Laws of Motion. One mark questions:- 1. Why an athlete runs some steps before taking a jump?

Ans- inertia of motion helps to take longer jump.

2. A man jumps from the upper storey of a house with a load on his back. What is the force of

the load on his back when the man is in air?

Ans-zero. Because weightlessness W=m(g-a)= M(g-g) =0

3. If a ball is thrown up in a moving train, it comes back to the person’s hand. Why?

Ans-ball will have same horizontal velocity as the train and covers same horizontal distance.

4. Glassware’s are wrapped in straw/paper before packing. Why?

Ans- to increase time of jerk and to decrease the effect of force.

5. Why buffers are provided between the bogies of a train?

Ans- Force=impuse/time , time increases and effect of force reduces.

6. Why does a heavy rifle not kick as strongly as light rifle using the same cartridge?

Ans-Vα 1/M, as the mass increases recoil velocity decreases.

7. A cricketer lowers his hand while catching a ball. Why?

Ans- he increases the time of catch which will reduce the effect of force. So he will not get

injured.

8. The two tends of a spring balance are pulled each by a force of 20kgwt. What will be the

reaing of the spring balance?

Ans- 20kgwt

9. A lift is going up with acceleration 2g. A man is inside he lift and his mass is m. What will

be the reaction of the floor on the man?

Ans- R-mg = 2mg , R=3mg

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

Ans- law of conservation of linear momentum

11. What is the angle of friction between two surfaces in contact if the coefficient of friction

is 1/√3?

Ans- µ=tanθ =1/√3 , θ = 30⁰

12. What is the relation between frictional force and instantaneous velocity of a body moving

over a rough surface?

Ans- θ = 180⁰

13. Which is easier? Pull or push?

Ans- Pull, because it apparently reduces the mass of the body.

14. What provides the centripetal force to a car taking a turn on a level road?

Ans- force of friction between the tyres and road.

15. What is the need of banking of circular road?

Ans- the horizontal component of normal reaction provides necessary centripetal force for the

vehicle to move in circular path.

16. A tennis ball of mass m strikes the massive wall with velocity v and traces the same path.

Calculate the change in momentum.

Ans- dp=p2-p1 = -mv-mv = -2mv

17. What will be the maximum velocity with which a vehicle can negotiate a turn of radius r

safely, when the coefficient of friction between the tyres and road is µ?

Ans- v=õrg

18. Two objects of masses 3kg and 6kg are having the same magnitude of momentum. Find

the ratio of their speeds?

Ans- v1/v2 = 6/3 = 2

19. Position time graph of a body of mass 2kg is shown in the fig. Calculate net force acting

on a body for the time interval 0<t<1 sec.

Ans- Straight line v= constant, a=0, F=0

20. Action and reaction are equal and opposite. Why cannot they cancel each other?

Ans- they act at two bodies . so they cannot cancel eachother.

3 marks

1. Give reasons for the following

(a)A cricket player lowers his hand while catching a ball.

(b)Action and reaction do not cacel each other.

(c)It is easier to maitntain a motion than to start it.

2. Why circular roads are banked? Deduce an expression for the angle of banking in

order to have an optimum velocity ‘v’.

HINT

derivation

3. Define the term impulse. Is it a vector or a scalar? Give its S.I unit. Give an example

of impulsive force

4. Define a) angle of friction b) angle of repose

Prove that angle of friction is equal to angle of repose.

HINT

5. Define coefficient of friction. Draw the graph showing the variation of force of

friction with applied force. Label the different region in the graph

6. State and prove impulse momentum theorem. Explain how impulse can be found

using a force –time graph for a variable force.

Impulse =change in momentum of the body.

Impulse=force x time =

momentuminchangevmtt

vmtma

Area under the force –time gives the impuse of the variable force.

7. Two masses m1 and m2 are connected at the two ends of an inextensible string. The

string passes over a smooth frictionless pulley. Calculate the acceleration of the

masses and tension in the string. Given m1>m2

m1g-T =m1a and T-m2g=m2a

On solving the two equations g

mm

mma )(

21

21

and gmm

mmT )

2(

21

21

8. A horizontal force F pulls two masses m and M lying on a frictionless table connected

by a light string as shown in the above figure. Calculate the tension on the string when

F acts on (a) m towards right b) M towards left

acceleration aMm

F

.

In case(a) F-T=ma and in case (b)F-T’=Ma (From FBD of the two

bodies) On solving T= )(mM

FM

and T’= )('

mM

FmT

9. A man of mass M kg stands on a weighing scale in a lift which is

(a)Moving upwards with a uniform velocity of v m/s

(b)Moving downwards with a uniform acceleration of a m/s2

(c) Falling freely under gravity.

Find out the reading of the weighing machine in each case.

Ans.(a)Apparent weight shown by the scale=Mg(no change)

(b)Mg-N=Ma apparent weight =N=M(g-a)

(c)apparent weight=0

10. State the laws of limiting friction. Why friction is called a necessary evil?.

11. It is easy to pull than to push a lawn roller. Why?

12. What is centripetal force? Derive the expression for it.

13. Newton’s second law is the real law. Why? Obtain the first and third law from the

second law.

14. State the law of conservation of linear momentum. Derive the law from Newton’s

third law. Give an example for the consequence of this law.

5 MARK QUESTIONS 1.a)State and prove law of conservation of momentum. b) How is the principle used in the launching

of rockets? c) Get an expression for the recoil velocity of a gun

Hint-statement& proof b)explanation c) V=-mv/M

2. State Newton’s second law of motion. Derive a mathematical expression for it.

Show that Newton’s first law and third law can be obtained from second law.

Hint-statement& proof b) Explanation

3.What is Banking of roads? B) What is the need for banking?c)Derive expression for the velocity of

a car on a banked circular road having coefficient of friction µ.

Hint-Definition b) explanation c) derivation of v= [rg(tanƟ+µ)/(1-µtanƟ)]1/2

4. Explain the terms 1. Angle of friction 2. Coefficient of friction 3. Angle of repose

Establish a relation between them.

Hint-Definitions Derivation of µ=tanƟ=tanα

5. Discuss the motion of a body in a vertical circle.Find expression for the minimum velocity at the

lowest point while looping a loop and tension in the string at the lowest and highest point.

Hint-Derivation of T L= m(u2+gr/r) 2) TH =m(u

2-5gr)/r 3) VL=(5gr)

1/2

6. State Newton’s third law of motion .Explain how a horse can pull a cart obeying Newton’s third

law.

A man of mass 70kg stands on a weighing machine in a lift ,which is moving a) upwards with a

uniform acceleration of 5ms-2

b) downwards with a uniform acceleration of 5ms-2

Hint-Statement and explanation II) a) R-m(g+a)=70x15=1050N b) R=m(g-a)= 70X5=350N

7.what are concurrent forces? Obtain a condition for the equilibrium of three concurrent forces.

Three bodies A,Band C each of mass m are hanging over a pulley as shown in figure.What are the

tensions in the strings connecting 1) A to B 2)B to C3m

Hint- Definition .Net force acting is zero .c) T1-mg=ma, 2mg-T1=2ma and mg-T2=ma

Solving a=2/3 T1=4/3mg T2=2/3mg

8.Distinguish between static friction, limiting friction and kinetic friction. How do they vary with

applied force,explain using a diagram

Why is friction called a necessary evil?

Hint-Definitions &graph ,Explanation

9. Define impulse. Give its SI unit. Prove that impulse of a force is equal to the change in

momentum.

Why do batsmen lower their hands while catching a cricket ball?

Hint- Definition &unit proof of Ft=mv-mu ,Explanation

10. Derive an expression for the work done when a body 1) slides down an inclined plane 2) is made

to slide up an inclined plane.

Hint- Derivation of a= g(sinƟ-µkcosƟ) for downward motion and a= g (sinƟ+µkcosƟ) for upward motion.

HOTS

S.No Questions Marks

Q1 A girl riding a bicycle along a straight road

with a speed of 5 m s–1

throws a stone of

mass 0.5 kg which has a speed of 15 m s–

1 with respect to the ground along her

direction of motion. The mass of the girl and bicycle is 50 kg. Does the speed of

the bicycle change after the stone is

thrown? What is the change in speed, if so?

A1 Yes, due to the principle of conservation

of momentum. Initial momentum = 50.5 × 5 kg m s–1

Final momentum = (50 v + 0.5 × 15) kg m s–1

(a) = 4.9 m s-1

, change in speed = 0.1 m s–1

Q2 A person of mass 50 kg stands on a weighing

scale on a lift. If the lift is descending

with a downward acceleration of 9 m s–2

,

what would be the reading of the

weighing scale? (g = 10 m s–2

)

A2

Q3

Let R be the reading of the scale, in newtons.

50g R

Effective downward acceleration g 50 R = 5g = 50N. (The weighing scale will show 5 kg).

The position time graph of a body of mass 2 kg

is as given in Fig.. What is the impulse on

the body at t = 0 s and t = 4 s

A3 Zero ; -3/2 kgm/s Q4 A person driving a car suddenly applies the brakes on seeing a

child on the road ahead. If he is not wearing seat belt, he falls

forward and hits his head against the steering wheel. Why?

A4 The only retarding force that acts on him, if he is not using a

seat belt comes from the friction exerted by the seat. This is

not enough to prevent him from moving forward when the

vehicle is brought to a sudden halt.

Q5 A block placed on a rough horizontal surface is pulled by a

horizontal force F. Let f be the force applied by the rough

surface on the block. Plot a graph of f versus F.

A5

Q6

f = F until the block is stationary.

f remains constant if F increases beyond this point and the block starts moving

f

F Why are porcelain objects wrapped in paper or straw before

packing for transportation?

A6 : In transportation, the vehicle say a truck, may need to halt suddenly. To

bring a fragile material, like porcelain object to a sudden halt means

applying a large force and this is likely to damage the object. If it is

wrapped up in say, straw, the object can travel some distance as the straw

is soft before coming to a halt. The force needed to achieve this is less,

thus reducing the possibility of damage

Q7 Why does a child feel more pain when she falls down on a hard

cement floor, than when she falls on the soft muddy ground in

the garden?

A7 The body of the child is brought to a sudden halt when

she/he falls on a cement floor. The mud floor yields and the

body travels some distance before it comes to rest , which

takes some time. This means the force which brings the child

to rest is less for the fall on a mud floor, as the change in

momentum is brought about over a longer period.

Q8 Why are mountain roads generally made winding

upwards rather than going straight up?

A8 : f = R = mg cos is the force of friction, if is angle

made by the slope. If is small, force of friction is high

and there is less chance of skidding. The road straight up

would have a larger slope.

Q9 A mass of 2kg is suspended with thread AB (Fig. ). Thread CD

of the same type is attached to the other end of 2 kg mass. Lower

thread is pulled

gradually, harder and harder in the downward directon so as to

apply force on AB. Which of the threads will break and why?

A9 AB, because force on the upper thread will be equal to sum of the weight

of the body and the applied force

Q10 In the above given problem if the lower thread is pulled with a

jerk, what happens?

A10 If the force is large and sudden, thread CD breaks

because as CD is jerked, the pull is not transmitted to AB

instantaneously (transmission depends on the elastic

properties of the body). Therefore, before the mass moves,

CD breaks.

Q11

A block of mass M is held against a rough vertical wall by

pressing it with a finger. If the coefficient of friction

between the block and the wall is µ and the acceleration

due to gravity is g calculate the minimum force

required to be applied by the fingerto hold the block

against the wall.

A11 Answer : If F is the force of the finger on the book, F

= N, the normal reaction of the wall on the book. The

minimum upward frictional force needed to ensure that

the book does not fall is Mg. The frictional force = N.

Thus, minimum value of F Mg

.

Q12

There are three forces F1, F2 and F3 acting on

a body, all acting on a point P on the

body. The body is found to move with

uniform speed.

(a) Show that the forces are coplanar. (b) Show that the torque acting on the body

about any point due to these three forces is

zero.

A12

(a) Since the body is moving with no acceleration,

the sum of the 3 forces is zero.

F 1 +F 2 +F3 =0

Let F1 , F2 , F3 be the 3 forces passing

through a point

Let F1 and F2 be in the plane A (one can

always draw a plane having two intersecting lines

such that the two lines lie on the plane)

. Then F1 F2 must be in the plane A.

Since F3 –F1 + F2 ,

F3 is also in the plane A

(b)Consider the torque of the forces about P. Since all the forces pass

through P, the torque is zero. Now consider torque about another point

0. Then torque about 0 is

Torque OPF1+ F2+ F3

Since F1 + F2 + F3 0 , torque = 0

Numericals from Laws of motion

Q.NO QUESTION MARK

Laws of motion

1 FInd the ratio of Inertia of two masses whose masses are in the ratio 2:5 Ans: Ratio of inertia=2:5

1

2 What will be the weight felt by a person of mass 50 kg in a lift accelerated at 2m/s2

downwards? Ans :50(10-2)N=400N

1

3 Find the mass of of a body which is accelerated at 4m/s2 when a force of 10N is applied. Ans :m=F/a=10/4 kg=2.5kg.

1

4 A body of mass 10kg is kept on a horizontal surface with μ=0.01.Find the limiting friction. Ans :Limiting friction=μmg=0.01x10x10n=1N

1

5 A bullet of mass=2g is accelerated at 500m/s2 from agun.Calculate the reaction on the

gun. Ans :Reaction on gun=action on bullet=ma=2/1000x500N=1N

1

6 A force of 5 N acts on a stationary body of mass 5kg.Find its speed after 10s. Ans :V=u+at=[0+ (1x10)] m/s=10m/s

2

7 A ball of mass 500g moving with a speed of 10m/s strikes a wall and rebounds with the same speed.Calculate the impulse imparted to the wall. Ans : J=m(v-u)=0.5x(-10-10)kgm/s=-10kgm/s

2

8 A body of mass 100 kg is kept on a plane inclined at an angle of 450with

the

ground.Calculate its downward acceleration if the if μ=0.05 2

Ans :a=g(1-μsinɵ)= 10(1-.05/√2)m/s2=.965m/s

2

9 A block of mass 50 kg is kept on a horizontal plane surface with μ=0.1. Calculate the limiting friction with the normal reaction. Ans :Fms=μmg=0.1x50x10N=50N

2

10 A car of mass 800 kg is negotiating a circular road radius 5m banked at an angle of 30

0.calculate the maximum speed allowed for the car to avoid an accident.(Neglect

friction) Ans :Vmax=√Rgtanɵ=√5x10x1/√3=5.38m/s

2

11 A cricket ball of mass 150g is moving with a speed of 12m/s and is hit by a bat, so thate ball is turned back with a speed of 20m/s.The force acts for 0.02seconds on the ball.Find the average force exerted by the bat on the ball. Ans :F=m(v-u)/t=0.15x(20+12)/0.02=240N

3

12 Two masses 8kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley.Find the acceleration of the masses and the the tension in the string when the masses are released. Ans :a=(12-8)x10/(12+8)m/s

2=2m/s

2

T=8(10+a)=12(10-a)=96N

3

13 a)What is banking of roads? b)A curved road of diameter 1.8km is banked, so that no friction is required at a speed 0f 30m/s.What is the banking angle? Ans :Definition tanɵ=Vmax/√Rg=30/(√1800x10)=.224 ɵ=12.6

o

3

Value Based Questions.

01. Rohit saw her grandmother trying to clean the carpet. She was finding difficulty in lifting

the carpet. Immediately Rohit helped her by beating the carpet with a stick.

(a) Name the scientific principle involved in Rohit’s action and explain?

(b) What are the values displayed by Rohit?

( c) Give one more such example.

Ans- ( a) inertia of rest - statement

(b) helping , caring and empathy

© when a bus suddenly starts from rest, passengers move backward due to inertia of rest

02. Akshay and Jonny both are the students of class11. On sports day they participated in

long jump event. Akshay took a long run before jumping from the starting point and could

finish as winner. But Jonny who jumped for the start mark could not cover more distance.

Jonny asked Akshay about his long distance jump.He explained the reason.

(a) What could be the scientific reason for Akshay’s long distance jump? Explain

(b) What are the values exhibited by Akshay?

© Give one more such example.

Ans-

(a) inertia of motion, momentum

(b)execution of knowledge , intelligent,smart

(c ) in putting the shot event, player need practice.

03. Reshmi was going to market with her grantfather. In a super market lift she coulod see a

man with weighing machine . With the permission of the owner of the machine , Rashmi

asked her grantfathe to stand on the machine.It showed 50kg. When the lift started

accelerating upward it showed 60 kg.Suddenly her grantfather said that the machine is

defective.

(a) What are the values showed by Rashmi?

(b) Why did the weighing machine showed 60 kg instead of 50kg while accelerating upward?

Ans-

(a) scientific thinking, awarenss, knowledge etc

(b) in upward acceleration, apparent weight = m(g+a) so it showed more.