02 Laws of Motion

SP/CCE/Physics/11

Chapter 2

Laws of Motion Newton's Laws of Motion

Newton's First Law: Every body continues to be in its state of rest or of uniform motion unless it is acted on by an external force to change its state. This law a.

explains inertia

b.

defines force

Newton's Second Law: The rate of change of linear momentum of a body is directly proportional to the external force acting on it and change takes place in the direction of the force. This law a.

gives the formula of force

b.

is known as law of motion

ur ur ur ur dp For a body of mass M moving with velocity v we have r  Ma Where a is instantaneous acceleration of dt centre of mass of body, where M is called inertial mass.

  

Points to Remember

ur r r r Fdt  I is called impulse. I  d p, the change in momentum. ur r In case M varies we cannot write F  M a . At very high velocities, very near to the velocity of light, M ur r ur r varies considerably with velocity. In such a case F  dp / dt is valid but F  Ma is not valid (only when u is comparable with c)



The same force produces accelerations a1 and a2 when acted on two different bodies separately. If the same force acts on the combined mass of those two bodies, it moves with acceleration ‘a’ then a =



If a ball of mass M hits a wall at an angle 1 with a velocity v1 and reflect by an angle 2 with

v1

a1a 2 a1  a 2

1

velocity v2. Then the average force acting on the ball

is

F

M t

v12  v 22  2v1v 2 cos  1  2 

v2

2

where t is the time of contact. 

A jet of water of density d from a tube of area of cross section, A comes out with a velocity v. Then a.

Average force exerted by tube on water is dAv2

b.

Force required to hold the tube in a fixed position = dAv2

c.

If the water travelling horizontally strikes a vertical wall normally and then flows down along the wall, the normal force exerted on the wall is dAv2.

d.

In the above case if water rebounds with the same speed, force exerted on the wall is 2dAv 2.

e.

In the above case if water strikes the surface at angle  with the normal and reflects with the same speed, force exerted on the wall in 2dAv2 cos .

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Motion of a body down an inclined plane A body starting from rest moves along a smooth inclined plane of length l, height h and having angle of inclination . (i)

Its acceleration down the plane is g sin .

(ii)

Its velocity at the bottom of the incline plane will be

2gh  2gl sin . 1

1

(iii) Time taken to reach the bottom will be t  (2l / g sin ) 2  (2l 2 / gh) 2  (2h /g sin 2 )1/ 2

1  2h    sin   g 

1 2

t1 sin 2  . t 2 sin 1

(iv) If angle of inclination is changed keeping the height constant, then (v)

If angle of inclination is changed keeping the length constant then the ratio of times to reach the bottom in

 sin 2 t the two cases are related to each other as : 1    t 2  sin 1

1/ 2

Motion of Connected Bodies (i)

Masses M1 and M2 are tied to a mass less

a

string, which goes over a frictionless pulley as shown in figure.

T

T

If M1 > M2 and they move with

a

acceleration a,

M

ur M  M ur uu r 2M M ur 2 1 2 a  1 g and T  g M1  M 2 M1  M 2 (ii)

If the pulley begins to moves downward with acceleration

r f



2

M

1

m

2

f  g  , then

ur M  M 2 ur ur uu r 2M1M 2 ur ur a  1 ( g  f ) and T  (g  f ) M1  M 2 M1  M 2

m

(iii) When three masses m1, m2 and m3 are placed in contact with one another on a smooth horizontal surface, and a push F produces an acceleration a in them, then

F a . m1  m 2  m3 As is clear from figure. Force on m1 is F1 = F, force on m2 is F2  Force on m3 is F3 

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m3 F m1  m 2  m3

(m 2  m3 )F (m1  m 2  m3 )

F

m

1

3

a

Chapter-2 / Laws of Motion

When three masses placed on a smooth horizontal table are connected with strings and mass m1 is pulled with a force F1, common acceleration produced is

a

F1 (m1  m 2  m3 )

m

Force on m1 is F = F1

3

(m 2  m 3 )F1 Force on m2 due to m1 is F2  m1  m 2  m3 Force on m3 due to m2 is F3 

m F

2

F

3

m

1

1

m3 F1 m1  m 2  m3

(iv) In the system shown in figure, if m 1 > m2, then

R2

R1

acceleration of the system is given by

[m1 sin   m 2 sin ]g m1  m 2

a

F

2

…1

m2 gsin β

m1 sinα

m2 gcos β m1 cos α  α

Tension in the string is given by

T (v)

m1m 2 (sin   sin )g (m1  m 2 )

…2

In the system shown m1 > m2

m

2

In this case a = 90° and b = 0° Using eq. (1) and (2)

a

m

m1g m1m 2 g and T  m1  m 2 m1  m 2

1

(vi) When the same system of two masses attached to a string passes over a frictionless pulley at the edge of an inclined plane, as shown in figure. Then, In this case take  = 90° and  = . Using eq. (1) and (2) we get a=

and T =

 m1  m 2 sin   g

m2g

sin 

mg  mg 2 2 cos

a mg 1

m1  m 2 m1 m 2  1  sin   g m1  m 2

Friction

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F o r c e o f f r ic tio n

If there are two bodies in contact with each other, the force which opposes relative motion between two bodies in contact is called the force of friction. The magnitude of the frictional force depends upon the nature of two surfaces in contact.

A

f m ax fk

S m o o th m o tio n

fs

O

F o r c e a p p lie d , F

The variation of force of friction with the applied force is shown in the graph when any block is moving over another surface. When the block is at rest, the force of static friction is equal to the force applied. It then reaches a maximum value at A. Once the motion starts, a smaller force is required for maintaining the motion. It is due to surface irregularities at molecular scales with the result that even a highly polished surface has irregularities. It results in producing interlocking of uneven surfaces. Once there is smooth motion of a body, the friction is less than the maximum force of static friction or limiting friction.

Points to Remember

 

The direction of friction between two bodies always opposite to their relative velocities.



The angle between the normal reaction and the resultant of limiting friction and normal reaction is called the angle of friction and is denoted by .

F tan   fr   N

N

R

 Ffr

Ffr =

Frictional force

or  = tan

–1

()



If a body of mass m is placed on an inclined plane which is gradually moved upwards, then the body just starts sliding down at a certain angle of inclination .  called angle of repose. tan  = µ



When  < , the body is in equilibrium and does not slide. If  > , the body starts sliding down with an acceleration.



If R denotes the magnitude of resultant contact force N denotes the normal reaction and f denotes the friction force then N – f < R < N + f



If a body of mass M is static under the action of a force F applied horizontally parallel to the floor then Mg  F  Mg



1  2 .

When a body is allowed to slide down a rough inclined plane its acceleration is given by a = g sin  – µg cos ,

it is independent of mass of body

Frame of Reference Any coordinate system with respect to which the motion of a body can be described. They are of two types (i) Inertial frame (Non accelerating): In this type of frame the acceleration of a body is only caused by actual or

r

r

real forces hence while applies Newton's second law we consider only real forces i.e., Freal  ma net -23-

Chapter-2 / Laws of Motion

(ii)

Non Inertial frame (Accelerating and Rotating frame): Those frame in which Newton’s law can not be applied as it is but required a little modification and hence the equation of acceleration can be written as

r r r r r r Fpseudo  Freal  ma net ,where a net = acceleration as observed by non inertial frame, where Fpseudo   ma r where a is acceleration of non inertial frame.



Points to Remember



Inertial frames of reference move with uniform velocity relative to each other.



All fundamental forces of nature are real.



Observers in all inertial frames, measure the same acceleration for a given object but might measure different velocities.



Earth by definition is a non-inertial frame, because of the rotation about its axis and revolution around the sun. However in many applications it can be approximated to an inertial frame because acceleration of the earth due to its rotation and revolution is negligible.



Pseudo Force: Force acting on an object relative to an observer in a non-inertial frame, without any interaction with any other object of the universe. Pseudo force examples are Centrifugal force, deflection of pendulum relative to accelerating car, gain or loss of weight experienced in an accelerating elevator etc.



Pseudo force does not possess a reaction counter part.



An object kept on a smooth inclined plane can be kept stationary relative to the incline by giving a horizontal acceleration of g tan  as shown.



An object of mass m is kept on a wedge of mass M. The wedge is moved with an acceleration ‘a’ such that the object moves down like a freely falling body. (no friction)



Here a = g cot . A pendulum is suspended from the roof a moving car.

(Accelerating Trolley)

(Retarding Trolly)

a) If the car is moving with uniform velocity, the position of the bob does not change b) If the car moves with acceleration or retardation, the bob moves in opposite direction to that of acceleration. For the equilibrium of bob, tan  = a/g A person of mass M climbs up a rope with acceleration a. The tension in the rope will be M(g+a) If the person slips down along the rope with acceleration a, the tension in the rope will be M(g–a) Motion of two bodies, one resting on the other

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(A) When a body A of mass m is resting on a body B of mass M and a force F is applied on A as shown in figure, three cases arise (i)

When there is no friction a A 

F , aB  0 m m A

F B

(ii)

M

When there is friction between A and B, the body A will not slide on B till F <

 s (mg)

(iii) When F > friction force, the two bodies will move in the direction of applied force, but with different accelerations. Force of dynamic friction, fk =  k mg This causes the motion of B.

f K  Ma B  a B 

f k  k mg  M M

For motion of A, F – fk = ma

 F  f k   m 

 aA  

(B) When a force F is applied to the lower body, as shown in figure, three cases arise : (i)

When there is no friction a B  F / M

a A  0, because there is no pulling force on A relative to B; A will remain there and will fall from it after some time. (ii)

When there is friction between A and B, and the two move together, then a = F/(m + M) In that case, force on A  F1  ma 

mF (m  M)

The two bodies will move together only when F1 < (iii) If F >

F

 s mg.

F r ic tio n le s s

 s (m + M)g, the bodies A and B will move with different acceleration.

Such that ma A   k mg, And for B:

a A  k g

F – fk = MaB

aB 

F  f k (F   k mg)  M M

*****

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A B

Chapter-2 / Laws of Motion

Exercise - I Multiple choice questions with only ONE option correct 1.

A body of mass 5 kg is accelerated from rest to 60 m/s in 1 s. What force (in newton) acts on it (a)

2.

5  60

(b)

(5/60)  981

(c)

602  52

(d) (5/2)  602  981

A particle is acted upon by two mutually perpendicular forces of 3 N and 4 N. In order that the particle remains stationary, the magnitude of the third force that should be applied is (a)

3.

12 N

(b)

5N

(c)

8N

(d) 7 N

The pendulum hanging from the ceiling of a railway carriage makes an angle 30° with the vertical when it is accelerating. The acceleration of the carriage is

3 g 2

(a) 4.

(b)

2 3

g

(c)

g

(d)

3

g 3

A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m. If a force P is applied at the free end of the rope, the force exerted by the rope on the block will be (a)

5.

P

(b)

Pm M m

(c)

PM (M  m)

(d)

Pm (M  m)

A man sitting on the top berth in a train compartment which is just going to stop on a railway station. The mass drops an apple aiming at the open hand of his friend sitting vertically below his hand at a distance of about 2 m. The apple will fall (a)

Precisely on the hand of his friend

(b)

Slightly away from the hand of this friend in the direction of motion of the train

(c)

Slightly away from the hand of his friend in the direction opposite to the direction of motion of the train

(d) None of these 6.

A cart of mass M has a block of mass m attached to it as shown in figure. The coefficient of friction between the block and the cart is µ. What is the minimum acceleration of the cart so that the block m does

a

M

m

not fall

7.

(a)

g

(b)

(c)

/g

(d) Mg/m

A block of mass m1 is resting on a rough

g/

µ

m1

horizontal plane, coefficient of kinetic friction between block and surface is µ. If m1 is connected to another mass m2 with the help of

m2

string and pulley as shown in figure. Then the

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SP/CCE/Physics/11

common acceleration when released from rest will be (a) 8.

m2 g m1  m2

(b)

 m2  m1    g  m1  m2 

(c)

(m2  m1 ) g m1  m2

(d)

 m2  m1    g  m1  m2 

In the figure, the blocks A, B and C of mass m each have accelerations a1, a2 and a3 respectively. F1 and F2 are external forces of magnitudes 2 mg and (a) a1 = a2 = a3

m

m

m

mg respectively.

2m

F1 = 2mg

(b) a1 > a3 > a2

m F2 = mg

(c) a1 = a2, a2 > a3

(A)

(B)

(C)

(d) a1 > a2, a2 = a3 9.

A block can slide on a smooth inclined plane of inclination  kept on the floor of lift. When the lift is descending with retardation a, the acceleration of the block relative to the incline is (a)

10.

(g +a) sin 

(b)

(g  a)

(c)

In the figure shown the relation between acceleration is (a)

a1 + a2 + 2a3 = 0

(b)

a1 + a2 = 2a3

(c)

a1 + a2 + a3

g sin 

(d) (g  a) sin

a1

a2

m1

m2

(d) a1 + a2 + a3 = 0 a3

11.

12.

When a force of constant magnitude always acts perpendicular to the motion of a particle then (a)

velocity is constant

(b)

(c)

speed is constant

(d) none of these

15 4

(2)

1 15

(3)

16 15

(4)

15 16

A smooth right-angled wedge of mass M is placed on a perfectly smooth horizontal surface. The wedge is given an acceleration a to the right horizontally such that a cart of mass m with smooth wheels placed on its sloping face (of slope angle ) does not roll up or down. The value of a is given by (1)

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acceleration is constant

A body of mass m takes 4 times as much time to slide down a 45° rough incline as it takes to slide down a perfectly smooth 45° incline, then the coefficient of kinetic friction between the object and incline will be (1)

13.

m3

mgsin  M

(2)

mg sin  (M  m)

(3)

g tan 

(4)

g sin 

Chapter-2 / Laws of Motion

14.

Two blocks of masses 3 kg and 2 kg are attached to the ends of a string passing over a smooth pulley fixed to the ceiling of the elevator. A man inside the elevator finds the acceleration of the system to be

g

acceleration of the elevator is (1) 15.

16.

17.

g

4

g

downward (2)

4

upward

g

(3)

A block of mass 15 kg is resting on a rough inclined plane as shown in figure. The block is tied up by a horizontal string which has tension of 50 N. The coefficient of friction between the surfaces of contact is (g = 10 m/s2) (1) 1/2 (2) 2/3 (3)

20

downward

(4)

g

20

4

. The

upward

T M

45° 3/4

(4)

1/4

For a car taking a turn on a horizontal surface, let N 1 and N2 be the normal reactions of the road on the inner and outer wheels respectively. (1) N1 is always greater than N2 (2) N2 is always greater than N1 (3) N1 is always equal to N2 (4) Either (1) or (2) depending on the speed of the car and the radius of curvature of the road A car of mass 500 kg accelerates at 1 m/s 2 when the force due to the engine is 600 N. The average frictional force on the car is (a)

18.

600 N

(b)

500 N

(c)

200 N

(d) 100 N

A small mass of 0.2 kg is whirled round in horizontal circle at the end of a string of length 0.5 m at a constant angular speed of 4 rad/s. The tension in the string is (a)

19.

0.2 N

(b)

1.6 N

(c)

0.8 N

(d) 1.0 N

If a car moves round a circular road of radius r at a constant speed v (a)

its velocity changes and the acceleration is v/r2.

(b)

there is no force on the car since its speed is constant

(c)

the force on the car is towards the centre and its magnitude mv2/r

(d) the force on the car is outwards from the centre and its magnitude is mv2/r 20.

A

A block of mass m is resting on a wedge of angle  as shown in the figure. The wedge is given an acceleration a. What is the value of a so that the mass m falls freely

21.

a

(a)

g

(b)

g cos 

(c)

g tan 

(d) g cot 

 B

C

A 30 kg block rests on a rough horizontal surface. A force of 200 N is applied on it. The block acquires a speed of 6 m/s in 2 second, starting from rest. What is the value of coefficient of friction (a)

22.

3 10

(b)

3 10

(c)

11 30

(d)

7 30

A certain moment of time velocity of ‘A’ is 2m/sec upwards and velocity of B is 2m/sec. upwards then the velocity of ‘C’ (masses of pulley’s & string’s are negligible).

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A B C

23.

(1) 2m/sec downwards (3) 6 m/sec downwards A block of mass ‘m’ is placed at the top of a

(2) 4 m/sec downwards (4) 2 m/sec upwards

A

smooth wedge ABC. The wedge is rotated about an axis passing through ‘C’ as in fig. (AC = l).

M

Then the minimum value of angular speed  such

24.







g sin   sec  (2) l   





g  cos  l 



B

that the block does not slip on the wedge will be (1)



m

(3)



g  cot     l cos 

(4)







C

g  sin  l 

Two identical frictionless and massless pulleys are arranged separately as shown in figure (i) and (ii). Assuming that the string has negligible mass, the acceleration of mass m, in the two cases will be 3m 3 /2 m 3 /2 m

m m (i) (1) (3) 25.

26.

(ii) (2) More in case (i) than in case (ii) (4) The same in the two cases and equal to g

A light string fixed at one end to a clamp on ground passes over a fixed pulley and hangs at the other side. It makes an angle of 30° with the ground. A monkey of mass 5 kg climbs up the rope. The clamp can tolerate a vertical force of 40 N only. The maximum acceleration in upward direction with which the monkey can climb safely is (Neglect friction and take g = 10 m/s2) (1) 2m/s2 (2) 4 m/s2 (3) 6 m/s2 A particle of mass m moves along the internal smooth surface of a vertical cylinder as shown in the figure. The force which acts on the wall of the cylinder if initially the velocity v0 of the particle makes an angle  with the horizontal (Assume that particle does not leave contact with the curved surface of the cylinder) is (1)

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The same but different from g More in case (ii) than in case (i)

mv02 R

(2)

mv02 cos 2  R

a

30° (4) 8 m/s2



v0

Chapter-2 / Laws of Motion

(3) 27.

30.

0.25

(2)

0.40

4m

m (3)

0.50

(4)

none of these

A flexible chain of weight w hangs between two fixed points A & B at the same level. The inclination of the chain with the horizontal at the two points of support is . Then the tension of the chain at the end point will be (1)

29.

(4) mg

In the system shown, the mass m moves in a circular arc of angular amplitude 60°. The minimum value of coefficient of friction between the mass 4 m and surface of table to prevent slipping is (1)

28.

mv02 sin 2  R

w sec  2

(2)

w cos 

(3)

w sin  3 M

A ball of mass 1 kg is released from position A inside a wedge with a hemispherical cut of radius 0.5 m as shown in figure. The force exerted by the vertical wall OM on wedge, when the ball is in portion B is (neglect friction everywhere) (1)

10 N

(2)

(3)

15 3 / 2 N

(4) 15 N

C

A

5 3N

The system shown in figure is released from rest. The spring gets elongated (1) if M > m (2) if M > 2m (3) if M > m/2 (4) for any value of M

w cos ec 2

(4)

B O

N

m

m

M

******

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Exercise - II More than one Correct 1. A body is kept on a smooth inclined plane having an inclination of 1 in x. Then (a) slope of inclined plane is (b) slope of inclined plane is

1 x

1 x - 1 2

(c) for the body of mass m to remain stationary relative to the incline, the incline must offer a normal reaction mg

x

x - 1 2

(d) for the body to remain stationary relative to the incline, the incline must be given a horizontal acceleration of

g

x - 1 2

2. Two masses of 10 kg and 20 kg are connected by a light spring as shown in the figure. A force of 200N acts on a 20 kg mass as shown in the figure. At a certain instant the acceleration of 10 kg mass is 12 ms–2

(a) At the instant the 20 kg mass has an acceleration of 12 ms–2 (b) At that instant the 20 kg mass has an acceleration of 4 ms–2 (c) The streching force in the spring is 120N (d) The collective system moves with a common acceleration of 30 ms –2 when the extension in the connecting spring is the maximum 3. A particle of mass M is moving with acceleration a0 as measured by an observer 1 stending in a frame of reference moving with a uniform velocity. Another observer 2 is standing in a frame of reference moving with acceleration a (a) observer 1 measures the force acting on the body as Ma0 (b) observer 2 measures the force acting on the body as Ma (c) observer 2 measures the force acting on the body as Ma0–Ma (d) obser 1 is standing in an inertial frame of reference and observer 2 is standing in non-inertial frame of reference 4. A light vertical chain is used to haul up an object of mass M attached to its lower end. The vertical pull applied has a magnitude F at t = O and it decreases at a uniform rate of Nm –1 over a distance S through which the object is raised.

æF - fy - Mg ö ÷ ÷ ÷when the object is raised through a distance y (