LAWS OF MOTION - Einstein Classeseinsteinclasses.com/Bluetooth Folder/(5)L_O_M.pdfi.e. its initial momentum p mv changes by p m v . According to the Second law,, t p or F k t p F

PLOM – 1

Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road

New Delhi – 110 018, Ph. : 9312629035, 8527112111

LAWS OF MOTION

5.3 The Law of Inertia :

Q. What is inertia ?

Solution : If the net external force is zero, a body at rest continues to remain at rest and a body in bodycontinues to move with a uniform velocity. This property of the body is called inertia.

5.4 Newton’s First Law of Motion :

Q. What is Newton’s first law of motion ?

Solution : Every body continues to be in its state of rest or of uniform motion in a straight line unlesscompelled by some external force to act otherwise. This law can be expressed in another ways as : If the netexternal force on a body is zero, its acceleration is zero. Acceleration can be non zero only if there is a netexternal force on the body.

Q. An athelete runs a certain distance before taking a jump. Why ?

Solution : An athelete runs a certain distance before taking a long jump. This is because velocity acquiredby running is added to the velocity of the athelete at the time of jump. Hence he can jump over a longerdistance.

Q. An astronaut accidentlly gets separated out of his small spaceship accelerating in inter stellerspace at a constant rate of 100 m s–2. What is the acceleration of the astronaut the instant after he isoutside the spaceship ? (Assume that there are no nearby stars to exert gravitational force on him)[NCERT Solved Example 5.1]

Solution : Since there are no nearby stars to exert gravitational force on him and the small spaceship exertsnegligible gravitational attraction on him, the net force acting on the astronaut, once he is out of the space-ship, is zero. By the first law of motion the acceleration of the astronaut is zero.

5.5 Newton’s Second Law of Motion :

Q. What is momentum or linear momentum ? Is it a vector or scalar quantity ?

Solution : Momentum of a body is defined to be the product of its mass m and velocity v

, and is denoted

by p

: vmp

. Momentum is clearly a vector quantity..

Q. Write down Newton’s second law of motion ?

Solution : The rate of change of momentum of a body is directly proportional to the applied force and takesplace in the direction in which the force acts.

Q. Express force in terms of rate of change in linear momentum. Hence prove that force equals toproduct of mass and acceleration.

Solution : If under the action of a force F

for time interval t, the velocity of a body of mass m changes

from v

to vv

i.e. its initial momentum vmp

changes by vmp

. According to the Second law,,

t

pkFor

t

pF

, where k is a constant of proportionality. Taking the limit t 0, the term

t

p

becomes dt

pd

. Thus dt

pdkF

.

For a body of fixed mass m, amdt

vdm)vm(

dt

d

dt

pd

. i.e., the second law can also be written as

akmF

which shows that force is propoetional to the product of mass m and acceleration a

. For

simplicity, we choose k = 1. The second law then is amdt

pdF

.

PLOM – 2


New Delhi – 110 018, Ph. : 9312629035, 8527112111

Q. A bullet of mass 0.04 kg moving with a speed of 90 m s–1 enters a heavy wooden block and isstopped after a distance of 60 cm. What is the average resistive force exerted by the block on thebullet ? [NCERT Solved Example 5.2]

Solution : 270 N

Q. The motion of a particle of mass m is described by 2gt2

1uty . Find the force acting on the

particle. [NCERT Solved Example 5.3]

Solution : mg

Q. Define impulse. Is it a vector or scalar quantity ? Write down its units and dimensions.

Solution : The product of force and time, which is the change in the momentum of body is called impulse.It is a vector quantity. It has the same unit and dimensions as the momentum. Its unit is kg m/s or Ns. Itsdimensional formula is [MLT–1].

Q. What is impulse momentum theorem ?

Solution : Impulse momentum theorem is given by : Impulse = Force × time duration = Change inmomentum. Hence a given change in linear momentum can be produced by applying a larger force for asmaller time or by applying a smaller force for a larger time.

Q. Give reason for the following : A cricket player lowers his hand while catching the cricket ball

Solution : The player draws in his hands backward in the act of catching the ball and he allows a longertime for his hands to stop the ball. As impulse = force × time = change in linear momentum. Therefore byincreasing the time of a catch, the player has to apply a smaller force against the ball in order to stop it. Theball in turn, exerts a smaller force on his hands and his hands are not injured.

Q. A batsman deflects a ball by an angle of 450 without changing its initial speed which is equal to54 km/h. What is the impulse imparted to the ball ? (Mass of the ball is 0.15 kg).

[NCERT Solved Example 5.4]

Solution : 3.6 N s

Q. Is the force necessary to change the direction of velocity (motion) if the speed is constant ? Giveexample.

Solution : If the direction will change keeping the speed constant then its velocity will change and hencethe motion is accelerated. That’s why from Newton’s second law, the force is necessary to change thedirection of motion if the speed is constant. Any particle in uniform circular motion will experience acentripetal force.

Q. Is the force necessary to change the direction of momentum, even if its magnitude is constant ?

Solution : If the direction will change keeping the magnitude is constant then its momentum will changeand hence the force is necessary.

5.6 Newton’s Third Law of Motion :

Q. Write down Newton’s third law ?

Solution : To every action, there is always an equal and opposite reaction.

Q. Prove that the second law is the real law of motion.

Solution : To establish it, we shall show that

(a) the first law is contained in the second law, and (b) the third law is contained in the second law.

(a) First law is contained in the second law : According to Newton’s second law of motion, amF

.

If no external force is applied on a body, F

= 0 m a

= 0. As m 0, therefore, a

= 0

Thus there will be no acceleration in the body if no external force is applied. This is Neuton’s first law ofmotion. Hence the first law is contained in the second law.

(b) Third law is contained in the second law : Consider an isolated system of two bodies A and B. Let

them collide. During collision, let the body A exert a force 1F

(action) on body B for a time t. Let the body

B exert a force 2F

(reaction) on body A for the same time t.

PLOM – 3


New Delhi – 110 018, Ph. : 9312629035, 8527112111

As, change in linear momentum = Force × time

change in linear momentum of tFB 1

and change in linear momentum of tFA 2

.

Total change in linear momentum of A and B = tFtF 21

.

As no external force is acting on the system, therefore according to Newton’s second law of motion, thetotal change in linear momentum of the system is zero. Thus

0tFtF 21

or tFtF 21

or 21 FF

It means action is equal and opposite to reaction. Thus third law is contained in the second law

Q. Two identical billiard balls strike a rigid wall with the same speed but at difference angles, and getreflected without any change in speed as shown in figure. What is (i) the direction of the force on thewall due to each ball ? (ii) the ratio of the magnitudes of impulse imparted to the balls by the wall ?


Solution : (i) towards positive x direction (ii) 1.2

5.7 Conservation of Momentum :

Q. Which laws of motion lead to the conservation of momentum ? Explain it.

Solution : The second and third laws of motion lead to the law of conservation of momentum.

Q. Prove that the total momentum of (bullet + gun) system is conserved.

Solution : A bullet is fied from a gun. If the force on the bullet by the gun is F

, the force on the gun by the

bullet is – F

, according to the third law. The two forces act for a common interval of time t. According to

the second law, Ft is the change in momentum of the bullet and – F

t is the change in momentum of the

gun. Since initially, both are at rest, the change in momentum equals the final momentum for each. Thus if

bp

is the momentum of the bullet after firing and gp

is the recoil momentum of the gun, bg pp

i.e.,

0pp gb

. That is, the total momentum of the (bullet + gun) system is conserved.

Q. What is law of conservation of momentum ?

Solution : The total momentum of an isolated system of interacting particles is conserved, i.e., if netexternal force acting on the system equals to zero, then linear momentum of the system remains conserve.

If 0F

0dt

pd

p

= constant (conserve)

Q. Explain conservation of Linear momentum by an example.

OR

Prove the following statement : “Momentum will be conserved in the collision of two bodies”.

Solution : Consider two bodies A and B, with initial momenta BA pandp

. The bodies collide, get apart,

with final momenta BA pandp

respectively. By the Second Law AAAB pptF

and BBBA pptF

PLOM – 4


New Delhi – 110 018, Ph. : 9312629035, 8527112111

(where we have taken a common interval of time t for both forces i.e., the time for which the two bodiesare in contact).

Since BAAB FF

by the third law,,

)pp(pp BBAA

i.e., BABA pppp

which shows that the total final momentum of the isolated system equals its initial momentum.

5.8 Equilibrium of a Particle :

Q. What is equilibrium of a particle ?

Solution : Equilibrium of a particle refers to the situation when the net external force on the particle is zero.

Q. A mass of 6 kg is suspended by a rope of length 2 m from a ceiling. A force of 50 N in the horizontaldirection is applied at the mid-point of the rope. What is the angle the rope makes with the verticalin equilibrium ? (Take g = 10 m s–2). Neglect the mass of the rope. [NCERT Solved Example 5.6]

Solution : 400

5.9 Common Forces in Mechanics :

Q. What is spring force ?

Solution : When a spring is compressed or extended by an external force, restoring force is generated. Thisforce is usually proportional to the compression or elongation (for small displacements). The spring forceF is written as F = – k x where x is the displacement (compression or extension) and k is the force constant.The negative sign denotes that the force is opposite to the displacement from the unstretched state.

Q. Draw the graph for the variation of spring force with compression or extension. What is thephysical meaning of slope of this graph ?

Solution :

The slope of this graph tan equals to –k, where k is the spring constant.

Q. What is the unit and dimensional formula of force constant ?

Solution : The unit of force constant is N/m and dimensional formula is [MT–2]

Q. A spring of force constant or spring constant k cuts into two equal parts. Find the spring constantof each part.

Solution : The force constant of spring is inversly proportional to length of spring. If it cuts into two equalparts then spring constant of each part is 2k.

Q. What is static friction or static frictional force ? Also define the limiting static friction ? Writedown the law of limiting static friction. Also define the coefficient of limiting static friction.

Solution : Static friction opposes impending motion. The term impending motion means relative motionthat would take place under the applied force if the surface is frictionless, but does not actually take place.i.e., there is a tendency of relative motion of point of contact between two surfaces.

As the applied force exceeds a certain limit, the body begins to move. The value of static friction just beforethe begining of relative motion of point of contact is known as limiting static friction.

Law of limiting static friction :

(i) It is found experimentally that the limiting value of static friction (fs)

max is independent of the area of

contact (ii) It varies with the normal force(N) approximately as : (fs)

max = µ

sN, where µ

s is a constant of

proportionality depending only on the nature of the surface in contact. The constant µs is called the

coefficient of static friction. The law of static friction may thus be written as fs µ

s N.

PLOM – 5


New Delhi – 110 018, Ph. : 9312629035, 8527112111

Q. What is kinetic or sliding friction ? Write down the law of kinetic friction. Also define thecoefficient of kinetic friction.

Solution : If the applied force F exceeds limiting static friction (fs)

max the body begins to slide on the

surface. It is found experimentally that when relative motion has started, the frictional force decreases fromthe static maximum value (f

s)

max. Frictional force that opposes relative motion between surfaces in contact

is called kinetic or sliding friction and is denoted by kf

.

Law of kinetic friction :

(i) Kinetic friction, like static friction, is found to be independent of the area of contact. (ii) Further, it is

nearly independent of the velocity. (iii) It is proportional to normal reaction ( N

) Nµf kk

, where µ

k is the

coefficient of kinetic friction.

Q. Does the static and kinetic friction depend on area of contact ?

Solution : The static and kinetic friction does not depend on area of contact.

Q. Does the kinetic friction depend on the velocity of body ?

Solution : The kinetic friction does not depend on the velocity of body.

Q. Is the law of friction are fundamental laws ?

Solution : The law of friction are not fundamental laws like conservation of linear momentum,conservation of energy etc.

Q. Which one is greater : µs or µ

k ?

Solution : µs is greater than µ

k

Q. Does coefficient of friction depend on nature of surfaces in contact ?

Solution : The coefficient of friction µs and µ

k depends on nature of surfaces in contact.

Q. Explain angle of friction. Relate angle of friction and coefficient of friction.

Solution : The angle of friction between any two surfaces in contact is defined as the angle which the netcontact force (P) makes with normal reaction R. The net contact force is the resultant of static friction andnormal reaction R.

Consider a body on which limiting static friction and normal force is shown in the figure :

Here is the angle of friction between the two bodies in contact.

The value of angle of friction depends on the nature of materials of the surfaces in contact and the nature ofthe surfaces.

Relation between µ and : ss µ

R

ftan ( f

s = µ

sR).

Hence µs = tan , i.e., coefficient of limiting friction between any two surfaces in contact is equal to tangent

of the angle of friction between them.

Q. Draw the graph for the variation of force of friction with applied force ?

Solution :

Q. What is angle of repose ? Prove that angle of repose equals to angle of friction ?

Solution : Angle of repose or angle of sliding is defined as the minimum angle of inclination of a planewith the horizontal, such that a body placed on the plane just begins to slide down.

PLOM – 6


New Delhi – 110 018, Ph. : 9312629035, 8527112111

The value depends on material and nature of the surfaces in contact. In figure, AB is an inclined plane suchthat a body placed on it just begins to slide down. BAC = = angle of repose. The various forces actingon the body is as shown in the figure (also known as force body diagram or free body diagram of the body).

In equilibrium, fs = mg sin , R = mg cos

Dividing both, we get

cosmg

sinmg

R

fs , i.e., µs = tan ( f

s = µ

sR).

Also µs = tan , where is the angle of friction between two surfaces.

Hence we obtain, tan = tan =

Q. What is rolling friction ?

Solution : The frictional force that comes into play when one body is actually rolling over the surface of theother body is called rolling friction. The frictional force acting on a disc, sphere or any type of roundedbody during rolling motion over a frictional surface is the rolling frictional force.

Q. Explain : Friction is a necessary evil.

Solution : Friction is called a necessary evil. It is a necessity because we cannot do work without it. At thesame time, it is an evil because it involves unnecessary waste of energy. The following facts make the pointclear :

(a) Friction is a Necessity (Advantages of friction) :

(i) Walking will not be possible without friction.

(ii) No two bodies will stick to each other if there is no friction, brakes of the vehicles will not work withoutfriction and many other mechanical phenomena cannot work without friction.

(b) Friction is an evil (Disadvantages of friction)

(i) Friction always opposes the relative motion between any two bodies in contact. Therefore, extra energyhas to be spent in overcoming friction. Thus friction involves unnecessary expense of energy.

(ii) Friction causes wear and tear of the parts of machinery in contact. Thus their life time reduces.

Hence we conclude that friction is a necessary evil.

Q. Write down some ways of reducing friction.

Solution : Some ways of reducing friction are (a) Ball bearing placed between moving parts of a machine.(b) Compressed cushion of air between surfaces in relative motion.

Q. A person starts his motion on a frictional ground towards east. What is the direction of frictionalforce acting on the person ?

Solution : The direction of friction force is directed towards east.

Q. Why is it easier to drive a bicycle when its tyres are fully inflated ?

Solution : When the tyres of a vehicle is fully inflated then rolling friction reduces. Therefore it is easier todrive a bicycle when its tyres are fully inflated.

Q. Comment on the following statement : Static friction is a self adjusting force.

Solution : Static friction adjusts itself so that its magnitude is equal to the magnitude of the applied forceand its direction is opposite to that of the applied force. Hence static friction is a self adjusting force.

Q. Determine the maximum acceleration of the train in which a box lying on its floor will remainstationary, given that the co-effecient of static friction between the box and the train’s floor is 0.15.[NCERT Solved Example 5.7].

Solution : 1.5 m s–2

PLOM – 7


New Delhi – 110 018, Ph. : 9312629035, 8527112111

Q. A mass of 4 kg rests on a horizontal plane. The plane is gradually inclined until at an angle = 150

with the horizontal, the mass just begins to slide. What is the coefficient of static friction between theblock and the surface ? [NCERT Solved Example 5.8]

Solution : 0.27

Q. What is the acceleration of the block and trolley system shown in the figure, if the coefficient ofkinetic friction between the trolley and the surface is 0.04 ? What is the tension in the string ?

(Take g = 10 m s–2). Neglect the mass of the string.


Solution : 0.96 m/s2, 27.1 N

5.10 Circular Motion :

Q. A stone of mass ‘m’ is rotated with speed v in a circle by a string of length l. Which force willprovide the centripetal force ? What is the value of this force ?

Solution : Tension in the string will provide centripetal force which equals to l

2mv.

Q. Which force will provide the centripetal force for the motion of planet around the sun ?

Solution : Gravitational force between planet and the sun will provide centripetal force.

Q. Find the maximum speed of a car on a level circular road of radius R ? The coefficient of staticfriction between the tyre of car and road is µ

s.

Solution : Consider a car which is moving with speed v on a circular level road, as shown in figure :

Three forces act on the car are : (i) The weight of the car, mg (ii) Normal reaction, N (iii) Frictional force,f. These force are shown in the figure :

As there is no acceleration in the vertical direction, N – mg = 0 N = mg.

The centripetal force rquired for circular motion is along the surface of the road is provided by staticfrictional force. Static friction opposes the impending motion of the car moving away from the circle.

Hence, R

mvf

2

R

mvNµf

2

s µsmg gRµvgRµv ss

2 .

PLOM – 8


New Delhi – 110 018, Ph. : 9312629035, 8527112111

which is independent of the mass of the car. This shows that for a given value of µs and R, there is a

maximum speed of circular motion of the car possible, namely Rgµv smax .

Q. What is banking of road ? Why the road is banked ?

Solution : By raising outer edge of the curved road above the inner edge, we can reduce the contribution offriction to the circular motion of the car. This phenomena is known as banking of road. By doing so, acomponent of normal reaction of the road will provide the centripetal force.

Q. Find the expression of the maximum speed of a car on a banked circular road of radius R ? Theangle of banking is and coefficient of static friction is µ

s.

Solution : The force body diagram of a car on a bank circular road is shown in figure :

Since there is no acceleration along the vertical direction, the net force along this direction muse be zero.Hence, N cos = mg + f sin ...(i)

N sin + f cos = R

mv2

...(ii)

But f µsN. Thus to obtain v

max we put f = µ

sN

Then eqs. (i) and (ii) become N cos = mg + µsN sin ...(iii)

N sin + µs N cos = mv2/R ...(iv)

Solving equation (iii), we obtain

sinµcos

mgN

s

Substituting value of N in eq. (iv), we get R

mv

sinµcos

)cosµ(sinmg 2max

s

s

or

2

1

s

smax

tanµ1

tanµRgv

.

Q. Write true or false : The maximum possible speed of a car on a banked road is greater than thaton a flat road.

Solution : True.

Q. Find the expression of the speed of a car on a frictionless banked circular road of radius R ? Theangle of banking is .

OR

Prove that gR

vtan

2

, where symbols have their usual meaning.

Solution : The F.B.D. of a car moving on a frictionless banked circular road is shown in the figure :

As there is no vertical acceleration, hence N cos = mg and N sin will provide centripetal force. i.e.,N sin = mv2/R.

PLOM – 9


New Delhi – 110 018, Ph. : 9312629035, 8527112111

Dividing mgR

mv

cosN

sinN

2

gR

vtan

2

tangRv .

Q. Why does a cyclist bend inwards from vertical while moving on a curved road or taking a turn ?Also find this angle with the vertical if the cyclist has the speed v and radius of curved road is R.

Solution : When a cyclist takes a turn, a centripetal force will be required which can be provided byfrictional force between the tyre and road and normal reaction by the road on the tyre. By bending inwardsthe cyclist reduces the contribution of frictional force.

The free body diagram of cyclist is as shown in figure :

As there is no vertical acceleration, hence N cos = mg and N sin will provide centripetal force. i.e.,N sin = mv2/R.

Dividing mgR

mv

cosN

sinN

2

gR

vtan

2

tangRv .

Q. A cyclist speeding at 18 km/h on a level road takes a sharp circular turn of radius 3 m withoutreducing the speed. The co-efficient of static friction between the tyres and the road is 0.1. Will thecyclist slip while taking the turn ? [NCERT Solved Example 5.10]

Solution : The cyclist will slip while taking the circular turn.

Q. A circular racetrack of radius 300 m is banked at an angle of 150. If the coefficient of frictionbetween the wheels of a racecar and the road is 0.2, what is the (a) optimum speed of the racecar toavoid wear and tear on its tyres, and the (b) maximum permissible speed to avoid slipping ?


Solution : (a) 28.1 m/s (b) 38.1 m/s

5.11 Solving Problems in Mechanics :

Q. A wooden block of mass 2 kg rests on a soft horizontal floor. When an iron cylinder of mass25 kg is placed on top of the block, the floor yields steadily and the block and the cylinder together godown with an acceleration of 0.1 m s–2. What is the action of the block on the floor (a) before and(b) after the floor yields ? Take g = 10 m s–2. Identify the action-reaction pairs in the problem.[NCERT Solved Example 5.12]

Solution : (a) 20 N (b) 267.3

PLOM – 10


New Delhi – 110 018, Ph. : 9312629035, 8527112111

NCERT EXERCISE

5.1 Give the magnitude and direction of the net force acting on

(a) a drop of rain falling down with a constant speed,

(b) a cork of mass 10 g floating on water,

(c) a kite skillfully held stationary in the sky,

(d) a car moving with a constant velocity of 30 km/h on a rough road,

(e) a high-speed electron in space far from all gravitating objects, and free of electric andmagnetic fields.

5.2 A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the netforce on the pebble.

(a) during its upward motion,

(b) during its downward motion,

(c) at the highest point where it is momentarily at rest. Do your answers change if the pebblewas thrown at an angle of say 450 with the horizontal direction ?

Ignore air resistance.

5.3 Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg,

(a) just after it is dropped from the window of a stationary train,

(b) just after it is dropped from the window of a train running at a constant velocity of36 km/h.

(c) just after it is dropped from the window of a train accelerating with 1 m s–2,

(d) lying on the floor of a train which is accelerating with 1 m s–2, the stone being at restrelative to the train.

Neglect air resistance throughout.

5.4 One end of a string of length l is connected to a particle of mass m and the other to a small peg on asmooth horizontal table. If the particle moves in a circle with speed v the net force on the particle

(directed towards the centre) is : (i) T, (ii) l

2mvT (iii)

l

2mvT (iv) 0

5.5 A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of15 m s–1. How long does the body take to stop ?

5.6 A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s–1 to 3.5 m s–1 in 25 s.The direction of the motion of the body remains unchanged. What is the magnitude and direction ofthe force ?

5.7 A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude anddirection of the acceleration of the body.

5.8 The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle ofthe road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the averageretarding force on the vehicle ? The mass of the three-wheeler is 400 kg and the mass of the driver is65 kg.

5.9 A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m s–2.Calculate the initial thrust (force) of the blast.

5.10 A particle of mass 0.40 kg moving initially with a constant speed of 10 m s–1 to the north is subjectedto a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is appliedto be t = 0, the position of the particle at that time to be x = 0, and predict its position at t = –5 s,25 s, 100 s.

5.11 A truck starts from rest and accelerating uniformly with 2.0 m s–2. At t = 10 s, a stone is dropped bya person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and(b) acceleration of the stone at t = 11 s ? (Neglect air resistance).

PLOM – 11


New Delhi – 110 018, Ph. : 9312629035, 8527112111

5.12 A bob of mass 0.1 kg hung from the ceiling of a room by a string 2m long is set into oscillation. Thespeed of the bob at its mean position is 1 m s–1. What is the trajectory of the bob if the string is cutwhen the bob is (a) at one of its extreme positions, (b) at its mean position.

5.13 A man of mass 70 kg stands on a weighing scale in a lift which is moving

(a) upwards with a uniform speed of 10 m s–1.

(b) downwards with a uniform acceleration of 5 m s–2,

(c) upwards with a uniform acceleration of 5 m s–2,

What would be the readings on the scale in each case ?

(d) What would be the reading if the lift mechanism failed and it hurtled down freely undergravity ?

5.14 Figure shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particlefor t < 0, t > 4 s, 0 < t < 4 s ? (b) impulse at t = 0 and t = 4 s ? (Consider one-dimensional motion only).

5.15 Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied tothe ends of a light string a horizonatal force F = 600 N is applied to (i) A, (ii) B along the direction ofstring. What is the tension in the string in each case ?

5.16 Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes overa frictionless pulley. Find the acceleration of the masses, and the tension in the string when themasses are released.

5.17 A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smallernuclei the products must be emitted in opposite directions.

5.18 Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 m s–1 collide andrebound with the same speed. What is the impulse imparted to each ball due to the other.

5.19 A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m s–1,what is the recoil speed of the gun ?

5.20 A batsman deflects a ball by an angle of 450 without changing its initial speed which is equal to54 km/h. What is the impulse imparted to the ball ? (Mass of the ball is 0.15 kg).

5.21 A stone of mass 0.25 kg tied to the end of a string in a horizontal plane is whirled around in a circleof radius 1 m with a speed of 40 rev./min. What is the tension in the string ? What is the maximumspeed with which the stone can be whirled around if the string can withstand the maximum tensionof 200 N ?

5.22 If, in Exercise 5.21, the speed of the stone is increased beyond the maximum permissible value, andthe string breaks suddenly, which of the following correctly described the trajectory of the stoneafter the string breaks :

(a) the stone moves radially outwards,

(b) the stone flies off tangentially from the instant the string breaks,

(c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of theparticle ?

5.23 Explain why

(a) a horse cannot pull a cart and run in empty space.

(b) passengers are thrown forward from their seats when a speeding bus stops suddenly.

(c) it is easier to pull a lawn mower than to push it.

(d) a cricketer moves his hands backwards while holding a catch.

PLOM – 12


New Delhi – 110 018, Ph. : 9312629035, 8527112111

5.24 Figure shows the position-time graph of a particle of mass 0.04 kg. Suggest a suitable physicalcontext for this motion. What is the time between two consecutive impulses received by theparticle ? What is the magnitude of each impulse ?

5.25 Figure shows a man standing stationary with respect to a horizontal conveyor belt that isaccelerating with 1 m s–2.

What is the net force on the man ? If the coefficient of static friction between the man’s shoes and thebelt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to thebelt ? (Mass of the man = 65 kg).

5.26 A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net forces atthe lowest and highest points of the circle directed vertically downwards are : [Choose the correctalternative]

Lowest Point Highest point

(a) mg – T1

mg + T2

(b) mg + T1

mg – T2

(c) mg + T1 – (mv

12)/R mg – T

2 + (mv

12)/R

(d) mg – T1 – (mv

12)/R mg + T

2 + (mv

12)/R

5.27 A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s–2. The crew and thepassengers weigh 300 kg. Give the magnitude and direction of the

(a) force on the floor by the crew and passengers,

(b) action of the rotor of the helicopter on the surrounding air,

(c) force on the helicopter due to the surrounding air.

5.28 A steam of water flowing horizontally with a speed of 15 m s–1 gushes out of a tube of cross-sectionalarea 10–2m2, and hits a vertical wall nearby. What is the force exerted on the wall by the impact ofwater, assuming it does not rebound ?

5.29 Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m kg. Give themagnitude and direction of

(a) the force on the 7th coin (counted from the bottom) due to all the coins on its top,

(b) the force on the 7th coin by the eight coin,

(c) the reaction of the 6th coin on the 7th coin.

5.30 An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 150. What isthe radius of the loop ?

5.31 A train rounds an unbanked circular bend of radius 30 m at a speed of 54 km/h. The mass of thetrain is 106 kg. What provides the centripetal force required for this purpose ? The engine or therails ? The outer or the inner rails ? Which rail will wear out faster, the outer or the inner rail ? Whatis the angle of banking required to prevent wearing out the rails ?

PLOM – 13


New Delhi – 110 018, Ph. : 9312629035, 8527112111

5.32 A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in figure.

What is the action on the floor by the man in the two cases ? If the floor yields to a normal force of700 N, which mode should the man adopt to lift the block without the floor yielding ?

5.33 A monkey of mass 40 kg climbs on a rope in figure, which can stand a maximum tension of 600 N. Inwhich of the following cases will the rope break if the monkey

(a) climbs up with an acceleration of 6 m s–2

(b) climbs down with an acceleration of 4 m s–2

(c) climbs up with a uniform speed of 5 m s–1

(d) falls down the rope nearly freely under gravity ?

(Ignore the mass of the rope)

5.34 Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigidpartition. The co-efficient of friction between the bodies and the table is 0.15. A force of 200 N isapplied horizontally at A.

What are (a) the reaction of the partition (b) the action-reaction forces between A and B ? Whathappens when the partition is removed ? Does the answer to (b) change, when the bodies are inmotion ? Ignore the difference between µ

s and µ

k.

5.35 A block of mass 15 kg is placed on a long trolley. The co-efficient of static friction between the blockand the trolley is 0.18. The trolley accelerates from rest with 0.5 m s2 for 20 s and then moves withuniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on theground. (b) an observer moving with the trolley.

5.36 The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end.The co-efficient of friction between the box and the surface below it is 0.15. On a straight road, thetruck starts from rest and accelerates with 2 m s–2. At what distance from the starting point does thebox fall off the truck ? (Ignore the size of the box).

PLOM – 14


New Delhi – 110 018, Ph. : 9312629035, 8527112111

5.1 (a) to (d) No net force according to the First Law (e) No force, since it is far awar from all materialagencies producing electromagnetic and gravitational forces.

5.2 The only force in each case is the force of gravity, (neglecting effects of air) equal to 0.5 N verticallydownwards. The answers do not change, even if the motion of the pebble is not along its vertical. Thepebble is not at rest at the highest point. It has a constant horizontal component of velocitythroughout its motion.

5.3 (a) 1N vertically downwards (b) same as in (a) (c) same as in (a); force at an instant depends on thesituation at that instant, not on history. (d) 0.1 N in the direction of motion of the train.

5.4 (i) T5.5 6.0 s5.6 0.18 N in the direction of motion5.7 2m/s2 at an angle of 370 with 8 N5.8 1.2kN5.9 F = 3.0 × 105N5.10 –50 km5.11 (a) 22.4 m s–1 at an angle of tan–1 (½) with the horizontal. (b) 10 m s–2 vertically downwards.5.12 (a) At the extreme position, the speed of the bob is zero. If the string is cut, it will fall vertically

downwards.(b) At the mean position, the bob has a horizontal velocity. If the string is cut, it will fall along aparabolic path.

5.13 (a) = 70 kg (b) 35 kg (c) 105 kg (d) zero5.14 (a) In all the three intervals, acceleration and, therefore, force are zero.

(b) 3 kg m s–1 at t = 0; (c) –3 kg m s–1 at t = 4 s.5.15 a = 20 m s–2, T = 200 N, a = 20 m s–2, T = 400 N5.16 a = 2 m s–2, T = 96 N5.17 By momentum conservation principle, total final momentum is zero. Two momentum vectors can-

not sum to a null momentum unless they are equal and opposite.5.18 Impulse on each ball = 0.05 × 12 = 0.6 kg m s–1 in magnitude. The two impulses are opposite in

direction.5.19 1.6 cm s–1

5.20 4.2 kg m s–1

5.21 35 m s–1

5.22 Alternative (b) is correct, according to the First Law

5.37 A long playing record revolves with a speed of 3

133 rev./min. and has a radius of 15 cm. Two coins

are placed at 4 cm and 14 cm away from the centre of the record. If the co-efficient of frictionbetween the coins and the record is 0.15, which of the two coins will revolve with the record ?

5.38 You may have seen in a circus a motorcyclist driving in vertical loops inside a ‘death-well’ (a hollowspherical chamber with holes, so the spectators can watch from outside). Explain clearly why themotorcyclist does not drop down when he is at the uppermost point, with no support from below.What is the minimum speed required at the uppermost position to perform a vertical loop if theradius of the chamber is 25 m ?

5.39 A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 mrotating about its vertical axis with 200 rev/min. The co-efficient of friction between the wall and hisclothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remainstuck to the wall (without falling) when the floor is suddenly removed ?

5.40 A thin circular wire of radius R rotates about its vertical diameter with an angular frequency .

Show that a small bead on the wire remains at its lowermost point for R/g . What is the angle

made by the radius vector joining the centre to the bead with the vertical downwards direction for

R/g2 ? Neglect friction.

ANSWERS

PLOM – 15


New Delhi – 110 018, Ph. : 9312629035, 8527112111

5.23 (a) The horse-cart system has no external force is empty space. The mutual forces between the horseand the cart cancel (Third Law). On the ground, the contact force between the system and theground (friction) causes their motion from rest.(b) Due to inertia of the body not directly in contact with the seat.(c) A lawn mower is pulled or pushed by applying force at an angle. When you push, the normalforce (N) must be more than its weight, for equilibrium in the vertical direction. This results ingreater friction f(f N) and, therefore, a greater applied force to move. Just the opposite happenswhile pulling.(d) To reduce the rate of change of momentum and hence to reduce the force necessary to stop theball.

5.24 8 × 10–4 kg m s–1 after every 2 s from the walls at x = 0 and x = 2 cm.5.25 2 m s–2

5.26 (a) is correct5.27 (a) 7.5 × 103N downwards (b) 3.25 × 104 N downwards (c) 3.25 × 104 N upwards5.28 2.25 × 103N5.29 (a) 3 m g (down) (b) 3 m g (down) (c) 4 m g (up)5.30 15 km5.31 370

5.32 750 N, 250 N; mode (b) should be adopted5.33 (a) 640 N (b) 240 N (c) 400 N (d) 0, the rope will break in case (a)5.34 (a) 200 N (b) 200 N, 1.3 × 102 N; opposite to motion, 1.3 × 102 N; in the direction of motion5.35 (a) Maximum frictional force possible for opposing impending relative motion between the block

and the trolley = 150 × 0.18 = 27 N, which is more than the frictional force of 15 × 0.5 = 7.5 N neededto accelerate the box with the trolley. When the trolley moves with uniform velocity, there is no forceof friction acting on the block.(b) For the accelerated (non-inertial) observer, frictional force is opposed by the pseudo-force of thesame magnitude, keeping the box at rest relative to the observer. When the trolley moves withuniform velocity there is no pseudo-force for the moving (inertial) observer and no friction.

5.36 20 m5.37 4 cm from the centre5.38 16 m s–1

5.39 5 s–1

5.40 600

PLOM – 16


New Delhi – 110 018, Ph. : 9312629035, 8527112111

ADDITIONAL QUESTIONS AND PROBLEMS

Q. Derive an expression for (i) acceleration of a body down a rough inclined plane (ii) acceleration of abody up a rough inclined plane. The coefficient of kinetic friction is µ

k and angle of inclination is .

A. (i) g sin – µk g cos (down the incline) (ii) g sin + µ

k g cos (down the incline).

Q. A lift is accelerated upward. Will the apparent weight of a person inside the lift increase, decrease orremain the same relative to its real weight ? If the lift is going with uniform speed, then what is theappearent weight.

A. The apparent weight will increase. If the lift is going with uniform speed then the aparent weight equals toreal weight.

Q. The distance travelled by a moving body is directly proportional to time. Is any external force actingon it ?

A. The external force is zero.

Q. A person sitting in the compartment of a train moving with uniform speed throws a ball in theupward direction. What path of the ball will appear to him ? What to a person standing outside ?

A. The path of the ball will be straight line w.r.t. the person in the comparement. The path of the ball will beparabolic w.r.t. the person standing outside.

Q. If force is acting on a moving body perpendicular to the direction of motion, then what will be itseffect on the speed and direction of the body ?

A. The speed will be constant but the direction of the body will change.

Q. A thief jumps from the roof of a house with a box of weight W on his head. What will be the weightof the box as experienced by the thief during jump ?

A. Zero

Q. A bullet fired from a gun is more dangerous than an air molecule hitting a person, though both ofthem have almost the same speed. Why ?

A. Mass of a bullet is much greater than the mass of an air molecule. Therefore, even when both have almostthe same speed, linear momentum of bullet is much larger than the linear momentum of an air molecule.Hence the force required to stop a bullet will be much bigger than the force required stop an air molecule.That is why a bullet fired from a gun is more dangerous than an air molecule.

Q. Rocket can move in air free space, but jet planes cannot. Why ?

A. A rocket is self contained, as both the fuel and the oxygen needed to burn the fuel are available inside therocket. Therefore, a rocket can move in air free space. A jet plane has fuel only. It needs to take oxygen fromthe atmosphere to burn the fuel. That is why a jet plane cannot operate in air free space.

Q. The distance travelled by a moving body is directly proportional to time. Is any external force actingon it ?

A. When s t, acceleration = 0. Therefore, no external force is acting on the body.

Q. The length of an ideal spring increases by 0.1 cm when a body of 1 kg is suspended from it. If thisspring is laid on a frictionless horizontal table and bodies of 1 kg each are suspended from its ends,then what will be the increase in its length ?

A. Increase in length will be 0.1 cm only.

Q. The two ends of a spring-balance are pulled each by a force of 10 kg. wt. What will be the reading ofthe balance ?

A. The reading of the balance will be 10 kg wt.

Q. A retarding force is applied to stop a motor car. If the speed of the motor car is doubled, how muchmore distance will it cover before stopping under the same retarding force ?

A. As s v2, therefore, motor car will cover a distance four times longer than before.

Q. A ball of 0.5 kg mass moving with a speed of 10 m/s rebounds after striking normally a perfectlyelastic wall. Find the change in momentum of the ball.

A. Change in linear momentum = – mv – mv = – 2 mv = – 2 × 0.5 × 10 = –10 kg ms–1.

PLOM – 17


New Delhi – 110 018, Ph. : 9312629035, 8527112111

Q. A body of 2 kg is suspended on a spring balance hung vertically in a lift. If the lift is fallingdownward under acceleration due to gravity g, then what will be the reading of the balance ? Ifgoing upward with the same acceleration, then ?

A. Reading of the balance = apparent weight W = m (g – a) = m (g – g) = zero.

If lift is going upwards with same acc ; W = m (g + a) = m (g + g) = 2 m g = 4 kg.

Q. A 5-kg body is suspended from a spring-balance, and an identical body is balanced on a pan of aphysical balance. If both the balances are kept in an elevator, then what would happen in each casewhen the elevator is moving with an upward acceleration ?

A. The reading of the spring balance will increase, but there will be no effect on the equilibrium of commonphysical balance.

Q. A body is dropped from the ceiling of a transparent cabin falling freely towards the earth. Describethe motion of the body as observed by an observer (a) sitting in the cabin, (b) standing on earth.

A. (a) The body will appear stationary in air.

(b) The body will appear to be falling freely under acceleration due to gravity.

Q. Action and reaction forces do not balance each other. Why ?

A. This is because force of action and reaction act always on two different bodies.

Q. What is the principle of working of a rocket ?

A. A rocket works on the principle of conservation of linear momentum.

Q. Why does a gun recoil when a bullet is fired ?

A. The recoiling of a gun is accounted for by the principle of conservation of linear momentum. The totallinear momentum of the gun and the bullet on firing is equal to their total linear momentum before firing,which is zero (as both are at rest initially).

Q. A soda water bottle is falling freely. Will the bubbles of the gas rise in the water of the bottle ?

A. Bubbles will not rise in water. This is because water in freely falling bottle is in the state of weightlessness.No upthrust acts on the bubbles.

Q. A bird is sitting on the floor of a closed glass cage and the cage is in the hand of a girl. Will the girlexperience any change in the weight of the cage when the bird (i) starts flying in the cage with aconstant velocity (ii) flies upwards with acceleration (iii) flies downwards with acceleration ?

A. In a closed glass cage, air inside is bound with the cage. Therefore,

(i) there would be no change in weight of the cage if the bird flies with a constant velocity.

(ii) the cage becomes heavier, when bird flies upwards with an acceleration.

(iii) the cage appears lighter, when bird flies downwards with an acceleration.

Q. When a ball is thrown upwards, its momentum first decreases and then increases. Is conservation oflinear momentum violated in this process ?

A. No, the momentum conservation principle applies only when no external forces act on the system.

Q. When a wheel is rolling on a level road, what is the direction of frictional force between the wheeland the road ?

A. As the wheel is moving forward, the portion of the wheel in contact with the road moves backwards. Hencethe force of friction must be acting in the forward direction, along the tangent to the surface of the road andthe wheel in contact.

Q. A stone tied to one end of a string is whirled in a circle. If the string breaks, the stone flies offtangentially. Why ?

A. The velocity of the stone at any instant is along the tangent to the circle at that instant. When the stringbreaks, centripetal force whirling the stone vanishes. Therefore, on account of inertia, the stone flies offtangentially.

Q. Why does a child in a merry-go-round press the side of his seat radially outward ?

A. This is because of centrifugal force acting radially outwards on the child rotating actually in the mettry goround.

PLOM – 18


New Delhi – 110 018, Ph. : 9312629035, 8527112111

Q. A bucket containing water is rotated in a vertical circle. Explain why water does not fall.

A. This is because weight of water in the bucket is spent in supplying the necessary centripetal force.

Q. Why does a pilot not fall down, when his aeroplane loops a vertical loop ?

A. This is because at the highest point of the (vertical) loop, weight of the pilot is spent in providing thenecessary centripetal force.

Q. A block is gently placed at the top of an inclined plane 6.4 m long. Find the time taken by the blockto slide down to the bottom of the plane. The plane makes an angle 300 with the horizontal.Coefficient of friction between the block and the plane is 0.2. Take g = 10 m/s2.

A. 1.98 s

Q. How does a lubricant help in reducing friction ?

A. When a lubricant is added to a machine, it spreads between the two surfaces rubbing each other, fills theirregularities present on the surface and forms a thin layer between the surfaces in contact. As a result of it,the contact between the two hard surfaces is replaced by the contact between the hard surface and lubricantlayer. Due to it, the force of friction is reduced considerably.

Q. What are the factors on which coefficient of friction depends ?

A. Coefficient of friction between any two surfaces in contact depends on (i) material of the surfaces. (ii)nature of the surfaces.

Q. Rubber tyres are preferred to steel tyres. Why ?

A. This is because coefficient of friction between rubber tyres and road is smaller than the coefficient offriction between steel tyres and road.

Q. Is friction a self adjusting force ?

A. No, all types of friction are not self adjusting. Only static friction is a self adjusting force.

Q. Why are wheels of an automobile made circular ?

A. Circular wheels roll on the road. Therefore, motion of the vehicle is opposed by rolling friction which ismuch smaller than the sliding friction.

Q. Automobile tyres are generally provided with irregular projections over their surfaces, why ?

A. Irregular projections over the surface of automobile tyres increase the force of friction between the tyresand the road.

Documents

LAWS OF MOTION - Einstein Classeseinsteinclasses.com/Bluetooth Folder/(5)L_O_M.pdfi.e. its initial momentum p mv changes by p m v . According to the Second law,, t p or F k t p F