# Newton's laws of motion

August 17, 2017 | Author: Sherif Yehia Al Maraghy | Category: Momentum, Force, Mass, Acceleration, Newton's Laws Of Motion

#### Short Description

This chapter contains Newton's first, second and third laws of motion, also it contains , impulse, change in momentu...

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 Mass :

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Newton’s law of motion

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  m2 v2 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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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  H1  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 of the 5 magnitude of its velocity before collision, Find the magnitude of the ball change in momentum due

by a vertical wall , And rebounds in a perpendicular direction to the wall after loosing

to collision with the wall .

Answer let the positive direction be in the direction of the 2nd velocity v1  135 m / sec .   v1  135 x

v1

The velocity after , Collision is opposite to x After loosing

ve

4 1 of its velocity  v2   135   27 x 5 5

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 u 0 Let the positive direction be    .  Before impact : u  0

s  16.9 m .

ve

v12  u 2  2 g s   v12  2  9.8  16.9 

v1

v 0 4.9 m

16.9 m v1  331.24   v1  18.2 x 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 . the direction changed 18.2  Momentum after impact : u  ?? v  0 " as the ball will become finally at rest" s  4.9 2

v2 2  u 2  2 g s  0  u 2  2  9.8  4.9   u2  9.8 m / sec .   u2  -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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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 x

5 x  362.5 x 18 H  m vPT  5  362.5 x  1812.5 x

 350 x  45  Its momentum

vP

vT

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

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Example (9) A car moves along a straight road with velocity 45 km / hr , Facing a sandy storm in the opposite direction to that of the car motion with velocity 36 km / hr , Find the momentum of the grain of the sand relative to the car , Knowing that the mass of the grain of sand is 7.5 milligram . Answer Let x be a unit vector in direction of the sand , Also vA  vcar & vB  vsand . vBA  vB  v A  36 x  45 x  81 x  The magnitude of the velocity of the grain of the sand relative to the car is 81 km / hr 250  2250 cm / sec . 9 7.5 H  m vBA   2250 1000  The momentum  16.875 gm . cm / sec . 81

x ve A

B

------------------------------------------------------------------------------------------------------Example (10) 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  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  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 . Water condineces at rate

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Dynamic – 3rd secondary

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Example (11) 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 .

x

 The velocity directly before collision  v1  14.7 x

t  1.5 sec

 Second  Magnitude of swimmer velocity directly after collision with water : v  u  2as 2

2

 u 2   5.4 

2

 

 0  u  2  2.7  5.4

u  ??

2

a  -2.7

 u  5.4 meter / sec .

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

V  ??

 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 .

------------------------------------------------------------------------------------------------------Example (12) 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

Dynamic – 3rd secondary

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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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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 In this case

FR

Or

Also , We can say that : N

F

R

" Resistance force " and it is denoted by R . F R0 "Normal reaction of the road "

W

" Weight "

W

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Case (2)

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

F

Motion F cos 

W

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Dynamic – 3rd secondary

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Case (3)

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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 

F cos 

R

W Sin

 

W cos 

W

------------------------------------------------------------------------------------------------------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 , Then : R  A V 2

2

R1 V1  R2 V2

R1 V12   R2 V2 2

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Dynamic – 3rd secondary

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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 R It moves with a uniform velocity : So we have to use Newton's 1st law  F  R  900   6  3n  15   900  90  45 n  45 n  810   n 

810  18 wagons . 45

------------------------------------------------------------------------------------------------------Example (3) A body is pulled along a horizontal straight road by a force of magnitude 1350 kgm . wt , 3 And inclined at an angle of sine to the horizontal , So the body moved in 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 rd Dynamic – 3 secondary

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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

------------------------------------------------------------------------------------------------------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   R

1 1  ton . wt 100 20

Motion

1  1000  50 kg . wt 20

So ,The resistance for each ton's is

5 sin  

50  10 kg . wt 5

5 cos 

5

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Dynamic – 3rd secondary

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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 sine , 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 car engine is stopped and the car moves downward N

R

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

Motion

1  90 kg . wt 50

 4500 sin

4500

4500 cos 

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

F

1  180 kg . wt 50

Motion

R  90 4500 sin  

 4500

4500 cos 

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Dynamic – 3rd secondary

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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 R1 R 1 So , R2  2500 sin   2500   78.125 kg . wt 32 N W  2500 3 And R2  R1 Motion 8  8 1 2500 sin  2500 cos   R1   78.125  208 kg . wt 3 3 2500  1 Ramp From  1 F1  R1  208 kg . wt 3 -------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

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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 . 20 N Answer

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

 1

 When the tension of the force reduced to 10 kg . wt , The body moved downwards and the 10 kg . wt Became another resistance . So , R  10  30 sin   R  30 sin   10     2 

   30

sin  

R

30 sin

30cos

30

R

10

N

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

Motion

Motion

1 2

 30 sin 

30

30cos

------------------------------------------------------------------------------------------------------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 30 o

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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

 F  2 T cos

 2

" Static "

2  F  2  200 cos 60   200 gm .wt

T

R

 The line of action of F bisects of the angle between the two ropes .  R  F  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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Chapter five – Newton’s law

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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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Chapter five – Newton’s law

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Example (14) A body of weight 7 Newton , Is suspended on a horizontal plane . Two forces F1  4iˆ  3 ˆj And F2  -iˆ  2 ˆj , ˆi and ˆj are two unit vectors , The first in the direction of the plane and the other in the perpendicular direction to the plane upwards , Act on the body . Find the resistance vector R and the reaction of the plane to the body N , Knowing that the body moves with a uniform velocity .

Answer The forces acting on the body are : F1  4iˆ  3 ˆj  F2  -iˆ  2 ˆj

N

W The weight of the body   -7 ˆj

F2

ˆj

F1

R N  N ˆj And R  -Riˆ The body moves with a uniform velocity :

Thus the resultant " Sum of forces "  Zero  F1  F2  W  R  N  0  4iˆ  3 ˆj  ˆi  2 ˆj  7 ˆj  N ˆj  Riˆ  0 4 1 R  0

W

 

 4  1  R ˆi   3  2  7  N  ˆj 

0

R  3 Newton .

------------------------------------------------------------------------------------------------------Example (15) A body moves in a straight line with a uniform velocity under the forces : F1 & F2  3iˆ  2 3 ˆj F3  -4iˆ  3 3 ˆj and F4  6 ˆi  10 3 ˆj . The magnitude of these forces are in Newton . Find The magnitude and direction of F1 Where ˆi and ˆj are two perpendicular unit vectors . Answer The body moves with uniform velocity Y  The sum of forces  0 F1 ˆj  F1  F2  F3  F4  0  F1  - F2  F3  F4 5 3 F1  - 3iˆ  2 3 ˆj  4iˆ  3 3 ˆj  6 ˆi  10 3 ˆj   X' o 5 ˆ ˆ ˆ ˆ F1  - 5i  5 3 j   F1  -5i  5 3 j

 

Then the magnitude of

F1 

 

 -5   5 3 2

2

 10 Newton

Y'

5 3  - 3    Tan 1 3  -60 o  neglect negative  -5 o o    180  60  120 o " we must start from The x - axis " And

Tan  

Dynamic – 3rd secondary

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Chapter five – Newton’s law

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Newton’s Second law 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. d The second law states that : mv  F   ma  K F dt

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 2nd 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

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Chapter five – Newton’s law

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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

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

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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Chapter five – Newton’s law

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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  106 gm .

wt  1000  9.8 N  980  106 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  4.8 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

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

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 five – Newton’s law

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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 And u  0 And t   60  30 sec 2  V  u  at  0  6  30  180 m / sec Magnitude of momentum  mv  70  180  12600 kgm . m / sec

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Dynamic – 3rd secondary

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Chapter five – Newton’s law

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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

acceleration appear

 retardation  and V final  0

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 car moves with velocity 7 m / sec , if it stopped by using brakes after it covered 10 meters , 1 Then prove that the brakes force is equal to the weight of the car . 4 Answer let x be a unit vector in the direction of F   F and a have the same direction u 7 m / s

V 0

a

S  10 m

V 2  u 2  2 a s   0  7   2 10  a 2

 And

a  -2.45 m / sec 2

R

F

 brakes 

F  R  m a  But there is no driving force mentioned  : -R  m a

 - R  -2.45 m  R  2.45 m     1 W And W  mg  m  9.8 W 1 Substitute in  1 :  R  2.45   W 9.8 4

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Chapter five – Newton’s law

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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 2nd 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 five – Newton’s law

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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 2nd 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 five – Newton’s law

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

A

U A  74

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

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

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Dynamic – 3rd secondary

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Chapter five – Newton’s law

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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 five – Newton’s law

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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 under the weight mg and a

acceleration appears in the

x

air when there is another

mg

force act on the body

F R

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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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 49 F  R  ma  F  R  m a  F  50  9.8  3  1000  150  F  490  980  1470 Newton  9.8  150 kgm . wt

x

------------------------------------------------------------------------------------------------------Example (14) A train of mass 5.6 ton moves from rest with a uniform acceleration along a horizontal straight 1 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 . road, it travelled a distance 1260 meter during

Answer We have two methods to get acceleration : 1260 2520 0  30  2520 m / min  V   42 m / sec and t  15 sec 0.5 60 2 x 42 2 V  u  at  Then : 42  15a  a   2.8 m / sec 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 five – Newton’s law

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Example (15) 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  6750  9.8  25  9.8  30  30  1000a  9.8  6750  750   3000a  9.8  6000  30000a

O

a R

u 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(16) 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 five – Newton’s law

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Example (17) 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

 -5  9.8  16  16  1000a 

a  - 0.049 m/s

2

 u  12.5 m/sec a  -0.049 m/s t  2  60  120 sec So , lets find the distance of the wagon after 2 minutes :

F

R

R

 -R  ma

u  12.5 m/sec S2

u  12.5 m/sec

S1

2

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 : S2  S1  1539.2  1147.2  392 meters u  12.5 m/s

a' 

Dynamic – 3rd secondary

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Chapter five – Newton’s law

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Example (18) 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 (19) 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 five – Newton’s law

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Example (20) 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 (21) 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   VB 2  125.44   VB  11.2 m/sec u 0

VB  11.2

S ?

B VB  ?

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

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Chapter five – Newton’s law

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Example (22) 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 (23) 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 rest. Given that magnitude of the resistance to its motion equals 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 Dynamic – 3rd secondary

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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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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 five – Newton’s law

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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 - 111 -

4  1    45o 4

Chapter five – Newton’s law

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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 five – Newton’s law

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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

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

Dynamic – 3rd secondary

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Chapter five – Newton’s law

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Email : [email protected] 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

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Dynamic – 3rd secondary

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Chapter five – Newton’s law

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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

- 111 -

And T  m g

Chapter five – Newton’s law

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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 - 118 -

a  0.49 m / sec 2 Chapter five – Newton’s law

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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 T  First  To find the pressure on the base of the box: a The box bear a mass 280 kgm Then the pressure: T  m  g  a   280  9.8  0.7   2940 Newton

 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

- 111 -

 m  m'  g Chapter five – Newton’s law

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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  m a

 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  - 111 2

a

9800  2.45 m / sec 2 4000

t 2  16

t  4 sec

Chapter five – Newton’s law

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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. u 0

So, from the begining:

 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: S  35  10  5  20 m -49 V 2  u 2  2 a S  0  49  40 a  a  40 49   By substituting in  1 : T  70  9.8    600.25 Newton  9.8   61.25 kg.wt 40  

Then

u  7 m / sec

V 0

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Dynamic – 3rd secondary

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Chapter five – Newton’s law

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Motion of a body on a smooth inclined plane Note : you have to know that “smooth plane” has no resistance

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 -------------------------------------------------------------------------------------------------------

Motion of a body on a non smooth inclined plane Note : if the word “smooth” is not mentioned in the problem, then we deal with a rough inclined plane Then :if the motion is upward  F  R  m g Sin  ma And : if the motion is downward  F  m g Sin  R  ma Dynamic – 3rd secondary

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Chapter five – Newton’s law

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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 60o , 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 Cos60o 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

12 3

60 o

m g Sin60o

60 o

12 3 Sin60 o mg Cos 60 o

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mg Chapter five – Newton’s law

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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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Example (5) A body of mass 24 kgm is projected with velocity of magnitude 14 m/s in the direction of the 1 line of the greatest slope upwards of a 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 "smooth" plane is not mentioned , so resistance appears 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 So,

R

S  ??

V 2  u 2  2a s

m g Sin

0   14   2  -0.392  S 2

mg

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

So,

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 : m g Sin  R  m a'

F

N

a

R

u 0

V 2  2  0.098  250 

V  7 m / sec

 m g Sin

m g Cos 

mg

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

Dynamic – 3rd secondary

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Chapter five – Newton’s law

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Example (6) 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 3m

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

uB  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 five – Newton’s law

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https://twitter.com/Mr_Sherif_yehia Email : [email protected] Impulse and Collision

(a) Impulse ‫ع‬

‫داف‬

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 Important note : we use this special case only when the problem mentioned the pressure or reaction of the body

Rball  F  m g

Rball  F  m g

Dynamic – 3rd secondary

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Chapter five – Newton’s law

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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 2nd 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 five – Newton’s law

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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

- 111 -

u  ???

2.725 m

t  0.5 sec

u  8.4

floor

Chapter five – Newton’s law

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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 five – Newton’s law

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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 five – Newton’s law

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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 18 m/s in the opposite direction as the first

 ii 

After impact :

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 - 111 -

Chapter five – Newton’s law

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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

motion of the body. Answer

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

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

I

 a-

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

Dynamic – 3rd secondary

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Chapter five – Newton’s law

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Example (5) Two smooth spheres of masses 8 and 4 kgm 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 before impact : v  6 1 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 :

4 kg

v

v

I

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

v2  3

8 kg

 m12 V12  m3 V3  m12 V12 ''  m3 V3 ''  3V3 ''  108

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 rebounded vertically upwards, and after 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 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

v ''

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

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Chapter five – Newton’s law

v2 '  100

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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

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Chapter five – Newton’s law

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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  S2  8  moving in opposite direction   7t 

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

t

A

65

8 sec 7

75

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

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Chapter five – Newton’s law

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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

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Chapter five – Newton’s law

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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

 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

0.25 m C

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

B

v 0

Chapter five – Newton’s law

a