------------------------------------Example (1) A particle of mass  m  is moving under the action of 3 coplaner forces F1  3m i  7m j , F2  5m i  2m j , F3  - m i  6m j , where i , j are 2 perpendicular vectors in the plane of the forces , Find the acceleration vector of this particle . Determine its magnitude and its direction . Answer The resultant force acting on the particle : F  F1  F2  F3

 F  3m i  7m j  5m i  2m j  m i  6m j  F  7m i  m j But

F  m a  7m  i   m  j  m a  m 7 i  j  m a

 a  7 i  j  a  49  1  50  5 2

 Tan 

y 1     8 o8' x 7

------------------------------------------------------------------------------------------------------Example (2) A body of variable mass moves along a straight line, where its mass: m  3t  2 and its displacement

1  vector S   t 3  2t  i where i is a unit vector in the direction of motion and t dentes time. 3  Find the force vector on the body and find its magnitude. Answer

ds 1  S   t 3  2t  i  V   t 2  2  i dt 3  The momentum vector  H  m V  H   3t  2   t 2  2  i   3t 3  2t 2  6t  4  i dH   9t 2  4t  6  i And its magnitude F  9t 2  4t  6 dt Note : we used this rule as mass is variable  m  f  t   Force  F 

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (3) A particle of mass 4.2 units , a force F acted upon this particle , its displacement due to this 2  force during a time "t" was S   t 2  5t  i where i is a unit vector in the direction of 3  motion . Find the momentum vector of this at any instant , and its magnitude when t  3 , find also the vector F and its magnitude when t  3. Answer ds  4 dv 4 2   S   t 2  5t  i  V    t  5 i  a   i dt 3 dt 3 3  

At the instant "t" , the momentum vector  H  m V 4  H  4.2   t  5  i   5.6 t  21 i 3  when t  3  H   5.6  3  21 i  37.8 i  F  m a Where:  F  4.2 

4 i 3

H  37.8

 F  5.6 i 

F  5.6

------------------------------------------------------------------------------------------------------Example (4) A particle of mass 6 units , A force F  24 i acted upon this particle , where i is a unit vector in the direction of motion , if the velocity vector of this particle at the end of the time "t" is   3k t  b  i where k , b are constants , find the momentum vector of this body when t  2,

V

if the initial velocity vector  9 i .

V   3k t  b  i

Initial velocity means the velocity at t  0  V   3k  0  b  i  b i But initial velocity vector  9 i   b i  9 i   b  9 F  m  a Where a 

d v  3k i dt

24 4    V   4t  9  i 18 3 H  mv  6  4t  9  i  At t  2 : H  mv  6  4  2   9  i  102 i

 24 i  6  3k i  24 i  18 k i  24=18 k  k  And

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (5) The mass of a particle one unit , a force F acted upon this particle so its displacement during 1 4  the time t is given by S   Lt 3  mt 2  i , where L , m are constants and i is a unit vector 2 3  in the direction of motion , If the momentum vector  69 i when t  3 and F  47 i at the same instant . Determine the displacement vector at the same instant . Answer 4 3 1 2 m  1 unit , S   Lt  mt  i , H  69iˆ And F  47iˆ When t  3 2 3  1 4  S   Lt 3  mt 2  i 2 3  When t  3 :

 V 

ds   4 Lt 2  mt  i dt

 a 

dv   8 Lt  m  i dt

V   4 L  9  m  3  i   36 L  3m  i

Momentum at that instant : H  m V  1   36 L  3m  i  69 i 36 L  3m  96 When t  3 :

12 L  m  23

   1

a   24 L  m  i

Equation of motion is : F  m a

47 i  1   24 L  m  i

24 L  m  47     2  Substracting  2    1 : From  2  :

12 L  24

L2

48  m  47

m  -1

1 1  4  8 Then by Substituting in : S    2t 3    -1  t 2  i   t 3  t 2  i 2 2  3  3 1 8  When t  3 : S    27   9  i  67.5 i 2 3 

------------------------------------------------------------------------------------------------------Example (6) A body of mass 1.5 kg moves such that the two algebraic components of its velocity in the horizontal and the vertical directions are Vx  4 m / sec and Vy   -9.8t  4  m/sec, determine the magnitude and the direction of the initial velocity of this body and also the force vector acting on it. Answer ˆ ˆ Vx  4  4i and Vy   -9.8t  4  j   V  4iˆ   -9.8t  4  ˆj

Initial velocity means the velocity at t  0   at t  0:  u  4iˆ  4 ˆj Its magnitude is u 

4 

2

dv  -9.8 ˆj  and dt – 3rd secondary Dynamic a 

  4   4 2 m / s  its direction : Tan  2

F  m a  1.5   -9.8 ˆj   -14.7 ˆj - 41 -

4  1    45 o 4

Chapter one – Newton’s laws of motion

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Example (7) A particle of mass one unit moves under 3 coplaner forces F1  L i  3 j , F2  -2 i  j F3  5 i  m j , Where i , j are 2 perpendicular unit vectors in the same plane of the forces where L , m are constants , given that the displacement vector of this particle given by : 3  S  2t i   t 2  t  j after a time "t", determine the values of the constant L, m . 2  Answer Resultant of the forces acting on the particle  F  F1  F2  F3

F1  L i  3 j  2 i  j  5 i  m j   L  3  i   m  2  j    1 3  S  2t i   t 2  t  j 2 

v 

d s  2 i   3t  1 j dt

Equation of motion of the particle is F  m a

a 

d v 0 i  3 j dt

From  1 , 2  :  L  3  i   m  2  j  1 0 i  3 j  0 i  3 j L30

and

m2 3

 L  -3

And m  5 .

------------------------------------------------------------------------------------------------------Example (8) A ball of mass 7 gm moves along a straight line inside air loaded with dust, such that dust accumulates on its surface at a rate 0.5 gm/sec and the ball displacement is given as a function 1  of the time  t  by S   t 3  4t  2  Cˆ , where Cˆ is a unit vector in the direction of the motion 3  of the ball, then find the force vector acting on the ball and find its magnitude at t  2 seconds. Answer After 2 seconds, the dust accumulates will be 0.5  2  1gm Then the mass of the ball  its mass  mass of dust accumulates after 2 seconds  7  1  8 gm

d s d v 1  S   t 3  4t  2  Cˆ  V    t 2  4  Cˆ  a   2t Cˆ dt dt 3  We can get the force vector  F  m a  F  8  2t  Cˆ  16 t Cˆ At t  2 seconds: F  16  2  Cˆ  32 Cˆ  its magnitude is 32 Dyne

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Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (9) 2 If the position vector P of a particle after  t  is given by P   2t  1 t  2  Cˆ , where Cˆ is a

 

constant unit vector , the magnitude of P in cm and  t  in seconds. Determine the velocity 1  t  2  gm after time  t  , then 2 find the force vector acting on the particle and calculate its magnitude when the particle stops. Answer We have two methods to get S : vector and if the mass of the body is variable and given by m 

 

1 The first method is to find: Then V 

S  P  Po where Po is the position when t  0

d S directly dt

d p directly which is offcourse the easiest method dt P   2t  1  t 2  4t  4    2t 3  7 t 2  4t  Cˆ

 2  The second method is to find: So V 

V 

d p   6 t 2  14t  4  Cˆ And dt

1 m  t  1  which is variable 2

dH dv 1  m   t  1  12t  14  Cˆ  6t 2  5t  14  Cˆ dt dt 2  To get the magnitude of the force vector when the particle stops : So we must used F 

So, particle stops when V  0  6t 2  14t  4  0

 2 

 3t 2  7t  2  0

1 So  3t  1 t  2   0  t  sec Or t  2 sec 3   1 2  1 -35 ˆ 35 1 When t  sec : F   6    5    14  Cˆ  C  Its magnitude is  3  3 3 3 3   2 When t  2 sec : F  6  2   5  2   14 Cˆ  20 Cˆ  Its magnitude is 20

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Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Case 4: Motion of a body on an inclined planes 1st : Smooth planes planes

Note : you have to know that “smooth plane” has no resistance and no friction

1

If the motion upward the plane : F  mg Sin  m a R  mg Cos

 2

    1

If the motion downward the plane :

 mg Sin   F  m a

   1

R  mg cos 

   2 

   2

Notes : From the above relations :

1 If F  m g Sin , Then the motion is upward with uniform acceleration  a  and is determined:

F  m g Sin  m a

 2  If F  m g Sin , Then the motion is downward with uniform acceleration  a  and is determined:

m g Sin  F  m a

 3  If F  m g Sin , Then the motion is at rest or with uniform velocity and a  0  4  If F  0 , and the motion is upwards the plane   a  - g Sin ‫ي ح ف ظ‬  5  If F  0 , and the motion is downwards the plane   a  g Sin ‫ي ح ف ظ‬

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (1) A body of mass 2.5 kg placed on a smooth plane inclined to the horizontal at an angle  where 3 Sin  moves under the action of force F acting along the line of the greatest slope. Find 5 the magnitude and the direction of the acceleration of the motion of the body and the normal reaction of the plane to the body if:

 a  F  9.7 Newton

b  F  1.5 Kgm.wt

 c  F  17.2 Newton

Answer 3 m g Sin  2.5  9.8   14.7 Newton  1.5 kg .wt 5  a  When F  9.7 Newton : F  m g Sin  then the motion is downward m g Sin  F  m a  14.7  9.7  2.5 a  a  2 m / sec 2 5 4 And R  m g Cos   R   9.8   19.6 Newton  2 kg.wt 2 5 F  m g Sin   b  When F  1.5 kg.wt : [

Then the body moves with a uniform velocity :  a  0 5 4 And R  m g Cos   R   9.8   19.6 Newton  2 kg.wt 2 5 F  m g Sin   c  When F  17.2 Newton : Then the motion is downward : F  m g Sin  m a  17.2  14.7  2.5 a  a  1m / sec 2 5 4 And R  m g Cos   R   9.8   19.6 Newton  2 kg.wt 2 5

------------------------------------------------------------------------------------------------------Example (2) A body of mass 4 kg placed on a smooth plane inclined to the horizontal at an angle 60 o , a horizontal force F of magnitude 12 3 kg .wt towards the plane acts on the body. Find the magnitude and the direction of the acceleration of the motion of the body and also find the normal reaction of the plane to the body . 12 3  9.8  Cos60

o

And m g Sin60 o  4  9.8 Sin60 o 

98 3 5

12 3 Cos60 o R

33.95 Newton

60 o

Then the motion is upward  12 3  9.8  Cos60 o  4  9.8 Sin60 o  4 a  a  16.97 m / sec 2 3 1 R  12 3  9.8   4  9.8   196 Newton 2 2  R  196  9.8  20 kg.wt Dynamic – 3rd secondary

60 o

m g Sin60 o

60 o

12 3

12 3 Sin60 o mg Cos 60 o

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mg Chapter one – Newton’s laws of motion

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Example (3) A body of mass 40 kg is placed on a smooth inclined plane whose inclination with the horizontal 3 is  , where Sin  . An inclined force whose inclination with the greatest slope to the plane is  5 5 where Sin   , acts on the body upwards the plane. The body moves upwards the plane, a 13 5 distance 225 cm in sec starting from rest. Then find the magnitude of the force and the normal 7 reaction of the plane to the body . Answer 5 S  225 cm  2.25 m t  sec u 0 7 1 1 25 S  u t  at 2  2.25=  a  a  8.82 m / sec 2 2 2 49 And F Cos  m g sin  m a 12 3  F  40  9.8   40  8.82   F  637 Newton 13 5 And N  F Sin   m g Cos  0 4 5 N  40  9.8   637   68.8 Newton 5 13

------------------------------------------------------------------------------------------------------Example (4 “important”) 1 A smooth plane inclined at an angle of Sin to the horizontal, a body is projected upwards with 20 velocity 196 cm/sec along the line of the greatest slope of this plane. Find the maximum distance which the body moves before it comes to rest, and the time taken to descend to the same point of projection Answer 1 F  0   a  - g Sin  -9.8   -0.49 m/sec 2 20 u  1.96 m / sec a  -0.49 V 0 a 2 2 2 V  u  2a S  0  1.96   2 0.49  S   S  3.92 m When the motion is downward :

m g Sin  1 2 F  0   a  g Sin  9.8   0.49 m/sec  20 u 0 a  0.49 S  3.92 1 1 And S  u t  at 2  3.92   0.49  t  t  4 seconds 2 2

------------------------------------------------------------------------------------------------------Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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2nd : Rough planes planes

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Note : In “rough planes” resistance Or friction force exists Example (1) A small wooden box of mass 5 kg is placed at the top of an inclined rough plane of 2 meters length 3 and height 1.2 m, the box slipped and arrived to the bottom of the plane after 1 sec, find the 7 magnitude of the acceleration of the body and the magnitude of the force of friction and also the coefficient of friction. Answer 1.6 R  5 g Cos   5  9.8   39.2 Newton u 0 2 a FF 10 1 2 cm u 0 S 2 t And S  u t  at 2 R 7 2 10 t sec 1  100  196 7  2  a   a   1.96 m / sec 1.2  2  49  100  So, m g Sin   FF  m a 5 g Sin 

 1.2    5  9.8     FF  5 1.96   2 

1.6

 29.4    39.2   9.8  39.2   19.6     And

FF   R 

5 g Cos 

1 2

5g

1  39.2  19.6 Newton  2 kg.wt 2

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Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (2) A body of mass 24 kgm is projected with velocity of magnitude 14 m/s in the direction of the line 1 of the greatest slope upwards of a rough plane inclined at an angle of Sin to the horizontal. 40 If the magnitude of the resistance to the motion of this body equals 360 gm.wt. Find the distance which this body covers before it comes to rest, and if this body returned along the same line under the same resistance. Then find the magnitude of the velocity of the body when it reaches the same point of projection. Answer In this problem, the "FF" is not mentioned , so we will use resistance instead So,

F  R  m g Sin   m a

a

motion

360  1   0  9.8   24  9.8     24 a 1000  40  -1176  24 a    a  -0.392 m / sec 2 125 So, u  14 m / sec a  -0.392 m / sec 2 V 0 S  ?? So,

V 2  u 2  2a s

R

m g Sin

0  14   2  -0.392  S 2

mg

1 360 m g Sin  R  m a'   9.8  24 a'  24  9.8      40  1000 N 2 a'  0.098 m / sec S  250 m

So,

V 2  u 2  2a s

a'  0.098 m / sec 2 

m g Cos 

 S  250 m When the body returns back :

So,

F

N

a

R

u 0 

V 2  2 0.098  250 

V  7 m / sec

m g Sin

m g Cos 

mg

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Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (3) A body starts sliding from the top of an inclined plane whose length is 260 cm, the plane is inclined 5 to the horizontal at an angle of Sin . Find its velocity when its reaches the bottom of the plane 13 1 given that the coefficient of friction between the body and the plane equals . And if the body is 4 projected upwards the plane from the lowest point on the plane, find the velocity that is required to make it reach the top of the plane. Answer 12 588 R  m g Cos   m  9.8   m u 0 13 65 a FF  5  So, m g Sin  FF  m a    9.8 m      R  m a R  13  260 cm 49 588 49 588  1   m m  ma     a  13 65 13 65  4  a u 0

98 m / sec 2 65

S  2.6

a

98 65

mg

 98  V 2  u 2  2 a S  2    2.6   7.84   V  2.8 m / s  65  To get the ascending velocity: - m g Sin   R  m a'   -

S  2.6

49 588  1  m   m  m a' 13 65  4 

a -

392 65

13

5

12 V 0

a

49 588  1  392  m / sec 2    a'   a'  13 65  4  65

u  ??

m g Cos 

m g Sin 

R 260 cm

V 0 u  ??

 392  V 2  u2  2 a S  0  u2  2   2.6   65 

m g Sin

FF

 m g Cos 

 u  5.6 m / s mg

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (4) 1 to the horizontal. A body 100 started motion from the top of the incline with velocity of magnitude 14 cm / sec, and when it An inclined plane of length 3 meters and inclined at an angle of Sin

reached the bottom of the incline, it continues its motion on a horizontal plane along the straight line. If the body is moving with a uniform acceleration in each of the two distances and the magnitude of the velocity of the body does not change when the body leaves the inclined to the 1 of the weight of the 200 body. Find the distance which the body moves on the horizontal plane before it comes to rest. Answer In this problem, the "smooth" plane is not mentioned , so resistance appears Motion from A  B : horizontal plane and magnitude of the resistance in each stage equals

u A  0.14 m / sec So

m g Sin  R  m a

1 1 W m g  0.049 m 200 200 N  1   9.8 m   0.049 m  m a  m  a   100 

S 3 m

R

motion

9.8  a  0.049  0.049 m / sec 2 100 So VB2  u A2  2 a S

 m g Sin

VB2   0.14   2  0.049  3   0.3136 2

R

VB  0.56 m / sec

 B

C

mg

u B  VB  0.56 m / sec

VC  0

 - R  m a'  -0.049 m  m a' V 2  u2  2 a S

A

m g Cos 

Motion from B  C :

So

R

 a'  -0.049 m / sec 2 0  0.56   2  -0.049  S 2

S  3.2 m

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (5) A body is left to slide from the top of an inclined plane that ends at a horizontal plane, the 1 coefficient of friction between the body and each of the planes equals , if the inclined plane is 3 4 150 cm long and the tangent of its angle of inclination to the horizontal equals . Find the 3 distance covered by the body on the horizontal plane when its velocity becomes 140 cm/sec, given that the velocity of the body doesn't change when it transferred from the inclined to the horizontal plane. Answer 1 3 Motion from A  B : FF   R  R where R  m g Cos   980  m  588 m 3 5 4 1 5 So m g Sin  FF  m a  980  m   588  m  m a   m  4 5 3  4 1  a  980    588   588 cm / sec 2 3 5 3 FF R u 0 S  150 a  588 So

V 2  u 2  2 a S  2  588  150   176400   V  420

a

motion

1 980 Motion from B  C : here FF  m g  m 3 3 R m g Sin 980  - FF  m a'  m  m a' 3 V  140 FF  980 2 B C  a'  m cm / sec mg 3 980 u  420 VC  140 a'  m cm/sec 2 3 980 2 2 So V 2  u2  2 a S   140    420   2  - m  S  3 

A

m g Cos 

mg

S  240 cm

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (6) A body slides over a rough plane inclined to the horizontal at 45o ,If the coefficient of friction is

3 4

show that the time required to cover a distance "S" is double the time required to cover the same distance if the plane is smooth.

Answer When the body is rough,  Resistance appears  R  m g Cos 45 o And

u 0

a

m g Sin45 o   R  m a

R

 2 3  2   9.8 m      9.8  m ma  2  4  2  

FF

45o

49 2 147 2 49 2  a   a m / sec 2 10 40 40

49 2 S s 40 1 1  49 2  2 49 2 2 S  u t  at 2   t1     1 t  2 2  40  80 When the plane is smooth  No resistance exists  u 0

S

 m g Sin45 o  m a   a  9.8 Sin45 o  u 0

m g Cos 45o

m g Sin 45o 45o

S s

a

mg

49 2 m / sec 2 10

49 2 10

a

R

1 1  49 2  2 49 2 2 S  u t  at 2   t2     2  t  2 2  10  20

45o

m g Cos 45o

m g Sin 45o 45o

49 2 2 t2 S 20 By dividing  1 and  2  :  S 49 2 2 t1 80

mg

t12 80 t  2 4   1 2 t2 20 t2

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Newton’s Third law Motion of a body in a lift moving vertically For every action, there is a reaction equal in magnitude and opposite in direction. The third law studies only the mutual effect between two connected bodies, where one of them

 

acts upon the other one by force F called the ACTION , Then the other body will react on the

 

first one by force -F called the REACTION .

 

This means that the resultant of the two forces vanishes F  -F  0

Explanation of the law Placing of a body on a plane :

T

 a  Action: Pressure of the body on the plane  P  vertically downward   b  Reaction: Opposite force acts on the body  T  vertically upward  1 P and T are equal in magnitude.  2  P and T are opposite in direction.  3  P and T have the same line of action.

P

Spring balance

Suspension of a body by a spring balance :

 a  Action: Tension force of the body on the plane  T  vertically upward   b  Reaction: Tension force acts on the body  T  vertically downward  1 P and T are equal in magnitude.  2  P and T are opposite in direction.  3  P and T have the same line of action.

T T

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

Rules

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Notes:  1 T  Apparent weight  Spring balance  Balance reading  records 

 2  m g  Common weight

 Pressure balance  Pressure reading  records 

If the lift is moving up with acceleration or moving down with retardation The equation of motion: T  m g  m a  T  m g  m a

T

a

 T  m g  a Then The apparent weight T   Common weight  m g 

mg

------------------------------------------------------------------------------------------------------If the lift is moving down with acceleration or moving up with retardation The equation of motion: m g  T  m a  T  m g  m a

T

a

 T  mg  a Then Common weight  m g   The apparent weight T 

mg

------------------------------------------------------------------------------------------------------If the lift is at rest or moving with a uniform velocity The equation of motion: T  m g  0

T

 T mg Then The apparent weight T   Common weight  m g 

mg

-------------------------------------------------------------------------------------------------------

Summary of rules

Lift is moving up

 Acceleration up  

T  m g  a

And T  m g

Lift is moving down

T  m g  a

And m g  T

Acceleration down   

Lift is at rest

T  mg

a  0 Dynamic – 3rd secondary

- 53 -

And T  m g

Chapter one – Newton’s laws of motion

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Example (1) A woman of mass 56 kgm stands on the floor of the lift . Calculate in kgm.wt the pressure of the floor of the lift in each of the following cases:

1 If the lift is moving vertically with uniform velocity.  2  If the lift is moving upwards with uniform acceleration of magnitude 1.05 m / sec 2 .  3  If the lift is moving downwards with uniform acceleration of magnitude 1.05 m / sec 2 .  4  If the lift is moving upwards with uniform retardation of magnitude 1.05 m / sec 2 . Answer

Question

1 If the lift is moving vertically with uniform velocity.  2  If the lift is moving upwards with uniform acceleration of magnitude 1.05 m / sec 2 .  3  If the lift is moving downwards with uniform acceleration of magnitude 2

Answer  T  m g  T  56  9.8  548.8 Newton  T  548.8  9.8  56 kg.wt  T  m  g  a   T  56  9.8  1.05   607.6 Newton  T  607.6  9.8  62 kg.wt  T  m  g  a   T  56  9.8  1.05   490 Newton  T  490  9.8  50 kg.wt

1.05 m / sec .  4  If the lift is moving upwards with

This is the same as lift is moving downwards with

uniform retardation of magnitude

uniform acceleration of magnitude 1.05 m / sec 2 ,

1.05 m / sec 2 .

which is the same as the third  50 kg.wt

------------------------------------------------------------------------------------------------------Example (2) A lift moves vertically with uniform acceleration, and during the motion a body is weighted by a spring balance fixed to the lift then by common balance . In the first case, the reading is 3.99 kgm.wt and in the sec ond case, the reading is 3.8 kgm.wt. Then find the direction and the magnitude of the lift acceleration Answer In the case of using the common balance: The reading  m g  3.8  9.8  37.24 Newton    1  also m  3.8 kg In the case of using the spring balance: The reading  T  3.99  9.8  39.102 Newton     2  From  1 and  2  :

T  m g   The lift is moving upwards

Then the rule will be in that case : T  m  g  a   m g  m a where m  real weight Dynamic – 3rd secondary

  39.102  37.24  3.8 a - 54 -

a  0.49 m / sec 2 Chapter one – Newton’s laws of motion

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Example (3) A body of mass 10 kgm is suspended to the hook of a spring balance fixed to the ceiling of a lift , If the lift is moving vertically with uniform acceleration, then find the acceleration and its direction when the apparent weight of the body is :

a

 b  10.5 kg.wt

8.5 kg.wt

 c  10 kg.wt

Answer Note : if the acceleration is unknown, find the apparant weight T  and the common weight  mg 

 a  T  8.5  9.8  83.3 Newton

& m g  10  9.8  98 Newton So, m g  T  Then the lift moves downward Its rule : T  m  g  a   m g  m a  83.3  98  10 a 14.7 10 a  98  83.3  10 a  14.7  a   1.47 m / sec 2 10  b  T  10.5  9.8  102.9 Newton & m g  10  9.8  98 Newton So, T  m g  Then the lift moves upward Its rule : T  m  g  a   m g  m a  102.9  98  10 a 4.9 10 a  102.9  98  10 a  4.9  a   0.49 m / sec 2 10  c  T  10  9.8  98 Newton & m g  10  9.8  98 Newton So,

T  mg

Then the lift is at rest

T

mg

 a 0

------------------------------------------------------------------------------------------------------Example (4) A body of mass 280 kgm is placed on the base of a box of mass 700 kgm and the box is lifted by a vertical rope to move with a uniform acceleration of magnitude 70 cm / sec 2 . Find the magnitude of the body pressure on the box base and the magnitude of the tension in the rope which liftes the box and if the rope is cut, find the pressure on the box base. Answer  First  To find the pressure on the base of the box:

The box bear a mass 280 kgm Then the pressure: T  m  g  a   280  9.8  0.7   2940 Newton

T

a

 Second  To find the tension on the rope:

The rope bear a mass 280 kgm and 700 kgm Then the tension: T  m  g  a   980  9.8  0.7   10290 Newton

Third 

T

mg

a

After cutting the rope:

After cutting the rope, the box will move downward according to acceleration 9.8 m / sec 2 Then the tension: T  m  g  a   280  9.8  9.8   0 Dynamic – 3rd secondary

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 m  m'  g Chapter one – Newton’s laws of motion

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Example (5) A body is suspended from the end of a spring balance fixed in the ceiling of an elevator, when

the lift is moving upwards with a uniform acceleration of magnitude  a  m / sec 2 . The balance reading was 7.5 kg.wt and when the lift is moving downwards with a uniform acceleration of magnitude  2a  m / sec 2 . The balance reading was 6 kg.wt. Find the mass of the body and magnitude of a. If the chain holding the elevator can not support a tension greater than the weight 1.5 tons , find the maximum load the elevator could raise while ascending with acceleration a given that the mass of the elevator when empty is 500 kg Answer Let the mass of the body is  m  kg and the balance reading is T 

T

The lift is moving upwards with a uniform acceleration  a  :  T  m g  a

a

7.5  9.8  m  9.8  a     1

The lift is moving downwards with a uniform acceleration  2a  :  T  m g  a

6  9.8  m  9.8  2a      2 

mg

5 9.8  a  4 9.8  2a By cross multiplication: 39.2  4a  49  10a 9.8  14a  9.8  a   0.7 m / sec 2 144 Then subtitute in  1 : 7.5  9.8  m  9.8  0.7   m  7 kg

 From  1 and  2  and by division:

T

a

mg

Let the mass of the lift containing the maximum cargo be m'  1.5  1000  9.8  m'  9.8  0.7   m'  1400 kg Then the maximum cargo can be loaded is 1400  500  900 kg

------------------------------------------------------------------------------------------------------Example (6) A crane lifts a mass of 4 tons and its chains bears a maximum tension of 5 tons.wt. Find the maximum acceleration which the crane can lift the body without breaking the chains, and find the time taken to lift the body with the acceleration a distance of 19.6 meters. Answer T  5  1000  5000 kgm.wt  9.8  49000 Newton & m g  4  1000  9.8  39200 Newton So, T  m g   The lift is upwards And :

T  m g  a  m g  ma

 4000a  49000  39200 1 S rd ut  at 2 Dynamic – 3 secondary 2

And

 

 49000  39200  4000 a

4000a  9800 19.6 

1  2.45t 2  - 56 2

a

9800  2.45 m / sec 2 4000

t 2  16

t  4 sec

Chapter one – Newton’s laws of motion

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Example (7) A lift has a balance on its floor is loaded with coal when it is at rest in the bottom of a mine 35 meters deep under the ground. The reading of the balance is 70 kg.wt, then the lift ascends with uniform acceleration 2.45 m / sec 2 a distance of 10 meters, then with uniform velocity a distance of 5 meters. Then with retardation till it stops at the top of the mine. Find the reading of the balance measured in kg.wt during the three distances. Answer * When the lift is at rest and its reading is 70 kg.wt :

Then

T  mg

 70  9.8  9.8 m

 m  70 kg

* When the lift ascends with uniform acceleration 2.45 m / sec 2 : Then

T  m g  a

 T  70  2.45  9.8   857.5 Newton  T 

857.5  87.5 kg.wt 9.8

* When the lift moves with uniform velocity : 686  70 kg.wt 9.8 * When the lift ascends with retardation : Means it moves downward with acceleration Then

T  mg

 70  9.8  686 Newton

Then

T  m  g  a      1

 T

So, in order to find  a  , we have to get the velocity in this part. So, from the begining:

u 0

 At rest 

S  10 m a  2.45 m / sec 2

And V 2  u 2  2 a S  2  2.45  10  49  V  49  7 m/s Then this velocity will be the initial velocity at the last part: Then u  7 m / sec V  0 S  35  10  5  20 m V 2  u2  2 a S

 0  49  40 a

 a

-49 40

49   By substituting in 1 : T  70  9.8    600.25 Newton  9.8   61.25 kg.wt 40  

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Applications on Newton’s laws – Connecting strings we are dealing with motion of bodies connected by strings. We assume that the length of a string remains constant (inextensible). We also assume that the weight of a string is too small compared to the weight of the connected bodies, so that we may neglect its weight. We have three types of applications st

1 application Motion of a system of two bodies hanging vertically from the ends of a string which passes over a pulley. To study this problem, we solve the motion of each body separately, this is because problems involving particles which are connected by strings are moving in different directions. So, since m1 is larger than m2 we expect m1 to move vertically downwards, and m2 to move vertically upwards. Let e1 , e2 be the unit vectors directed respectively along the downward and upward verticals. Let a be the algebraic measure relative to e1 of the acceleration vector of body of mass m1 , then : Rule  1

T  m2  g  a 

when the movement is upward

T  m1  g  a 

when the movement is downward

From that, we can say that: m1  g  a   m2  g  a  Then the acceleration of the whole system : Rule  2  a  where m1 : greatest mass

 m1  m2   g  m1  m2 

g  9.8 m / sec 2 or 980 cm / sec 2

m2 : lowest mass

------------------------------------------------------------------------------------------------------Important remarks (1) The distance between the two bodies is S  2 (2) The maximum height from the original position is Smax

U2  S 2g

(3) When the bodies are moving on the pulley, then we are dealing with

and Timemax 

U g

 g  a  , but when the

strings are cut, then we are dealing with gravity (g) only. (4) If the pulley is holded, then T  m g only (5) The pressure on the pulley : The string exerts two forces on the pulley each of magnitude T, which are directed vertically downwards, as in the opposite figure. The resultant of these two forces is called the pressure on the pulley, its magnitude is: Dynamic – 3rd secondary

P  2T - 58 -

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Example (1) Two bodies of masses 420, 280 gm are tied at the two ends of a light string which passes over a fixed smooth pulley, the system starts motion vertically from rest, find:

 a The magnitude of the acceleration of the system.  b The magnitude of the tension of the string in gm.wt.  c The pressure on the pulley.  d The velocity of the two bodies after 3 seconds from the begining of the motion.  e The vertical distance between the two bodies after 3 seconds. Answer

 a  To get the acceleration:

a

m1  m2 g m1  m2

420  280  980  196 cm/sec 2 a 420  280  b  To get the tension,we can use the rule of the body of mass  420 gm: a 

body is downward  descending    T  m  g  a   T  420  980  196   329280 dyne  980  336 gm.wt

 c To get the pressure on the pulley: P  2T  P  2  336   672 gm.wt  d  V  u  at   V  196  3  588 cm / sec

a

T t 3 u 0

T

t 3

280

1 1 2  e  S  u t  at 2   S  196  3   882 cm 2 2 Then the vertical distance between the two bodies is S  2  882  1764 cm

u 0

420

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (2) Two scale pans each of them is of mass 20 gm are hanged at the two ends of a light string passes over a smooth pulley, a body of mass 140 gm is placed on one of the two pans, and the another body of mass 100 gm is placed on the other pan, calculate:

 a The magnitude of the common acceleration of the system.  b The magnitude of the tension of the string.  c The magnitude of the pressure on each scale pan. Answer the first body is decending downward under the action of mass  20  140  160 gm and the second body is ascending upward under the action of mass  20  100  120 gm

 a  To get the acceleration:

a

m1  m2 g m1  m2

160  120  980  140 cm/sec 2 a 160  120  b  To get the tension, we can use the rule of the body of mass  160 gm: a 

body is downward  descending    T  m  g  a   T  160  980  140   134400 dyne  980 

 c To get the pressure on each pan:

a

T

960 gm.wt 7

T 20 + 100

for the 1st pan

for the 2 nd pan

Downward

upward

T1  m1  g  a 

T2  m2  g  a 

T1  140  980  140 

T2  100  980  140 

20 + 140

800 gm.wt 7 ------------------------------------------------------------------------------------------------------T1  117600  980  120 gm.wt

Dynamic – 3rd secondary

T2  112000  980 

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Chapter one – Newton’s laws of motion

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Example (3) Two bodies of masses 140, m gm are connected at the two ends of a string which passes over a smooth pulley, if the system starts motion from rest when the two bodies are on the same horizontal level and the magnitude of the pressure of the pulley is 240 gm.wt, find the magnitude of  m  and the vertical distance between the two bodies after 2 seconds from the begining of motion.

Answer Let the body whose mass is  m  be the higher mass so to descend downward And

P  2T   240  2T   T  120 gm.wt

T  m  g  a  is the rule of the body whose mass is 140 gm  120  980  140  980  a    140   840  980  a   a  -140 cm/sec

a

2

Then substitute in the rule of mass  m  :

T  m g  a

a

T t2

120  980  120  980  m  980   -140     m   105 gm u  0 980  140 To get the vertical distance: u  0 t 2 a  140

T

u 0 t2

140

1 1 2 S  u t  at 2   S  140  2   280 cm 2 2 Then the vertical distance between the two bodies is S  2  280  560 cm

m

------------------------------------------------------------------------------------------------------Example (4) Two bodies are hanged at the two ends of a string passing over a smooth pulley, the string holds

a light spring balance  neglect its weight  . If we hold the string at the pulley, the reading of the balance is 200 gm.wt and if we left the string and the system started motion the reading became 300 gm.wt, find the mass of each of the two bodies. Answer Let the spring balance be put at the string of the body of m  m1 The string is holded, then a  0 " rest state"   T  m1 g  200 gm.wt

a

 T  m1  980  200  980 dyne   m1  200 gm When the string is moving  T  300  980  294000 dyne And T  m1  g  a   294000  200  980  a 

a

T

T

 980  a  1470   a  980  1470  -490 sm/sec 2 To get the body of mass m2 :

T  m2  g  a 

m2

 294000  m2  980   -490     m2  600 gm Dynamic – 3rd secondary

- 61 -

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Chapter one – Newton’s laws of motion

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Example (5) Two bodies of masses of 500 gm,  m  gm are tied to the ends of a string passing over a smooth pulley, the system starts motion with velocity 40 cm/sec, such that the mass  m  descends with this velocity vertically downwards, and after 4 seconds, it is found that this mass returned to its starting position, prove that the magnitude of the acceleration equals 20 cm/sec 2 , and find the value of  m  and the magnitude of the tension in the string in Newton. Answer The mass return back to its position means S  0 : 1 1 2 S  u t  at 2   0  40  4    4  a   a  -20 cm/sec 2 2 2 m  m2 m  500 a a 1  g   -20   980 m1  m2 m  500 m  500 -20 -1      49m  24500  - m  500 m  500 980 49

t 4

 50 m  24000  m  480 gm

u  40

To get the tension:

a

T

T

u  40 t 4

500

T  m1  g  a   480  980   - 20  

 480000 Dyne   10 5   4.8 Newton

m

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (6) Two bodies of masses 42,  m  gm are hanged from the ends of a light string which passes over a smooth pulley, the system starts motion from rest when the two bodies are on the same horizontal level, and the mass  m  moved vertically downwards till the vertical distance between it and the other body became 1260 cm after 3 sec. from the begining of motion, find the magnitude of the acceleration and the value of  m  . And if the string is cut at this instant, find the maximum height that the mass of 42 gm can reach.

Answer Let the body whose mass is  m  be the higher mass so to descend downward The system before cutting:

2 S  1260   S  630 cm

1 1 2 S  u t  at 2   630   a  3   a  140 cm / sec 2 2 2 m  m2 m  42 And a 1  g   140   980   140  m1  m2 m  42  1

m  42  7   m  42  7m  294 m  42

a a

T

S  630 t 3 u 0

 6m  336   m  56 gm

T 42

u 0 t 3 S  630

2

u where u is the velocity after 3 seconds 2g m So in order to get the velocity after 3 sec: u  0 t 3 a  140 S  630

The system after cutting:

S max 

V  u  at   V  140  3   420 cm / sec u 2  420     90 cm  the body can reach is 90  630  720 cm from its 2g 2  980 original position. 2

Smax

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (7) A light string is passing over a smooth small pulley and holds at its two ends two bodies of masses of 500, 300 gm vertically below the pulley, write down the equation of motion of each of the two masses, then determine the magnitude of the acceleration of the system, and the magnitude of the tension in the string, and if the system starts motion from rest then the string is cut after 2 seconds. Mention the maximum height that the mass of 300 gm ascended from its original position, and the time taken to inverse its motion. Answer The system before cutting: T  m1  g  a  is the rule of the body whose mass is 500 gm T  m2  g  a  is the rule of the body whose mass is 300 gm

And

a m1  m2 500  300 2 And a g  a  980  245 cm/sec m1  m2 500  300 T Then substitute in T  m1  g  a   T  500  980  245   367500   980   375 gm.wt

S  ?? t2 u 0

The system after cutting:

T 300

u 0 t2 S  ??

2

u where u is the velocity after 2 seconds 2g So in order to get the velocity after 2 sec: u  0 t 2 Smax 

a

500

a  245

V  u  at   V  245  2   490 cm / sec u 2  490  Smax    122.5 cm , now we have to get the distance  S  from the starting point 2g 2  980 So, u 0 t 2 a  245 2

1 1 2 S  u t  at 2   245  2   490 cm 2 2 Then the body can reach is 122.5  490  612.5 cm The time of the maximum height: T

u 490 1   sec. g 980 2

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (8) Two bodies of masses 45, 25 gm are attached to the two ends of a light string passes over a smooth pulley, the system started its motion when the two bodies were on the same horizontal 1 level at a hight of 65 cm from the floor surface and the string is cut after seconds from the 2 begining of the motion, find:

 a  The magnitude of the acceleration of the system.  b  The velocity of the system immediately before cutting the string.  c  The velocity of each of the two bodies when they reach the floor surface. Answer The system before cutting: a

m1  m2 45  25 g  a  980  280 cm/sec 2 m1  m2 45  25

We have to find the velocity just before cutting: u 0

t  0.5

a

a  280

V  u  at   V  280  0.5   140 cm / sec

v  140

So in order to get the velocity of the system when the body reaches the floor, we have to find the distance of each one

t  0.5 u 0

0.5 seconds before. So,

U 0

a

T

T 25

t  0.5

a  280

u 0 t  0.5 v  140

65 cm

1 1 2 S  u t  at 2   280  0.5   35 cm 2 2 The system after cutting:  use the rules of gravity 

45

floor

Note : when they reach the floor means that the rules of gravity is positive The body of m2  25

The body of m1  45

S 2  65  35  100 cm

S1  65  35  30 cm

U  140

U  140

V2  ?

S 2  100

V1  ?

S1  30

V 2  u2  2 g S V2 2   140   2  980  100 

V12   140   2  980  30 

2

2

 215600

 78400

V2  215600  464.32 cm

V1  78400  280 cm

------------------------------------------------------------------------------------------------------Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (9) Two bodies of masses 280, 210 gm are attached to the two ends of a light string passes over a smooth pulley, the system started its motion when the two bodies were on the same horizontal level, if the string is cut after 3 seconds from the begining of the motion, find the distance between the two bodies after one second from cutting the string. Answer The system before cutting: a

m1  m2 280  210 g  a  980  140 cm/sec 2 m1  m2 280  210

We have to find the velocity just before cutting: u 0

t 3

a

a  140

V  u  at   V  140  3   420 cm / sec

v  480

So in order to find the distance of each body after 1 sec

t 3 u 0

from cutting, we have to find their distance before cutting. So,

U 0

t 3

a

T

a  140

T 210

1 1 2 S  u t  at 2   140  3   630 cm 2 2 The system after cutting:  means to use the rules of gravity 

u 0 t 3 v  480

280

Note We won't use the rule of maximum height, as the problem didnt mention it The body of m2  210 U  420

g  980

t 1

S2  ??

U  420

g  980

t 1

S 2  ??

1 2 gt 2 1 1 2 2  S  420  1   980  1  -70 cm  S  420  1   980  1  910 cm 2 2 Then the total distance done by one body from the begining of motion: S  ut 

1 2 gt 2

The body of m1  280

 -ve asbody is above 

S  ut 

 S   630  2   910  70  2100 cm

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (10* Excellent students*) Two pans of an ordinary balance scale  Common balance  the mass of each pan is 100 gm. they are connected by a light string passes over a smooth pulley that's vertically fixed. A body of mass 150 gm is placed on one pan, a body of mass 350 gm is placed on the other pan, the motion of the system starts from rest , find the magnitude of the acceleration of the system and the distance that each mass will cut after 3 seconds from the begining of the motion, and if the greatest mass is dropped from the pan in that instant , then prove that the time taken from the begining of the motion till the system is back to its first position is 5  10 sec onds. Answer m  m2  450    250   980  280 cm/sec 2 To get the acceleration: a 1  g  a  m1  m2  450    250  To get the distance of each body will cut after 3 sec.: 1 1 2 S  u t  at 2  S   280  3   1260 cm 2 2 When the body of m  350 gm dropped

a a

Then the acceleration and the tension of the string changed T

So, first we have to find the final velocity just before dropping the body,

V  u  at  280  3  840 cm / sec

To find the new acceleration: 

a' 

V  840 t 3 s  1260 u 0

m1  m2 g m1  m2

100    250   980  -420 cm/sec 2 a'  100    250 

u 0

T

t 3

100 + 150

s  1260 V  840

To get the total time from the begining of the motion: U  840

a  -420

S  -1260

 don' t say S  0 

100 + 350

1 1 S  u t  at 2  -1260  840t   -420  t 2 2 2 210t 2  840t  1260  0   210    t 2  4t  6  0 -b  b 2  4 a c By using formula: t   2  10 sec 2a Then the time from the begining of the motion is 3  2  10  5  10 sec onds

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (11) A thread passes over a smooth pulley, a body of mass 80 gm is attached to one of its end and on the other end 2 bodies of masses 180 gm, 60 gm . The system starts motion from rest for one second. Find the magnitude of the acceleration of the system, the magnitude of the tension in the thread, if the body of the mass 180 gm is separated at this instant, prove that the system will come to rest again after 3.5 seconds more. Answer a

m1  m2 240  80 g  a  980  490 cm/sec 2 m1  m2 240  80

T  m1  g  a  is the rule of the body whose mass is 500 gm  T  240  980  490   117600 dyne   980   120 gm.wt

a

When the body of m  180 gm is separated Then the acceleration and the tension of the string changed So, first we have to find the final velocity just before separation,

V  u  at  490  1  490 cm / sec

To find the new acceleration:

a' 

a

T

v  490 t 1 u 0

T

t 1 v  490

80

m1  m2 g m1  m2

u 0

180

60  80  980  -140 cm/sec 2 60  80 To get the total time from the begining of the motion:

 a' 

U  490

a  -140

60

t  3.5

V  u  at   V  490   -140  3.5   0 Then the velocity becomes to rest after 3.5 seconds from separation

a a

T t  3.5

T

u  490

u  490 t  3.5

80

60

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (12 * Excellent students*) Two bodies of masses 44, 36 gm are tied at the two ends of a string passes round a smooth pulley the 36 gm is attached to another string that holds at its end a body of mass 30 gm vertically, the system starts motion when the mass of 30 gm was at height 12.5 cm above the ground, prove that the 36 gm mass will rest when its height is at 35 cm above the ground given that the length of the string connecting the two masses 36 and 30 gm is 60 cm. Answer The acceleration of the system: a

m1  m2  36  30   44  980  196 cm/sec 2 g  a m1  m2  36  30   44

a

The body of mass 36 gm make a distance 12.5 cm when the body of mass 30 gm reaches the ground

T

V  u  2 a s  V  2 196 12.5   4900 2

2

a

T

2

 V  70 cm / s is the velocity of the body when m  30 touched floor

60 cm

36

44

After that: The string is relaxing T  0  and then acceleration changed 36  44  980  -98 cm/sec 2 is the new acceleration of the system 36  44 To find the distance when the body of m  36 becomes to rest

12.5 cm

 a' 

U  70

a  -98

30

floor

V 0

V 2  u 2  2a S   0  70   2  -98  S  2

4900  196 S  4900   S   25 cm 196 Then its distance from the ground is S  60  25  35 cm

a a

T u 0

44

T

u 0 S  12.5

V1  u  70

S  25 cm 36 T 0

30 V  0

floor

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (13) Two bodies of equal masses are tied by a light string passes over a smooth pulley, the two bodies were on the same horizontal level, and the length of each side of the string is 30 cm, a body of mass 9 gm is placed over one of the two bodies and the system starts motion from rest. It is found that the vertical distance between the two bodies is 20 cm after one second, and then the 9 gm mass is separated, find the value of the two masses, and the time taken after separating the 9 gm mass so that the ascending mass hit the pulley, and find the difference between the two magnitudes of the tension of the string in the two cases in dyne. Answer To find the acceleration u  0 t 1 2S  20  S  10 1 1 2 S  u t  at 2  10   a 1  a  20 cm / sec 2 2 2 To find the masses of the bodies: m  m2 m9m a 1  g   20   980 m1  m2 m9m

S  20

s  10

9 20 1      2m  9  441   2m  432 2m  9 980 49

a a

T

S  30

t 1 u 0

 m  216 , then the first body is of mass  9  216  225 gm

T

u 0 t 1

m

s  10

The system after separating the 9 gm: Then the acceleration and the tension of the string changed

m9

So, first we have to find the final velocity just before separation,

V  u  at  20  1  20 cm / sec

To find the new acceleration:

a' 

m1  m2 g m1  m2

216  216  980  0  uniform motion  216  216 When the ascending body hit the pulley: S  30  10  20 cm

 a' 

1 S  20   S  u t  at 2   20  20t   t  1 sec 2 To get the tension in the two casses U  20

a 0

Before separation: After separation:

T  m1  g  a   T  225  980  20   216000 dyne T  m1  g  a   T  216  980  0   211680 dyne

Then the difference between the two tensions: 216000-211680  4320 dyne Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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nd

2 application The motion of a system of two bodies, one of them moving on a smooth/ rough horizontal plane, the other moving vertically. To study this problem, we consider separately the motion of each of the bodies, this is because problems involving particles which are connected by strings passes through pulley (s)are moving in different directions. Case  1 Two bodies, one of them is horizontal and the other is vertical connected by a string To find the acceleration * Body at B: mB g  T  mB a    1 By adding:  mB g  mA a  mB a   mB g  a  m A  mB 

* Body at A: T  mA a    2  a A

mB g a  mA  mB

T

T

If resistance  FF  appears  plane is rough 

mA g

* Body at A: T  R  mA a    2  a 

T

a

B mB g

mB g  R m g  FF Or a  B where FF   R and R  m A g mA  mB mA  mB

Remark If the string is cut T  0  , then we have one out of two options * If there is no resistance  FF  given

* If there is a resistance  FF  given

- R  m a  a  -ve

The body is moving with uniform motion  a  0 

1 And S  u t  at 2 2

And S  u t

For the vertical body: The body is moving under the action of gravity g  9.8 m / sec 2 Or g  980 cm / sec 2

------------------------------------------------------------------------------------------------------Important rule The pressure on the pulley:

P  2T

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Case  2  Three bodies, two of them is horizontal and the other is vertical connected by 2 strings To find the acceleration * Body at B: mB g  T2  mB a   1

a

* Body at A: T2  T1  m A a    2 

C

T2

A

T1 T1

T2

* Body at C: T1  mC a    3 

T2

By adding:  mB g  mB a  mA a  mC a

mC g

  mB g  a  mA  mB  mC  a 

mA g

a

B mB g

mB g mA  mB  mC

If resistance  FF  appears  plane is rough  * Body at C: T1  R  FF   mC a   1 * Body at A: T2  T1  R  FF   m A a    2 

Important rule The pressure on the pulley:

P  2 T2

------------------------------------------------------------------------------------------------------Case  3  Three bodies, one of them is horizontal and two are vertical connected by 2 strings

To find the acceleration

a

* Body at A: T2  m A a

A

* Body at C: mC g  T1  mC a   1

T2

T2

* Body at B: mB g  T1  T2  mB a    2 

T2

By adding:  mC g  mB g  T2  mB a  mC a

mA g

B

T1

m g  mC g a B mA  mB  mC

a

T1

C

If resistance  FF  appears  plane is rough 

mC g

* Body at A: T2  R  FF   mA a

Important rule The pressure on the pulley:

P  2 T2

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Case  4  Three bodies, one of them is horizontal and two are vertical connected by 2 strings Assume the acceleration is moving in the direction of body B a

Before cutting: to find the acceleration * Body at B: mB g  T2  mB a    1 * Body at A: T2  T1  mA a    2  * Body at C: T1  mC g  mC a    3 

T1

a

T2

T2

a

mA g

B

mc g

  mB g  mC g  a  mA  mB  mC  a 

T2

T1 C

By adding:  mB g  mC g  mB a  mA a  mC a

A

T1

mB g

mB g  mC g mA  mB  mC

------------------------------------------------------------------------------------------------------Example (1) A body of mass 600 gm rests on a smooth horizontal table and tied to a string passing over a smooth pulley fixed at the edge of the table and its other end hangs vertically from it, a scale pan of mass 100 gm, a body of mass 50 gm is placed on the pan, find the magnitude of pressure on each of the pulley and the pan. Answer To find the acceleration * Body at B: mB g  T  mB a    1 * Body at A: T  mA a    2  By adding:  mB g  mA a  mB a   mB g  a  m A  mB  a 

mB g 150  980 a  196 cm / sec 2 mA  mB 150  600

Substitute in  2  :

a A

T

T

600 T

T  mA a  600  196  117600 dyne   980   120 gm.wt Then the pressure on the pulley is P  2 T  120 2 gm.wt

mA g

B

a

100 + 50

mB g

To find the pressure on the pan T  mB  g  a   50  980  196   39200 dyne  40 gm.wt

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (2) A body of mass 350 gm is placed on a smooth horizontal table, then it is connected by a light string passing over a smooth pulley at the edge of the table whose other end holds a body of mass 140 gm, find the magnitude of the acceleration of the system and the magnitude of the tension in the string. If the string is cut after just one second from the begining of the motion, 1 find the distance that each of the two bodies will cut after second from the moment of cutting 2 the string. Answer To find the acceleration * Body at B: mB g  T  mB a    1 By adding:  mB g  mA a  mB a   mB g  a  m A  mB 

* Body at A: T  mA a    2  a A

mB g 140  980 a  a  280 cm / sec 2 mA  mB 140  350

350

Substitute in  2  :

mA g

T

T T

B

T  mA a  350  280  98000 dyne   980   100 gm.wt

a

140

mB g

In order to find the distance done by each body after cutting the string, we have to find the final velocity just before cutting Before cutting: Get the final velocity just before cutting So V  u  at  0  280  1  280 cm / sec After cutting:

 Horizontal body  is moving with a uniform motion  as T  a  0  Body A

u  280

a 0

t

1 2

1 1 S  u t  at 2  280    140 cm 2 2

Body B Vertical body  is moving under the action of the gravity u  280

g  980

t

1 2

1 1 1 1 S  u t  g t 2  280     980    2 2 2 2  262.5 cm

2

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (3) Two bodies of masses 125 gm and 175 gm are placed on a smooth horizontal table and connected by a light string, then the greatest mass is tied to another string passing over a smooth pulley at the edge of the table, and a body of mass  m  is hanged from the other end of the string, if the motion of the system starts from rest and the magnitude of the tension in the string that connects the two bodies on the table is 50 gm.wt, find the magnitude of the acceleration of the system and the magnitude of the tension of the string, also find the value of  m  .

a

* Body at B: mB g  T2  mB a   1

C

* Body at A: T2  T1  mA a    2 

T1 T1

125

* Body at C: T1  mC a    3 

A

T2

T2

175

T2 mC g

By adding:  mB g  mB a  m A a  mC a

mA g

B

  mB g  a  mA  mB  mC  a 

a

a

m mB g

mB g mA  mB  mC

m  980    4 m  175  125

From  3  :

T1  mC a  50  980  125a   a  392 cm / sec 2

980m   117600  392 m  980m   m  200 gm m  300 Substitute in  2  :  T2  T1  m A a  T2  50  980  175  392 Substitute in  4  :  392 

 T2  117600 dyne   980   120 gm.wt

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (4) A body of mass 700 gm is placed on a smooth horizontal table and tied by a string passing over a smooth pulley fixed at the edge of the table and holds at its end a body of mass 280 gm hangs vertically from it, and it is connected by another string below it to hang another body of mass  m  gm. , if the magnitude of the pressure on the pulley equals 300 2 gm.wt, then find the magnitude of the acceleration of the system and the magnitude of the tension in both strings, also find  m  . Answer Pressure on the pulley is P  2 T2  2 T2  300 2   T2  300 gm.wt

To find the acceleration * Body at A: T2  m A a

a A

 300  980  700 a   a  420 cm/sec 2

P T2

T2

700

* Body at C: mC g  T1  mC a   1

T2

* Body at B: mB g  T1  T2  mB a    2 

mA g

By adding:  mC g  mB g  T2  mB a  mC a

B

280

T1

980 m  280  980  300  980  280  420  420m

C

a

T1 m

560 m  137200   m  245 From  1 :

mC g

mC g  T1  mC a

 245  980  T1  245  420  T1  137200 dyne  140 gm.wt

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (5) A & B are two bodies of masses 344 , 48 gm respectively, the body A is placed on a horizontal table whose height is 60 cm at a distance 180 cm from the edge of the table, it is connected by a light string of length 180 cm, while the body B is connected by the other end of the string that is passing over a smooth pulley that is fixed at the edge of the table, if the body B is moved softly so that it is dropped from the edge of the table, find the time taken to hit the ground, and the time that the body A will take after that to reach the edge of the table Answer To find the acceleration * Body at B: mB g  T  mB a    1 * Body at A: T  mA a    2  By adding:  mB g  mA a  mB a   mB g  a  m A  mB 

a 180 cm

mB g 48  980 a  a  120 cm / sec 2 mA  mB 344  48

A

T

T

344

To find the time taken by B to hit the ground: u 0

S  60

T

mA g

a  120

B

a

48

60 cm

1 1 mB g S  u t  at 2   60  120  t 2 floor 2 2 120  t2   1   t  1 sec 120 When body B reached the ground, body A have made a distance 60 cm, after that body A moves uniformly  a  0  , so we need to know the final velocity done. So,

V  u  at  120 1  120 cm / sec

 u  120

a 0

S  180  60  120 cm

S  ut

 120  120t

  t  1 sec

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (6) Two bodies of masses 1300 gm and 600 gm are placed on a smooth horizontal plane and connected together by a tight string of length 50 cm, the body of mass 600 gm is then tied by another string on the same level of the first and passing over a smooth pulley fixed at the edge of the plane that is near to the second mass, and another body of mass 100 gm is hanged to the other end of the string, find the magnitude of the acceleration of the system and the magnitude of the tension in each of the two strings. And if the string that connects the first two bodies is cut after two seconds from the begining of the motion, what is the distance between the two bodies after 1 second from cutting the string. Answer Before cutting: to find the acceleration * Body at B: mB g  T2  mB a    1 a 50 cm * Body at A: T2  T1  mA a    2  C

* Body at C: T1  mC a    3 

1300

By adding:  mB g  mB a  mA a  mC a

A

T1 T1

T2

T2

600

T2

  mB g  a  mA  mB  mC 

mc g

mA g

B

mB g a  mA  mB  mC

a

100

mB g

100  980  49 cm/sec 2 100  1300  600 From  3  : T1  mC a  T1  1300  49   63700 dyne  65 gm.wt a

Substitute in  2  :  T2  T1  mA a   T2  63700  600  49    T2  93100 dyne  95 gm.wt After cutting the string connected between AC, there are two parts separated from each other First : we have to find the final velocity just before cutting: u  0 a  49 t  2  V  u  at  49  2   98 cm / sec a 1st part 2 nd part C A Body B is moving under T T 1300

the action of new acceleration Body C is moving with a

* Body at B: mB g  T  mB a'   1

uniform motion  as T  a  0 

* Body at A: T  mA a'    2 

u  98

a 0

t 1

1 S  u t  at 2  98 1  98 cm 2

600

T

a

B 100 600 g mB g

mB g 100  980 By adding:  a'    140 cm / sec 2 mA  mB 100  600 u  98

a'  140

t 1

1 1 2 S  u t  a' t 2  98 1  140 1  168 cm 2 2 Then the distance between the two parts is 168 - 98 + 50 = 120 cm Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (7) A body of mass 300 gm is placed on a smooth horizontal table and connected by two light strings each of them is passing over a smooth pulley where the two pulleys are fixed at the table's opposite edges, so that the two pulleys on the table are collinear, the first string holds a body of mass 250 gm and the second string holds a body of mass 150 gm and both bodies are hanged vertically, find the magnitude of the common acceleration of the system, and if the string which holds the greatest mass is cut after 3 seconds from the begining of the motion, prove that the rest of the system will move a distance of 270 cm before it comes to rest instantaneously. Answer Assume the acceleration is moving in the direction of body B a Before cutting: to find the acceleration * Body at B: mB g  T2  mB a    1 * Body at A: T2  T1  mA a    2  * Body at C: T1  mC g  mC a    3 

T1

T2

T2

300

a

T1

T2 250

B

mc g

  mB g  mC g  a  mA  mB  mC 

a

mA g

C 150

By adding:  mB g  mC g  mB a  m A a  mC a

a 

A

T1

mB g

mB g  mC g mA  mB  mC

250  980  150  980  140 cm/sec 2  then our assumption is true  150  300  250 After cutting the string connected between AB, a

The bodies separated and new acceleration appears First : we have to find the final velocity just before cutting:

a

a  140 t  3  V  u  at  140  3   420 cm / sec

u 0

T

* Body at C: T  mC g  mC a'   1

T

A 300

* Body at A: -T  mA a'    2 

a

- mC g By adding:  - mC g  mA a'  mC a'   a'  mA  mC

T

C

150

a

a

300 g

150 g

-150  980 -980  cm/sec 2 The bodies change their directions  150  300 3 -980 u  420 a'  V 0 3 1960 2  -980  V 2  u 2  2a' S  0   420   2  S  176400   S  270 cm S  3  3 

 a' 

So,

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (8) A body of mass 800 gm is placed on a smooth horizontal plane and is tied to two threads, the first passes over a smooth pulley at the edge of the table 150 cm from the body, and a body of mass 400 gm hangs vertically downwards, the second string passes over a smooth pulley at the opposite edge of the table 80 cm from the body and hangs vertically a body of mass 200 gm, the two pulleys and the body between them are collinear and the system starts its motion from rest for one second then the string that holds the 400 gm mass is cut, calculate when the 800 gm mass will arrive to the edge of the table. Answer Assume the acceleration is moving in the direction of body B Before cutting: to find the acceleration * Body at B: mB g  T2  mB a    1 * Body at A: T2  T1  mA a    2 

a

* Body at C: T1  mC g  mC a    3 

T1

By adding:  mB g  mC g  mB a  m A a  mC a   mB g  mC g  a  mA  mB  mC  a 

A

T1

T2

T2

800

a

mB g  mC g mA  mB  mC

T1

T2 mA g

C 200

mB g

a

After cutting the string connected between AB, The bodies separated and new acceleration appears First : we have to find the final velocity just before cutting: u  0 a  140 t  1  V  u  at  140 1  140 cm / sec

* Body at C: T  mC g  mC a'   1

By adding:  - mC g  mA a'  mC a'   a'   a' 

a T

T

C

- mC g mA  mC

T

A 800

a

* Body at A: -T  mA a'    2 

400

B

mc g

400  980  200  980  140 cm/sec 2 800  200  400  then our assumption is true 

a

200

a

a

800 g

200 g

-200  980  -196 cm/sec 2 The bodies change their directions  200  800 u  140 a'  -196 S  70  80   -150  opposite direction 

1 15 S  u t  a' t 2  -150  140  t  2  -196  t 2  98t 2  140t  150  0   t  sec 2 7 15 1 Then the total time done from cutting the string is  1  3 sec. 7 7 So,

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (9) A body of mass 90 kg is placed on a rough horizontal table far from the table's edge 80 cm, it is then tied to a string passing over a smooth pulley that's fixed at the edge of the table, and hangs vertically from the other end of the string a body of mass 50 kg, if the magnitude of the resistances 1 to the motion of the body that's placed on the table equals of its weight, and the system starts 5 1 motion from rest, then the string is cut after a second from the begining of the motion. Find the 2 distance that the body on the table will move to come to rest from the begining of the motion. Answer Note when we are dealing with rough planes, then we use either Resistance force or FF   R 1 1 R  W   90  980  17640 dyne 5 5 a To find the acceleration * Body at B: mB g  T  mB a    1 First, we have to find the resistance:

80 cm

* Body at A: T  R  mA a    2 

A

R

By adding:  mB g  R  mA a  mB a

m gR 50  980  17640 a  B a  224 cm / sec 2 mA  mB 90  50 Before cutting the string, we have to find the final velocity u 0 So,

a  224

T

T

90 T

mA g

B

floor

a

50

mB g

t  0.5

V  u  at  224 0.5   112 cm / sec

1 1 2 And S  u t  at 2   224 0.5   28 cm    1 2 2 After cutting the string, body A is moving under the action of resistance only So,

T  R  mA a'  - R  mA a'  -17640  90 a'   a'  -196 cm/sec 2

 u  112

a'  -196

V 2  u 2  2 a' S

V 0

 0   112   2  -196  S   S  32 cm     2  2

Then from  1 and  2  : the total distance is S  28  32  60 cm

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (10) A body of mass 126 gm is placed on a rough horizontal table and it is connected by a horizontal string passing over a smooth pulley fixed at the edge of the table, at the other end of the string a body of mass 70 gm is hanged and at a height of 70 cm over the floor, if the coefficient of friction 1 is , find the magnitude of the acceleration of the system and the velocity when the mass hanged 3 arrives to the floor, then determine the distance that will be covered by the body on the table until it stops. Answer Note when we are dealing with rough planes, then we use either Resistance force or FF   R 1 FF   R , where R  126  980  123480 dyne   FF   123480  41160 dyne 3 To find the acceleration * Body at B: mB g  T  mB a    1 * Body at A: T  FF  mA a    2 

A

By adding:  mB g  FF  mA a  mB a a 

mB g  FF mA  mB

a

a

R

FF

70  980  41160  140 cm / sec 2 126  70

T

126

So,

a  140

floor

S  70

T

1

mA g   3

The final velocity of body B when reached the floor u 0

T

B

a

70 mB g

70 cm

V 2  u 2  2a S  2  140 70   19600  V  140 cm / sec

The distance done by body A till it stops When body B reached the ground, then body A moves under the action of resistance  FF  only So,

T  FF  mA a'  - FF  mA a'  -41160  126 a'   a'  -

 u  140

a' 

V 2  u 2  2 a' S

-980 3

980 cm/sec 2 3

V 0

2  -980   0  140   2   S   S  30 cm  3 

ThenThen the total distance is S  70  30  100 cm

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (11) A body of mass 500 gm is placed on a rough horizontal table and it is connected by a horizontal string passing over a small smooth pulley fixed at the edge of the table, hanged at its other end a scale pan of mass 80 gm, it is noticed that if we placed on the pan a body of mass 70 gm, the body on the table is about to move. Find the coefficient of friction. And if we increase the mass of the body on the pan to 120 gm, and the system moved with a uniform. Find the magnitude of the system's velocity after 5 seconds from the begining of motion. Answer Note when we are dealing with rough planes, then we use either Resistance force or FF   R The body is about to move  then the body is in equilibrium position  a  0 and FF   R where R  500  980  490000 dyne * Body at B: mB g  T  0    1 * Body at A: T  FF  0    2 

A

FF

By adding:  mB g  FF  0

 FF  150  980     490000   147000 

a

R T

500

mA g  

147000 3  490000 10

T

1 3

T

a

B 80 + 70

floor

mB g

When increasing the mass on the pan to 120 gm Then the body is moving with a uniform acceleration Then: * Body at B: mB g  T  mB a    3 

* Body at A: T  FF  mA a    4 

By adding:  mB g  FF  mA a  mB a   a 

mB g  FF mA  mB

3  200  980    490000   10   70 cm / sec 2 a 500   120  80  To find the velocity of the system after 5 seconds u 0 So,

a  70

t 5

V  u  at  70  5   350 cm / sec

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

- 83 -

Chapter one – Newton’s laws of motion

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Example (12) Two bodies of masses 210 gm and 90 gm are connected by a light string and placed on a rough horizontal plane, the mass 90 gm is tied with another light string passing over a smooth pulley at the edge of the table and hangs from its free end a body of mass 120 gm. If the magnitude of the resistances to the motion of the two bodies 210 gm and 90 gm equal 50 gm.wt and 30 gm.wt respectively. The system started motion from rest then the string that connects the two bodies on the table is cut after 3 seconds from the begining of the motion, find the distance that the 210 gm mass covers after cutting the string to come to rest then prove that the ratio between the magnitude of the pressure on the pulley before and after cutting the string equals 19:12. Answer Before cutting: to find the acceleration * Body at B: mB g  T2  mB a    1 a 30980 * Body at A: T2  T1  30  980  mA a    2  * Body at C: T1  50  980  mC a    3 

50980

A

T2

T2

90

T2

  mB g  78400  a  mA  mB  mC 

a

T1 T1

210

By adding:  mB g  30  980  50  980  mB a  m A a  mC a a 

C

mC g

mA g

B

mB g  78400 mA  mB  mC

a

120

mB g

120  980  78400 280  cm/sec 2 120  90  210 3

From  1 :

mB g  T2  mB a  T2  120  980  120 

280  106400 dyne 3

Then the pressure on the pulley before cutting is P1  106400 2 dyne     4  After cutting the string connected between AC, there are two parts separated from each other First : we have to find the final velocity just before cutting: 280 280 u 0 a  t  3  V  u  at   3   280 cm / sec 3 3 1st part Body C is moving under the action of resistance only  as T  0  -700  T1  50  980  mC a'   0  49000  210 a'   a'  cm / sec 2 3 -700 u  280 a'  V 0 3 2  -700  V 2  u 2  2a' S  0   280   2   S   S  168 cm  3  Dynamic – 3rd secondary

- 84 -

50980

C

T1

210

Chapter one – Newton’s laws of motion

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2

nd

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part Body B is moving under the action of resistance and new acceleration

a

* Body at B: mB g  T  mB a''    5  * Body at A: T  30  980  m A a''   6 

30980

mB g 120  980  30  980 By adding:  a''    420 cm / sec 2 mA  mB 120  90 From  5  :

mB g  T  mB a''  T  120  980  120  420  67200 dyne

A

T

90

T T

a

B 120 mA g mB g

Then the pressure on the pulley after cutting the string is P2  67200 2 dyne    7  Then the ratio between the pressures on the pulley before and after the cutting is: P1 106400 2 19   P2 67200 2 12

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Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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rd

3 application Motion of a system of two bodies, one of them is moving on a smooth/rough inclined plane and the other is hanged vertically. If body of mass mA moves up the plane, body of mass mB will move vertically downwards  a  0  and if body of mass mA moves down the plane body of mass mB will move vertically upwards  a  0  . The acceleration of the system: * Body at B: * Body at A:

T T

ma

T  mA g Sin  m A a    2 

mA g Sin 

By adding  1 ,  2  : mB g  mA g Sin  a  m A  mB  mB g  mA g Sin mA  mB

The pressure on the pulley

T

A

mB g  T  m B a    1 

a 

T

a

B mb

 mA g Cos 

mB g

mA g

 90

  P  2T Cos  45o   2 

o

 

T

To get the vertical distance between the two bodies T

S  S A  S B  S B  S B Cos 

 P  the pressure on the pulley bisects this angle

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Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (1) A body of mass 600 gm is placed on the surface of smooth plane inclined at angle of 30 o to the horizontal and is connected by a string passing over a small smooth pulley at the top of the plane and a body of mass 380 gm is hanging vertically, if the string coincides on the line of the greatest slope and the system starts motion from rest when the two bodies were on the same horizontal level then find:

 a  The magnitude of the acceleration of the system.

 b  The magnitude of the pressure on the pulley.  c  The vertical distance between the two bodies after 2 seconds from the begining of the motion. * Body at B:

mB g  T  mB a   1

By adding  1 ,  2  : mB g  mA g Sin   a  m A  mB  a 

T  mA g Sin  mA a    2  T

a

mB g  mA g Sin mA  mB

T

A

T

600

380  980  600  980  Sin30 o  80 cm/sec 2 600  380 Then from  2  : T  m A g Sin   mA a a

mA g Sin 

o 30

T  600  980  0.5  600  80   T  342000 dyne

T

B 380

 mA g Cos 

mB g

mA g

  Then the pressure on the pulley: P  2T Cos  45o    2  342000  Cos30 o  342000 3 dyne 2  The distance done by body B: u  0 a  80 t 2 1 1 2 S B  u t  at 2   80  2   160 cm 2 2 The distance done by body A: S A  S B Sin   160 Sin30 o  80 cm Then the vertical distance between the two bodies is 160  80  240 cm

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Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (2) A body of mass 630 gm is placed on a smooth plane inclined to the horizontal at an angle 30 o is connected by a light string passes over a smooth pulley fixed at the top of the plane, while another body of mass 350 gm is hanged vertically from other end of the string, find the magnitude of the acceleration of the system and the magnitude of the tension in the string, and if the string is cut after 2 seconds from starting motion, then find the magnitude of the velocity of the body on the plane and its direction after 1 second from cutting the string, then calculate the distance that the mass 630 gm will make before inverting its direction. Answer * Body at B: mB g  T  mB a   1 * Body at A: T  mA g Sin  mA a    2  By adding  1 ,  2  : mB g  m A g Sin   a  m A  mB 

T

a

m g  mA g Sin a  B mA  mB

T

T

A

T

630

350  980  630  980  Sin30 o a  35 cm/sec 2 630  350 Then from  2  : T  m A g Sin   mA a

mA g Sin 

30 o

T  630  980  0.5  630  35   T  330750 dyne  337.5 gm.wt Before cutting the string: u  0 a  35 t2

B 350

 mA g Cos 

mB g

mA g

V  u  at  35  2  70 cm/sec After cutting the string: Then new acceleration appears  and T  0  - mA g Sin  mA a'  a'  - g Sin  -980 Sin30 o  -490 cm/sec 2 To find the velocity after 1 second: u  70

a  -490

t 1

V  u  at   V  70  490 1  -420 cm/sec

The word "before inverting its direction" means that the body comes to rest then chages its direction, so to get the distance of the body at this instant: u  70 a  -490 V 0 V 2  u 2  2 a S   0  70   2  490  S   S  5 cm 2

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Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (3) A body of mass 3 kg is placed on the surface of a smooth inclined plane of 210 cm length and height of 140 cm, and this body is connected to another body of mass 4 kg by a string of length 210 cm coincides on the line of the greatest slope, the second body is hanged vertically at the top edge of the plane, if the system starts motion from rest till the biggest mass arrived to the ground and comes to rest, find the distance that the small mass move on the plane before it stops * Body at B:

mB g  T  mB a   1

T  mA g Sin  mA a    2 

By adding  1 ,  2  : mB g  m A g Sin   a  m A  mB  a 

mB g  mA g Sin a mA  mB

4  9.8  3  9.8  34

T

a

2 3  2.8 m/sec 2

T

T

A

T

3 2.1 m

mA g Sin 

When body B reached the ground, body A made a distance S  1.4 m u 0 S  1.4 a  2.8

Sinθ =

4

B

mA g Cos  mA g

V 2  u 2  2 a S  2  2.8  1.4   7.84  V  2.8 m / sec

1.4 m

mB g

1.4 2 = 2.1 3

After body B reached the ground, the string relaxed T  0  , so new acceleration appeared 2 98 m/sec 2 3 15 To find the distance of A after relaxing the string till it stops: - mA g Sin  mA a'

 a'  - g Sin   -9.8 

u  2.8

a-

98 15

V 0

2  98  V 2  u 2  2 a S   0   2.8   2   S   S  0.6 m  15  Then the total distance done by body A is 1.4  0.6  2 m

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Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (4) Two bodies of masses 175, 105 gm are connected to the two ends of a string placed on the surface 1 of a smooth plane inclined to the horizontal by angle whose Sin  , the body 105 gm is tied by 7 another string passing over a snooth pulley fixed at the top of the plane, and another body of mass 70 gm is hanged vertically from the other end of the second string, if the motion of the system starts from rest, find:  a  The magnitude of the accelerration of the system.

 b  The magnitude of the tension in each string.  c  If the string connecting the two bodies on the plane is cut after 2 seconds from the begining of

1 the motion, find the velocity of the three bodies after second from the moment of cutting. Answer 2  * Body at B: mB g  T2  mB a    1 T2 a T2 T2 * Body at A: T2  T1  mA g Sin   mA a    2  A * Body at C: T1  mC g Sin  mC a    3  mA g Sin 

By adding  1 ,  2  ,  3  :

C

mB g  mA g Sin  mC g Sin  a  mA  mB  mC  a 

mB g  mA g Sin  mC g Sin mA  mB  mC

T1

105

T1

T2

 mA g Cos 

175 mC g Sin 

1 1 70  980  105  980   175  980  7 7  84 cm/sec 2 a 70  105  175

70 B mB g

mA g

mC g Cos  mC g

Then from  1 : mB g  T2  mB a   70  980  T2  70  84   T2  62720 dyne  64 gm.wt 1 Then from  3  : T1  175  980   175  84   T1  39200 dyne  40 gm.wt 7 Before cutting the string: u  0 a  84 t  2  V  u  at  84  2  168 cm/sec After cutting the string connecting AC, two parts are separated: mB g  T  mB a   1 Body C : T  0  - m g Sin   m a' * Body at B:

 

* Body at A: T  mA g Sin  m A a    2  1 2  a'  - g Sin  -980   -140 cm/sec By adding  1 ,  2  : m g  m g Sin  a  m  m  B A A B 7 1 1 70  980  105  980  u  168 a'  -140 t m g  mA g Sin 7 2 a  B  mA  mB 70  105 1  V  u  at  168  140   98 cm/sec 1 2  a  308 cm/sec 2 u  168 t 2 1  V  u  at  168   308   322 cm/sec 2 rd 1

Dynamic – 3 secondary

C

C

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Chapter one – Newton’s laws of motion

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Example (5) A body of mass  m  gm placed on a smooth plane inclined at angle  to the horizontal and is connected by a light string passing over a smooth pulley fixed at the top of the plane whose other end hangs a body of mass 140 gm, if the system starts motion from rest and covered a distance of 140 cm in two seconds, find the mass  m  and the value of  given that the magnitude of the pressure on the pulley is 150 3 gm.wt and the system starts motion downwards. Answer u 0 S  140 t 2

1 1 2 S  u t  at 2  140   2  a  a  70 cm /sec 2 2 2 * Body at B: T  mB g  m B a    1 

T T

m mA g Sin 

   P  2T Cos  45 o    150 3 2 

T

A

 T  140  70  140  980  147000 dyne  150 gm.wt The pressure on the pulley

T

a

B 140

mA g Cos 

mB g

mA g

  3  3    2  150  Cos  45 o    150 3  Cos  45 o     45o   Cos 1  30 o 2 2 2 2 2    * Body at A:

 45 o  30 o  15 o     30 o

2 mA g Sin  T  m A a  980 m Sin30 o  147000  70 m

 490 m  70 m  147000   420 m  147000   m  350 gm

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Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (6) A body of mass  m1  gm is placed on a smooth plane inclined at an angle of 30 o to the horizontal and is connected by a light string passing over a smooth pulley fixed at the top of the plane, while the other end of the string holds a scale pan of mass 190 gm placed on it a body of mass  m2  gm If the system starts motion from rest, so that the body which moves upwards the plane cuts a distance of 490 cm in 2 seconds while the magnitude of the pressure of the body on the pan is 225 gm.wt, find the value of  m1  and  m2  . Answer u 0 S  490 t 2 1 1 2 T S  u t  at 2  490   2  a  a  245 cm /sec 2 a 2 2 T T A The pressure on the Pan T m1

m2 g  T  m2 a  T  m2  g  a 

 225  980  m2  980  245    m2  300 gm To find the magnitude of m1 * Body at B:

mA g Sin 

 30

B 190 m2

 o

mA g Cos 

mB g

mA g

mB g  T  m B a

  300  190   980  T   300  190  245   T  360150 dyne  367.5 gm.wt * Body at A:

T  m A g Sin   m A a  360150  980 m1 Sin30 o  245 m1

 735 m1  360150   m1  490 gm

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Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (7) Two inclined smooth planes equal in height, one of them is inclined to the horizontal at an angle 3 whose Sin is , and the other plane is inclined at an angle 30 o and fixed at its top a small smooth 7 pulley, a body of mass 140 gm is placed on the first plane and a body of mass 210 gm on the second plane, if the two bodies are connected by a light string passing over a smooth pulley and the system starts motion from rest, then the string is cut after 5 seconds from the start of the motion, find the velocity of the second mass when the first mass rests instanteously. Answer  o a * Body at B: mB g Sin30  T  mB a   1 a T

T  mA g Sin  m A a    2 

* Body at A:

T

T

By adding  1 ,  2  :

T

A

 mB g Sin30  mA g Sin   a  m A  mB 

mA g Sin 

mB g Sin30 o  mA g Sin  a  mA  mB

B

140

o

3 210  980  0.5  140  980  7  126 cm/sec 2 a 140  210 Before cutting the string: u  0 a  126 t 5

210

mA g Cos 

mB g Sin30

30 o

30

mA g Cos 30

o

mB g

mA g

7 3

2 10 V  u  at  126  5  630 cm/sec After cutting the string: The two bodies are separated and new acceleration appears

* Body at A: u  630 * Body at B:

- mA g Sin   m A a' a  -420

V 0

mB g Sin30 o  mB a' u  630

a  490

3  -420 cm/sec 2 7 V  u  at  0  630  420t  t  1.5 sec

 a'  - g Sin   -980  

 a'  g Sin30 o  980  0.5  490 cm/sec 2 t  1.5

V  u  at  V  630  490 1.5   V  1365 cm / sec

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (8) A body of mass 250 gm rests on a rough plane inclined to the horizontal by an angle whose tangent 4 is , the body is connected by a string passes over a smooth pulley fixed at the top of the plane, 3 another weight is hanged at the other end of the string, if the least value of that weight hanged at this end that is necessary to keep the body in state of equilibrium on the plane equals 150 gm.wt. 1 Prove that the coefficient of friction is . And if the hanged weight equals 350 gm.wt, find the 3 acceleration of motion of the system then. Answer Note :  1 Least weight means that the movement is down

 2  Equilibrium state means that a  0 * Body at B: T  mB g  0    1 

T

FF   R , where R  m A g Cos  3  R  250  980   147000 dyne 5 mA g Sin  FF  T  0    2 

T T

250 mA g Sin 

B 150

mA g Cos 

By adding  1 ,  2  : m A g Sin   FF  mB g  0  FF  250  980 

FF

A

* Body at A:

Also

T

a

mB g

mA g

4  150  980  49000 dyne 5

5

4

49000 1   147000 3 3 When the weight of the body increased to be 350 gm.wt, then acceleration appeared

 FF   R   49000  147000     

And the direction of movement changed to upward * Body at B: mB g  T  m B a    1  * Body at A:

T

T  m A g Sin   FF  m A a    2 

A

By adding: mB g  mA g Sin   FF  a  m A  mB  a 

a

FF

mB g  mA g Sin  FF mA  mB

mA g Sin 

4  49000 490 5  cm/sec 2 250  350 3

350  980  250  980 

Dynamic – 3rd secondary

T

a

- 94 -

T T

250

B 150

 mA g Cos 

mB g

mA g

Chapter one – Newton’s laws of motion

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Example (9) 4 A body of mass 28 kg is placed on a rough plane inclined to the horizontal at an angle of Sin , it 5 is connected to a string passes over a smooth pulley fixed at the top of the plane, another body of mass 42 kg is hanged vertically at the other end of the string, if the system starts motion from rest, and the magnitudes of the resistances to the motion of the body on the plane equal 9.6 kg.wt, the string between the two bodies is cut after 2 seconds from the begining of the motion, find the distance covered by the body placed on the plane to come to instantaneous rest. Answer * Body at B: mB g  T  m B a    1 

* Body at A: FF  9.6  9.8  94.08 Newton Also T  mA g Sin  FF  mA a    2  m g  mA g Sin  FF a  B mA  mB

FF mA g Sin 

T

B 42

 mA g Cos 

mB g

mA g

V  u  at  1.4  2  2.8 m/sec 1 1 2 and S  u t  at 2  1.4  2   2.8 m 2 2 After cutting the string: Then new acceleration appears  and T  0 

a  -11.2

T

28

4 42  9.8  28  9.8   94.08 5 a  1.4 m/sec 2 42  28 Before cutting the string: u  0 a  1.4 t 2

u  2.8

T

A

By adding  1 ,  2  : mB g  m A g Sin   FF  a  m A  mB 

- mA g Sin  FF  mA a'

T

a

5

4

 3

4  94.08  28 a'   a'  -11.2 m/sec 2 5  V 2  u2  2 a S

 - 28  9.8  V 0

 0   2.8   2  11.2  S   S  0.35 m 2

Then the total distance done by body A is 2.8  0.35  3.15 m

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (10) 1 An inclined rough plane of coefficient of friction is inclined to the horizontal at an angle of 2 3 1 measure 30 o is connected at its top with a rough horizontal plane of coefficient of friction . A 4 body of mass 60 gm is placed on the horizontal plane and is tied by one end of a thin string passes over a smooth pulley at the edge of connecting the two planes, and the other end of the string is attached to a body of mass 100 gm that is placed on the inclined plane. If each of the two branches of the string is perpendicular to the line of intersection of the two planes, find the acceleration of the system. And if the string is cut after 4 seconds from starting the motion, find the total distance the mass of 60 gm will move till it comes to rest. Answer R1 a * Body at A: FF1  1 R1 where R1  mB g  60  980  58800 dyne  T  FF1  m A a   1

* Body at B:

T

FF2  2 R2

3 where R2  mB g Cos 30 o  100  980  2 Then R2  49000 3 dyne Also

T

R2

B 100

mA g Sin 30 o

mB g Sin30 o  T  FF2  mB a    2 

A 60



1 4

FF1

FF2 T 

1 2 3

mB g

mA g Cos 30 o

30 o

By adding  1 ,  2  : mB g Sin30  FF1  FF2  a  mA  mB    a  o

T

mB g Sin30 o  FF1  FF2 mA  mB

1 1 49000 3  58800   245 4 2 3 a  cm/sec 2 100  60 4 245 Before cutting the string: u  0 a t 4 4 245 1 1  245  2 V  u  at   4  245 cm/sec and S  u t  at 2     4   490 cm 4 2 2 4  After cutting the string: Then new acceleration appears  and T  0  100  980  0.5 

1 -  58800 - FF1  mA a'   a'  4  -245 cm/sec 2 60 u  245 a  -245 V 0  V 2  u2  2 a S  0   245   2  245  S   S  122.5 cm 2

Then the total distance done by body A is 490  122.5  612.5 cm Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Impulse and Collision (a) Impulse ‫دددد‬ Definition of Impulse

If. a constant force acts on a particle for an interval of time , then the change in the momentum during the interval is equal to the impulse of the forces and is denoted by I :

I = F ×t

• Impact : the force  F  acting on an object usually when it hits something.

Impulse may be I  F t Or I  F t so there is a difference between Impact and Impulse --------------------------------------------------------------------------------------------------------

Proof of the rule change in momentum : H 2  H 1  m V2  V1   I   And

H 2  H1 I  F   F    I  F t t t

H 2  H1 V  V  I m 2 1  t t t

I

I  m V2  V1   F  t

from what we have proved above :

F T

From all the definitions of forces we have taken till now , we can get the following relation:

I m V  V  dH  t dt t

F ma  

2

1

when the motion is horizontal  resistance is neglectd 

I m V  V  dH  t dt t

F R ma  

2

1

when the motion is horizontal  resistance exists 

Special case :  1 when a ball is projected downward to the ground surface, then its pressure on the floor is R  F  m g

 2  when a ball is projected

upward to ceilling ,

then its pressure on the ceilling is R  F  m g

Rball  F  m g

Important note : we use this special case only when the problem mentioned the pressure or reaction of the body Dynamic – 3rd secondary

- 97 -

Rball  F  m g

Chapter one – Newton’s laws of motion

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The unit of measurement : Newton I

F Dyne

 kg.m / sec

second ×

t

 m V2  V1 

second

 gm.cm / sec

Important note : the magnitude of impact must be positive , so let the positive direction be always in the direction of the 2nd velocity

-------------------------------------------------------------------------------------------------------Example (1) 1 kg moves horizontally with velocity 7.3 m / sec, this sphere impinges 2 with a smooth vertical wall and rebounds after impact with velocity 9 km / h . Find the magnitude of the impulse of the wall on the sphere . If the time of contact is 0.01 sec . Then find the average force with which the ball acts on the wall . Answer V let the positive direction be in the direction of the 2 nd velocity 5  V1  7.3 m / sec V2  9km / hr  9   2.5 m / sec I 18  ve 1 V2  I  m V2  V1    2.5  7.3   4.9 Newton.sec 2 490 I  Ft   4.9  F  0.01  F  490 Newton   F   50 kg.wt . 9.8 A smooth sphere of mass

-------------------------------------------------------------------------------------------------------Example (2) An airplane throws out fuel with velocity 6300 km / hr at rate of 3.5 kg / sec find in unit of kg.wt the impulsive force by which airplane moves. Answer here the mass is 3.5 kg per each second F

V  6300 

5  1750 m / sec 18

m V2  V1  3.5  F 1750  0   6125 Newton  625 kg.wt T 1

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (3) A sphere of mass 50 gm falls from height of 6.4 meters above the surface of a horizontal ground and it rebounds vertically upwards, given that the force of the ball on the ground is 1.05 kg.wt 1 and the time of contact with the ground is sec, find the maximum height which the ball 10 attained after collision . Answer We calculate the velocity of the sphere before impact V  : u 0 V12  U 2  2gS   V12  0  2  9.8  6.4 ve v 0 14  8  V12    V1  11.2 m / sec . 10 6.4 m v 1 Smax We calculate the velocity of the sphere after impact  u  0.05 V2  11.2  I m V2  V1  u  ??   1.05  9.8  u  Not the same as V 1 T T the direction changed 10 1.029  0.05 V2  11.2   1.029   V2  11.2   20.58  V2  u  20.58  11.2  9.38 m / sec 0.05 2 u 2  9.38  To find the maximum height : S max    4.489 m 2g 2  9.8 F

------------------------------------------------------------------------------------------------------Example (4) A smooth ball of mass 30 gm is projected vertically upwards with velocity of magnitude 840 cm/sec from a point at a distance 110 cm below the ceilling of a room. It collides with the ceilling and rebounds to the floor of the room during half of a second after the collison, find the magnitude of the total pressure on the ceilling given that the height of the room equals 272.5 cm and the time of 1 sec Answer 10 V  ?? We calculate the velocity of the sphere before impact V  :

collision is

V12  U 2  2gS

  V12   8.4   2  9.8  1.1

ve

2

 V12  49   V1  7 m / sec . We calculate the velocity of the sphere after impact  u 

1.1 m

1 2 S  U t  g t 2   2.725  0.5u  0.5  9.8  0.5  2  0.5u  1.5   u  3 m / sec . 0.03  3  7  I m V2  V1  F   F  3 Newton 1 T T 10 And Rupwards  F  m g  3   0.03  9.8   2.706 Newton Dynamic – 3rd secondary

- 99 -

u  ???

2.725 m

t  0.5 sec

u  8.4

floor

Chapter one – Newton’s laws of motion

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Example (5) 1 A smooth ball of mass 700 gm is placed at the top of a smooth plane inclined at an angle of Sin 5 to the horizontal, the ball is left to descend from rest along the line of the greatest slope, after descending a distance 2 meters, it collides with a barrier fixed to the plane in a position perpendicular to the line of motion of the ball, after collision the ball rebounded a distance 50 cm along the line of the greatest slope upwards before it comes to rest. Find the total pressure 1 of the ball on the barrier given that the time of contact of the ball with the barrier is sec. 20 Answer We calculate the velocity of the sphere before impact V  : u 0

we have to get acceleration : 1 so a  g Sin  9.8    1.96 m / sec 2 5 V  ??? S 2 u 0 a  1.96

2m v  ??

V 2  U 2  2a S   V 2  2  1.96  2  7.84  V  2.8 m / sec . We calculate the velocity of the sphere after impact  u  we have to get acceleration : so u  ???

S  0.5

V 0

m g Sin

0.5 m

1 a  -g Sin   -9.8    -1.96 m / sec 2 5 a  -1.96

V 2  U 2  2a S   0  u 2  2  1.96  0.5  u  1.4 m / sec . I  m V2  V1   0.7  1.4  2.8   2.94 Newton.sec And

P  F  m g Sin  where F 

I  2.94  20  58.8 Newton T

1  P  58.8   0.7  9.8     60.172 Newton   9.8   6.14 kg.wt 5

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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(b) Collision If two spheres move on one straight line and their velocity vectors are parallel to their line of centers and they collide , then this kind of collide is called the direction impact .

V1

V2

M1

M2

Theorem If two spheres collide with each other, then the sum of their momentum before impact is equal to the sum of their momentum after impact

Rule of Collision

m1 V1  m2 V2  m1 V1 '  m2 V2 ' before collision

after collision

Some types of collisions: (1) Elastic collision : bodies are separated from each other after collision with different velocities

Or

'

(2) Inelastic collision : after collision, bodies are moving with common velocities

Here the Rule of Collision Very important notes :

'

after collision

'

m1 V1  m2 V2  V '  m1  m2  before collision

after collision

1 The Impulse of the first ball on the second is equal to the change of momentum of the second ball.  2  The Impulse of the second ball on the first is equal to the change of momentum of the first ball.    3  Take care of the direction of the velocities, so let the  direction be   4  we use the law of impulse : I  F  t  m V2  V1  to know what happened to a single body due to impact But we use the law of collision : m1 V1  m2 V2  m1 V1 '  m2 V2' to know what happened to the two bodies  or more  due to impact

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (1) Two smooth spheres of masses 5 kg and 8 kg are moving on a horizontal ground in the same straight line, if the velocity of the first is 11 m/sec and the second is 5 m/sec and in the same direction as the first. The two spheres collide and the velocity of the first after impact is 7 m/sec in the same direction . find : (i) The velocity of the second after impact . (ii) The magnitude of the impulse of the first on the second . Answer  Let the  direction be   The sum of the momentum before impact  The sum of the momentum after impact For momentum    :  m1 V1  m2 V2  m1 V1 '  m2 V2 '  5  11  8  5  5  7  8 V2 '

before impact : v 1  11

 8 V2 '  55  40  35  60   V2 '  7.5 m / sec

I

v2  5

5 kg

I

8 kg

Then the velocity of the second sphere is in the same as the first

 ii 

After impact : v 1  7

The Impulse of the first ball on the second

v 2  ?

is equal to the change of momentum of the second ball     I  m2 V2 '  V2   8 7.5  5   20 kg.m / sec Or 20 Newton sec .

------------------------------------------------------------------------------------------------------Example (2) Two smooth spheres of mass 40 gm. and 120 gm are moving on a smooth horizontal surface in an opposite direction . the velocity of the first is 9 m/sec and the second is 6 m/sec. the two spheres collide and the second sphere rebounds after impact with velocity 3m/sec. find : (i) The velocity of the first sphere after impact. (ii) The averaged force, which the second sphere acts on the first , knowing that the time 1 during which they are in contact is second. 18 Answer  Let the  direction be  

The sum of the momentum before impact  The sum of the momentum after impact For momentum    :  m1 V1  m2 V2  m1 V1 '  m2 V2 '  40  9  120   -6   40  V1 '  120  3

before impact : v1  9

 V1 '  -18 m / sec

I

Then the velocity of the first sphere after collision is

After impact :

18 m/s in the opposite direction as the first

 ii 

v2  6

40 gm

120 gm

v1  ??

v2  3

I

The Impulse of the first ball on the second is equal to the change of momentum of the

 

second ball 

F Dynamic – 3rd secondary

I  m1 V1 '  V1   0.04  18  9   1.08 Newton.sec

I  1.08  18  19.44 Newton T - 112 -

Chapter one – Newton’s laws of motion

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Example (3) Two smooth spheres are projected on a smooth horizontal plane such that they move in the same straight line and in the same direction. If the front sphere has a mass of 500 gm. and velocity 20 cm/sec. and the back sphere has a mass of 200 gm and velocity 50 cm/sec, and the impulse of the back sphere to the front one due to impact equals 15  10 3 dyne. Sec. Find the magnitude and the direction of the velocity of each sphere just after impact. Answer 3    I  m1 V1 '  V1    15  10  500 V1 '  20   30  V1 '  20   V1 '  50 cm / sec 

which is in the same direction before impact  Let the  direction be  

before impact : v2  50

For momentum    :  m1 V1  m2 V2  m1 V1 '  m2 V2 '

I

200 gm

 500  50  200V2 '  500  20  200  50  200V2 '  -5000

v1  20

500 gm

After impact : v2  ??

 V2 '  -25 cm / sec

I

v1  ??

 the second sphere moves after impact with velocity 25 cm/sec in the opposite direction of its motion before impact.

------------------------------------------------------------------------------------------------------Example (4) A smooth sphere of mass 70 gm moves in a straight line on a horizontal plane with velocity 30 cm/sec collides with another sphere being at rest and of mass 210 gm. and lying on the same straight line and they move after impact as one body. Find their common velocity , and give that this body covers a distance of 2 1 cm till it stops, find the resistance of the plane to the 7

Let the  direction be  For momentum    :  m1 V1  m2 V2  V '  m1  m2   70  30  210  0  V ' 70  210    280 V '  2100  V '  7.5cm / sec in the same direction before impact 2

V  U  2a S 2

2

15  15   0     2a  7  2

 before impact : v1  30

I After impact :

v2  0

70 gm

210 gm

v

v

I

105 -105 cm / sec  - R  ma  280  dyne 8 8  R  the resistance of the table   3675 dyne   980   3.75 gm.wt  a-

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (5) Two smooth spheres of masses 8 and 4 kg move in the same straight line and in the same direction. The first moves with velocity 6 m /sec. and the second with velocity 3 m/sec. the faster sphere catches the slower one and collides with it and moves together as one body , then they collide with a third sphere at rest and of mass 3 kg then the body rebounded with velocity 4 m/sec , find the velocity of the third sphere after collision and the impulse acting on it. Answer  Let the  direction be  

For momentum    :  m1 V1  m2 V2  V '  m1  m2    8  6  4  3  V '  8  4  

 12 V '  60   V '  5 m / sec

Let m12  m1  m2 and v12 is the velocity of the body m12  m12  8  4  12 kg

I

V12  5 m / sec

m3  3 kg

V3  0

After impact :

8 kg

4 kg

v

v

 m12 V12  m3 V3  m12 V12 ''  m3 V3 '' 

I

 12  5  3  0  12  - 4   3V3 ''  3V3 ''  108

v2  3

before impact : v  6 1

before impact : v12  5

 V3 ''  36 m / sec

I

4 8 kg

3 kg

After impact : v ''12  4

v3 ''

The impulse of the third sphere is : I  m3 V3 '  V3   3  36   108 Newton.sec

v3  0

I

------------------------------------------------------------------------------------------------------Example (6) A body A of mass 10 gm moves vertically downwards, it collides with another body B of mass 4 gm

moving vertically upwards when the velocity of A was 200 cm/sec and the velocity of B is 800 cm/sec After impact, the body B rebounded vertically downwards with velocity 100 cm/sec, while A 1 sec, the body A collided with a third body C of mass 7 100 gm moving vertically downwards with velocity 13 cm/sec, and they form one body, find the rebounded vertically upwards, and after

magnitude and the direction of their common velocity after impact. Answer For momentum :  m1 V1  m2 V2  m1 V1 '  m2 V2 '

 10  200  4  800  -10V1 '  4  100   V1 '  160 cm / sec 1 sec : VA  u  g t vC  13 7 1  VA  160  980    20 cm / sec 7  So sphere A after

mA VA  mC VC  V ''  mA  mC 

vA  20

C 100 gm

A 10 gm

v1 ' 

v '' 

A 10 gm

v1  200

v2  800

B 4 gm

v2 '  100

v ''

 -10  20  100  13  V '' 10  100    V ''  10 cm / sec dow nwards Dynamic – 3rd secondary

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Chapter one – Newton’s laws of motion

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Example (7) A body of mass one kg is projected vertically upward from the ground surface with velocity of magnitude 11.9 m/s and after one second of the instant of its projection another body of mass 1.5 kg is projected vertically upwards from the same place with velocity of magnitude 16.1 m/s if these two bodies formed one body after impact , find the maximum height which this body reaches from the ground surface . Answer Let the time taken by the second body to reach the first be  t  seconds from its projection instant  The time taken by the 1st body is  t  1 seconds for sphere A u  11.9 m/s

T t 1

vA  2.8

1 2 2 g t  11.9  t  1  4.9  t  1 2  S A  11.9t  11.9  4.9t 2  9.8t  4.9

A 1 kg 

SA  u t 

for sphere B u  16.1 m/s

vB  11.2

T t

B 1.5 kg

1 2 1 g t  16.1t   9.8  t 2  16.1t  4.9t 2    1 2 2 S1  S 2  0 "No distance mentioned"

SB  u t  And

1 seconds is the time they collide 2 1 1 This means that the time of second body  seconds and the time of first body  1 seconds 2 2 Velocity of sphere A before impact : u  11.9 m / sec t  1.5 sec

 S1  S 2 

t

VA  U  gt  11.9   9.8 1.5   -2.8 m/s  downward  Velocity of sphere B before impact : u  16.1 m / sec

t  0.5 sec

VB  16.1   9.8  0.5   11.2 m/s  upward  After impact, both spheres have common velocity  1   2.8   1.5  11.2   1  1.5   V '

m1V1  m2V2   m1  m2 V '

  V '  -5.6 m/sec   V '  5.6 m/sec upward 

The maximum height from the impact instant : U 2  5.6     1.6 m from the instant they meet each other 2 g 2  9.8  2

 Smax

from  1 :

S B  16.1 0.5   4.9 0.25   6.825 m

Then the maximum height from the ground is 1.6  6.825  8.425 m

Dynamic – 3rd secondary

- 115 -

Chapter one – Newton’s laws of motion

01009988836 – 01009988826

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Example (8) AB is the line of the greatest slope of a smooth plane of length 8 m and inclined at an angle of measure 30 o to the horizontal. A body of mass 65 gm at the bottom of the plane B is projected with velocity 7 m/sec along BA , and at the same instant another body of mass 75 gm at the top of the plane A is left to slide along AB, state when and where the two bodies will collide and if the two bodies form one body after collision, find when it reaches the bottom of the plane . Answer The distance done by sphere B u  7 m/s a  - g Sin  -9.8 Sin30 o  -4.9 m/sec 2

1 1 S B  u t  at 2  7t   4.9  t 2    1 2 2 The distance done by sphere A u  0 a  g Sin  4.9 m/sec 2 1 1 S A  u t  at 2   4.9  t 2     2  2 2 And S1  S 2  8  moving in opposite direction   7t 

1 1  4.9  t 2   4.9  t 2  8  2 2

t

65

8 sec 7

75

A

30 o

B

Substitute in  1 : S B  4.8 m from B When the both spheres collide : Velocity of sphere B before impact : u  7 m / sec

t

8 sec 7

a  -4.9 m/sec 2

8 VB  u  at  7   4.9     1.4 m/s 7  Velocity of sphere A before impact : u  0

t

8 sec 7

8 VA   4.9     -5.6 m/sec   V '  5.6 m/sec 7  After impact, both spheres have common velocity

a  4.9 m/sec 2

m1V1  m2V2   m1  m2 V '

 65  1.4  75  5.6   65  75   V '   V '  -2.35 m/s   V '  2.35 m/s  downward  To find when the body reaches the bottom of the plane u  2.35

S  4.8 m

a  4.9 m/sec 2

1 1 S  u t  at 2  4.8  2.35t   4.9  t 2   2.45t 2  2.35t  4.8  0   t  1 sec 2 2

Dynamic – 3rd secondary

- 116 -

Chapter one – Newton’s laws of motion

01009988836 – 01009988826

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Example (9) Two smooth spheres A and B of masses 300 and 250 gm are moving along the same horizontal line with velocities of magnitudes 15 m/sec and 9 m/sec respectively in the same direction, if they collide and move in the same direction after impact such that the ratio between the magnitude of their velocities immediatelty after impact is 2 : 3, find the magnitude of each. If the sphere A continues its motion with constant velocity while B moves with constant deceleration 1 of its weight, find the distance between the two spheres after 5 5 seconds from the instant of their separation after collision . Answer  After impact m1 V1  m2 V2  m1 V1 '  m2 V2 ' due to a constant resistance equal

 300  15  250  9  300  2x  250  3x  x5 m/ s

  V1 '  10 m / sec and V2 '  15 m / sec

v1  15

v2  9

A 300 gm

B 250 gm

The distance done by the two spheres after 5 seconds for sphere A S A  V t  10  5  50 m

v1 '  2 x

v2 '  3x

1 1  250  R  m g=    9.8   0.49 Newton 5 5  1000  250 - R  m a   -0.49  a   a  -1.96 m/sec 2 1000 t  5 sec a  -1.96 m/sec 2 u  15 m / sec

for sphere B And So

1 1 S B  ut  at 2   S B  15  5    -1.96  25   50.5 m 2 2 Then the distance between the two spheres S B  S A  moving in the same direction   0.5 m

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

- 117 -

Chapter one – Newton’s laws of motion

01009988836 – 01009988826

Email : [email protected]

Example (10) Two smooth spheres each of mass 240 gm move along a straight line on a smooth horizontal plane each with velocity 15 m/sec in the same direction, and there is a distance between them . A barrier is fixed to the plane such that it perpendicular to the line of motion of the two spheres if the front sphere collides with the barrier and rebounds along the same straight line and collides with the back sphere and it rebounds once more with velocity 8 m/sec, find the magnitude of the velocity of the back sphere after impact given that the magnitude of the impulse of the barrier to the front sphere is 6.48 Newton.sec Answer  240 I  m2 V2 '  V2    6.48  V2 '  15  1000 v2  15 v1  15  V2 '  12 m / sec m1 V1  m2 V2 '  m1 V1 ''  m2 V2 ''

After impact

 240  15  240  12  240V1 ''  240  8  15  12  V1 ''  8

240 gm

  240 

  V1 ''  -5 m / sec

 V1 ''  5 m / sec in the opposite direction

240 gm

v1  15

v2 '  12

240 gm

240 gm

v1 ''

v2 ''  8

------------------------------------------------------------------------------------------------------Example (11) A hammer of mass 2.5 tons falls from a height of 2.5 meters on a body of mass 1 tons and form one body after impact and are imbedded in the ground at a distance of 25 cm. Find the average resistance of the ground, measured in units of ton .wt. Answer we don't have the velocity of the hummer before impact V 2  U 2  2g S   V 2  0  2  9.8  2.5  49   V 7 m / sec R For momentum    :  m1 V1  m2 V2  V '  m1  m2 

So

 2.5  7  1  0  V '  2.5  1 

 V '  5 m / sec

u 0

Imbedded in the ground means that :  m g  R  m a

A

2.5 m

then we have to get the acceleration

mg

V  0 when S  0.25 m and u  5 m / sec So

V  U  2as   0  25  2a  0.25 2

2

0.25 m

 a  -50 m / sec 2    2.5  1  1000   9.8   R    2.5  1  1000   -50   R  209300 Newton   9.8   Dynamic – 3rd secondary

C

149500 299 kg.wt   1000   ton.wt 7 14 - 118 -

B

v 0 a

Chapter one – Newton’s laws of motion