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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 = 3012. What is the relation between frictional force and instantaneous velocity of a body moving over a rough surface? Ans- θ = 18013. 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

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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 = 30
12. What is the relation between frictional force and instantaneous velocity of a body moving
over a rough surface?
Ans- θ = 180
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=√µrg
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
(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 =

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

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 a Mm
.
In case(a) F-T=ma and in case (b)F-T’=Ma (From FBD of the two
bodies) On solving T= )( mM
F M
and T’= )('

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/s 2
(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(tan+µ)/(1-µtan)]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=tan=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(u 2 +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(sin-µkcos) for downward motion and a= g (sin+µkcos) for upward motion.
HOTS
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
,
)
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.
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
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
.
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
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/s 2
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/s 2 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 45 0 with
the
ground.Calculate its downward acceleration if the if μ=0.05 2
Ans :a=g(1-μsin)= 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=√Rgtan=√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
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 tan=Vmax/√Rg=30/(√1800x10)=.224 =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.
(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-
(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

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