1) An object experiences a net zero

external unbalanced force. Is it possible for the object to be travelling with

a non-zero velocity? If yes, state the conditions that must be placed on the

magnitude and direction of the velocity. If no, provide a reason.

external unbalanced force. Is it possible for the object to be travelling with

a non-zero velocity? If yes, state the conditions that must be placed on the

magnitude and direction of the velocity. If no, provide a reason.

ANSWER:-Yes, an object may travel with a

non-zero velocity even when the net external force on it is zero. A rain drop

falls down with a constant velocity. The weight of the drop is balanced by the

up thrust and the velocity of air.

non-zero velocity even when the net external force on it is zero. A rain drop

falls down with a constant velocity. The weight of the drop is balanced by the

up thrust and the velocity of air.

2) When a carpet is beaten with a

stick, dust comes out of it. Explain.

stick, dust comes out of it. Explain.

ANSWER:-When the carpet is beaten, it is

suddenly set into motion. The dust particles tend to remain at rest due to

inertia of rest, therefore the dust comes

out of it.

suddenly set into motion. The dust particles tend to remain at rest due to

inertia of rest, therefore the dust comes

out of it.

3) Why is it advised to tie any luggage

kept on the roof of a bus with a rope?

kept on the roof of a bus with a rope?

ANSWER:-When a bust starts suddenly, the

lower part of the luggage kept on the roof being in contact with the bus

begins to move forward with the speed of bus, but the upper part tends to

remain at rest due to inertia of rest. Therefore, the upper part is left behind

and hence luggage falls backward. So, it is advised to tie any luggage kept on

the roof of a bus with a rope.

lower part of the luggage kept on the roof being in contact with the bus

begins to move forward with the speed of bus, but the upper part tends to

remain at rest due to inertia of rest. Therefore, the upper part is left behind

and hence luggage falls backward. So, it is advised to tie any luggage kept on

the roof of a bus with a rope.

4) A batsman hits a cricket ball which

then rolls on a level ground. After covering a short distance, the ball comes

to rest. The ball slows to a stop because

then rolls on a level ground. After covering a short distance, the ball comes

to rest. The ball slows to a stop because

(a) The batsman did not hit the ball

hard enough.

hard enough.

(b) Velocity is proportional to the

force exerted on the ball.

force exerted on the ball.

(c) There is a force on the ball

opposing the motion.

opposing the motion.

(d) There is no unbalanced force on

the ball, so the ball would want to come to rest.

the ball, so the ball would want to come to rest.

ANSWER:-The ball slows down and comes to

rest due to opposing forces of air resistance

and frictional force on the ball opposing its motion. Therefore the choice (c) there is a force on the ball opposing the

motion is correct.

rest due to opposing forces of air resistance

and frictional force on the ball opposing its motion. Therefore the choice (c) there is a force on the ball opposing the

motion is correct.

5) A truck starts from rest and rolls

down a hill with a constant acceleration. It travels a distance of 400 m in 20

s. Find its acceleration. Find the force acting on it if its mass is 7 metric

tonnes (Hint: 1 metric tonne = 1000kg).

down a hill with a constant acceleration. It travels a distance of 400 m in 20

s. Find its acceleration. Find the force acting on it if its mass is 7 metric

tonnes (Hint: 1 metric tonne = 1000kg).

ANSWER:-Initial velocity, u = 0 Distance travelled, s =

400 m Time taken, t = 20 s

400 m Time taken, t = 20 s

We know, s = ut + ½ at2

Or, 400 = 0 + ½ a (20) 2

Or, a = 2 ms –2

Now, m = 7 MT = 7000 kg, a = 2 ms–2

Or, F = ma = 7000 × 2 = 14000N

6) A stone of 1 kg is thrown with a

velocity of 20 m s −1 across the frozen surface of a lake and comes to rest

after travelling a distance of 50 m. What is the force of friction between the

stone and the ice?

velocity of 20 m s −1 across the frozen surface of a lake and comes to rest

after travelling a distance of 50 m. What is the force of friction between the

stone and the ice?

ANSWER:-Initial velocity of the stone,

u= 20 m/s

u= 20 m/s

Final velocity of the stone, v=0

Distance covered by the stone,

s = 50 m

Since, v 2 – u 2

2 = 2as,

Or, 0 – 202 = 2a × 50,

Or, a = – 4 ms-2

Force of friction, F = ma = – 4N

7) A 8000 kg engine pulls a train of 5

wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force

of 40000 N and the track offers a friction force of 5000N, then calculate:

wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force

of 40000 N and the track offers a friction force of 5000N, then calculate:

(a) The net accelerating force;

(b) The acceleration of the train; and

(C) the force of wagon 1 on wagon 2.

ANSWER 🙁 a) Force exerted by the engine, F = 40000 N

Frictional force offered by the track, Ff = 5000 N

Net accelerating force, Fa = F − Ff = 40000 − 5000 =

35000 N

35000 N

Hence, the net accelerating force is 35000 N.

(b) Acceleration of the train =a

The engine exerts a force of 40000 N on all the five

wagons.

wagons.

Net accelerating force on the wagons, Fa = 35000 N

Mass of the wagons, m = Mass of a wagon x Number of

wagons

wagons

Mass of a wagon = 2000 kg

Number of wagons = 5

∴ m = 2000 × 5 = 10000 kg

Total mass, M = m = 10000 kg

From Newton’s second law of motion:

F a = Ma

a=Fam

= 35000 10000 = 3.

= 35000 10000 = 3.

5 ms-2

Hence, the acceleration of the wagons and the train is

3.5 m/ s 2.

3.5 m/ s 2.

(c) Mass of all the wagons except wagon 1 is 4 ×

2000 =8000 kg

2000 =8000 kg

Acceleration of the wagons = 3.5 m/s 2

Thus, force exerted on all the wagons except wagon 1

= 8000 × 3.5 = 28000 N

8) An automobile vehicle has a mass of

1500 kg. What must be the force between the vehicle and road if the vehicle is

to be stopped with a negative acceleration of 1.7 m s −2?

1500 kg. What must be the force between the vehicle and road if the vehicle is

to be stopped with a negative acceleration of 1.7 m s −2?

ANSWER:-Mass of the

automobile vehicle, m= 1500 kg

automobile vehicle, m= 1500 kg

Final velocity, v= 0 (finally the automobile stops)

Acceleration of the automobile, a = −1.7 ms−2

From Newton’s second law of motion:

Force = Mass x Acceleration = 1500 x (−1.7) = −2550 N

Hence, the force between the automobile and the road

is −2550 N, in the direction opposite to the motion of the automobile.

is −2550 N, in the direction opposite to the motion of the automobile.

9) What

is the momentum of an object of mass m, moving with a velocity v ?

is the momentum of an object of mass m, moving with a velocity v ?

(a) (mv)

2 (b) 1/2mv2 (c)mv

2 (b) 1/2mv2 (c)mv

ANSWER:-Mass

of the object = m

of the object = m

Velocity =

v

v

Momentum =

Mass x Velocity

Mass x Velocity

Momentum =

mv

mv

10) Using a horizontal force of 200 N,

we intend to move a wooden cabinet across a floor at a constant velocity. What

is the friction force that will be exerted on the cabinet?

we intend to move a wooden cabinet across a floor at a constant velocity. What

is the friction force that will be exerted on the cabinet?

ANSWER:-The cabinet will

move with constant velocity only when the net

force on it is zero. Therefore, force of

friction on the cabinet = 200 N, in a

direction opposite to the direction of motion of the cabinet.

move with constant velocity only when the net

force on it is zero. Therefore, force of

friction on the cabinet = 200 N, in a

direction opposite to the direction of motion of the cabinet.

11) Two objects, each of mass 1.5 kg are

moving in the same straight line but in opposite directions. The velocity of

each object is 2.5 ms−1 before the collision during which they stick together.

What will be the velocity of the combined object after collision?

moving in the same straight line but in opposite directions. The velocity of

each object is 2.5 ms−1 before the collision during which they stick together.

What will be the velocity of the combined object after collision?

ANSWER:-Mass of one of the objects, m1=

1.5 kg

1.5 kg

Mass of the other object, m2 =1.5 kg

Velocity of m1 before collision,

u1 = 2.5 m/s

Velocity of m2, moving in opposite direction before

collision, u 2 = −2.5 m/s

collision, u 2 = −2.5 m/s

Let v be the velocity of the combined object after

collision.

collision.

By the law of conservation of momentum,

Total momentum after collision = Total momentum before

collision,

collision,

Or, (m1 + m2) v = m1u1 +m2u2

Or, (1.5 + 1.5) v = 1.5 × 2.5 +1.5 × (–2.5) [negative

sign as moving in opposite direction]

sign as moving in opposite direction]

Or, v = 0 ms–1

12) According to the third law of

motion when we push on an object, the object pushes back on us with an equal

and opposite force. If the object is a massive truck parked along the roadside,

it will probably not move. A student justifies this by answering that the two

opposite and equal forces cancel each other. Comment on this logic and explain

why the truck does not move.

motion when we push on an object, the object pushes back on us with an equal

and opposite force. If the object is a massive truck parked along the roadside,

it will probably not move. A student justifies this by answering that the two

opposite and equal forces cancel each other. Comment on this logic and explain

why the truck does not move.

ANSWER:-When we push a massive truck,

the force of friction between its tyres and the road is very large and so the

truck does not move.

the force of friction between its tyres and the road is very large and so the

truck does not move.

13) A hockey ball of mass 200g

travelling at 10 m s −1 is struck by a hockey stick so as to return it along

its original path with a velocity at 5 m s -1. Calculate the change of momentum

occurred in the motion of the hockey ball by the force applied by the hockey

stick.

travelling at 10 m s −1 is struck by a hockey stick so as to return it along

its original path with a velocity at 5 m s -1. Calculate the change of momentum

occurred in the motion of the hockey ball by the force applied by the hockey

stick.

ANSWER:-Mass of the hockey ball, m =200

g = 0.2 kg

g = 0.2 kg

Hockey ball travels with velocity, v1 =

10 m/s

10 m/s

Initial momentum = mv 1

Hockey ball travels in the opposite

direction with velocity, v2 = −5 m/s

direction with velocity, v2 = −5 m/s

Final momentum = mv2

Change in momentum = mv 1 − mv 2 = 0.2 [10 − (−5)] =

0.2 (15) = 3 kg m s −1

0.2 (15) = 3 kg m s −1

Hence, the change in momentum of the hockey

ball is 3 kg m s −1.

ball is 3 kg m s −1.

14) A bullet of mass 10 g travelling

horizontally with a velocity of 150 m s −1 strikes a stationary wooden block

and comes to rest in 0.03 s. Calculate the distance of penetration of the

bullet into the block. Also calculate the magnitude of the force exerted by the

wooden block on the bullet.

horizontally with a velocity of 150 m s −1 strikes a stationary wooden block

and comes to rest in 0.03 s. Calculate the distance of penetration of the

bullet into the block. Also calculate the magnitude of the force exerted by the

wooden block on the bullet.

ANSWER:-Initial velocity, u= 150 m/s

Final velocity, v= 0 (since the bullet finally comes to

rest)

rest)

Time taken to come to rest, t=0.03 s

According to the first equation of motion, v = u + at

Acceleration of the bullet, a 0 = 150 + (a × 0.03 s) a

= -150 / 0.03 = -5000 m/s2

= -150 / 0.03 = -5000 m/s2

(Negative sign indicates that the velocity of the

bullet is decreasing.)

bullet is decreasing.)

According to the third equation of motion:

v 2= u 2+ 2 as

0 = (150) 2+ 2 (-5000)

= 22500 / 10000

= 2.25 m

Hence, the distance of penetration of

the bullet into the block is 2.25 m.

the bullet into the block is 2.25 m.

From Newton’s second law of motion:

Force, F = Mass × Acceleration

Mass of the bullet, m = 10 g =0.01 kg

Acceleration of the bullet, a = 5000 m/s 2

F = ma = 0.01 × 5000 = 50 N

15) An object of mass 1 kg travelling

in a straight line with a velocity of 10 m s −1 collides with, and sticks to, a

stationary wooden block of mass 5 kg. Then they both move off together in the

same straight line. Calculate the total momentum just before the impact and

just after the impact. Also, calculate the velocity of the combined object.

in a straight line with a velocity of 10 m s −1 collides with, and sticks to, a

stationary wooden block of mass 5 kg. Then they both move off together in the

same straight line. Calculate the total momentum just before the impact and

just after the impact. Also, calculate the velocity of the combined object.

ANSWER:-Mass of the object, m1 = 1 kg

Velocity of the object before collision, v 1 = 10 m/s

Mass of the stationary wooden block, m2 = 5 kg

Velocity of the wooden block before collision, v 2 = 0

m/s

m/s

∴ Total momentum before collision = m1 v1 + m2 v

2

2

= 1 (10) + 5 (0) = 10 kg m s −1

It is given that after collision, the object and the wooden

block stick together.

block stick together.

Total mass of the combined system = m1 + m2

Velocity of the combined object = v

According to the law of conservation of

momentum:

momentum:

Total momentum before collision = Total momentum

after collision

after collision

m1 v1 + m2 v 2 = (m1 + m2) v

1 (10) + 5 (0) = (1 + 5) v

v = 10 / 6

= 5 / 3

The total momentum after collision is

also 10 kg m/s.

also 10 kg m/s.

Total momentum just before the impact =

10 kg m s −1

10 kg m s −1

Total momentum just after the impact = (m1 + m2) v = 6

× 5 / 3 = 10 kg ms-1

× 5 / 3 = 10 kg ms-1

16) An object of mass 100 kg is

accelerated uniformly from a velocity of 5 m s −1 to 8 m s −1 in 6 s. Calculate

the initial and final momentum of the object. Also, find the magnitude of the

force exerted on the object.

accelerated uniformly from a velocity of 5 m s −1 to 8 m s −1 in 6 s. Calculate

the initial and final momentum of the object. Also, find the magnitude of the

force exerted on the object.

ANSWER:-Initial velocity of the object,

u= 5 m/s

u= 5 m/s

Final velocity of the object, v = 8 m/s

Mass of the object, m = 100 kg

Time take by the object to accelerate, t = 6 s

Initial momentum = mu = 100 × 5 = 500 kg m s −1

Final momentum = mv = 100 × 8 = 800 kg m s −1

Force exerted on the object, F

= mv – mu / t

= m (v-u) / t

= 800 – 500

= 300 / 6

= 50 N

17)Akhtar, Kiran and Rahul were riding

in a motorocar that was moving with a high velocity on an expressway when an

insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran

started pondering over the situation. Kiran suggested that the insect suffered

a greater change in momentum as compared to the change in momentum of the motorcar

(because the change in the velocity of the insect was much more than that of

the motorcar). Akhtar said that since the motorcar was moving with a larger

velocity, it exerted a larger force on the insect. And as a result the insect

died. Rahul while putting an entirely new explanation said that both the motorcar

and the insect experienced the same force and a change in their momentum.

Comment on these suggestions.

in a motorocar that was moving with a high velocity on an expressway when an

insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran

started pondering over the situation. Kiran suggested that the insect suffered

a greater change in momentum as compared to the change in momentum of the motorcar

(because the change in the velocity of the insect was much more than that of

the motorcar). Akhtar said that since the motorcar was moving with a larger

velocity, it exerted a larger force on the insect. And as a result the insect

died. Rahul while putting an entirely new explanation said that both the motorcar

and the insect experienced the same force and a change in their momentum.

Comment on these suggestions.

ANSWER:-The suggestion made by Kiran

that the insect suffered a greater change in momentum as compared to the change in

momentum of the motor car is wrong.

that the insect suffered a greater change in momentum as compared to the change in

momentum of the motor car is wrong.

The suggestion made by Akhtar that the

motor car exerted a larger force on the insect because of large velocity of

motor car is also wrong. The explanation put

forward by Rahul is correct. On collision of insect

with motor car, both experience the same force as action and reaction are

always equal and opposite. Further, changes in their momenta are also the same.

Only the signs of changes in momenta are

opposite.

motor car exerted a larger force on the insect because of large velocity of

motor car is also wrong. The explanation put

forward by Rahul is correct. On collision of insect

with motor car, both experience the same force as action and reaction are

always equal and opposite. Further, changes in their momenta are also the same.

Only the signs of changes in momenta are

opposite.

18) How much momentum will a dumbbell

of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take

its downward acceleration to be 10 m s −2.

of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take

its downward acceleration to be 10 m s −2.

ANSWER:-Mass of the dumbbell, m = 10 kg

Distance covered by the dumbbell, s = 80 cm = 0.8

m

m

Acceleration in the downward direction, a = 10 m/s 2

Initial velocity of the dumbbell, u = 0

Final velocity of the dumbbell (when it was about to hit

the floor) = v

the floor) = v

According to the third equation of motion:

v 2 = u2 + 2as

v 2 = 0 + 2 (10) 0.8

v = 4 m/s

Hence, the momentum with which the dumbbell hits

the floor is

the floor is

= mv = 10 × 4 = 40 kg m s −1