Q1 
An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a nonzero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason. 
Ans: 
Yes. It is possible.
(i) The object should already be moving at uniform speed in a straight line path.
(ii) There is no change in speed.
(iii) No direction change.
(iv) Friction force must be zero.
(v) Air resistance must be zero. 

Q2 
When a carpet is beaten with a stick, dust comes out of it. Explain. 
Ans: 
When a carpet is beaten with a stick, the dust comes out of it because of the law of inertia. Initially the dust particles are at rest along with the carpet but, the dust particles try to resist their state of rest. According to Newton’s first law of motion, the dust particles stay in a state of rest, while the carpet moves. Hence, the dust particles come out of the carpet. 

Q3 
Why is it advised to tie any luggage kept on the roof of a bus with a rope? 
Ans: 
When a moving bus suddenly stops, the luggage on the roof tends to continue its state of motion and may fall. Also, when the bus suddenly starts from rest, luggage maintains its rest position and may fall backward. So, it is advised to tie any luggage kept on the roof of a bus with a rope. 

Q4 
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
(a) the batsman did not hit the ball hard enough.
(b) velocity is proportional to the force exerted on the ball.
(c) there is a force on the ball opposing the motion.
(d) there is no unbalanced force on the ball, so the ball would want to come to rest. 
Ans: 
(c) there is a force on the ball opposing the motion. 

Q5 
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 = 1000 kg.) 
Ans: 
Initial velocity, u = 0
Distance travelled, s = 400 m
Time taken, t = 20 s
According to equation of motion:
a = 2 m/s^{2}
1 metric tonne = 1000 kg (Given)
∴ 7 metric tonnes = 7000 kg
Mass of truck, m = 7000 kg
Force, F = Mass × Acceleration
F = ma = 7000 × 2 = 14000 N
The acceleration of the truck is 2 m/s2 and the force acting on the truck is 14000 N. 

Q6 
A stone of 1 kg is thrown with a velocity of 20 m s1 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? 
Ans: 
Mass of the stone = 1 kg
Initial velocity of the stone, u = 20 m/s
Final velocity of the stone, v = 0
Distance travelled by the stone, s = 50 m
Using the third equation of motion:
v_{2} = u_{2} + 2as
Where,
Acceleration, a
(0)_{2} = (20)_{2} + 2 × a × 50
a = −4 m/s^{2}
Force, F = Mass × Acceleration
F = ma
F = 1 × (− 4) = −4 N
The negative sign shows that force of friction is in the opposite direction of motion , the force of friction between the stone and the ice is −4 N. 

Q7 
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 5000 N, then calculate:
(a) the net accelerating force;
(b) the acceleration of the train; and
(c) the force of wagon 1 on wagon 2. 
Ans: 
Mass of engine = 8000 kg
Mass of 1 wagon = 2000 kg
Mass of 5 wagons = 2000 x 5 = 10000 kg
Force exerted by the engine, F = 40000 N
Frictional force offered by the track, F_{f} = 5000 N
(a) Net force, F_{a} = Force of engine  Force of friction = F − F_{f}
= 40000 − 5000 = 35000 N
Hence, the net accelerating force is 35000 N.
(b) Total mass of the train, M = mass of 5 wagons + mass of engine
= 10000 + 8000 = 18000 kg
From Newton’s second law of motion:
Force = Mass x acceleration
35000 = 18000 x a
a = 35000 / 18000 = 1.94 m/s^{2}
Hence, the acceleration of the train is 1.94 m/s^{2}. 

Q8 
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 s2? 
Ans: 
Mass of the automobile vehicle, m = 1500 kg
Acceleration of the automobile, a = −1.7 ms^{−2}
We know that
Force = Mass × Acceleration = 1500 × (−1.7) = −2550 N
The negative sign shows that force is in the opposite direction of motion, the force between the automobile and the road is −2550 N. 

Q9 
What is the momentum of an object of mass m, moving with a velocity v?
(a) (mv)2 (b) mv2 (c) 1⁄2 mv2 (d) mv 
Ans: 


Q10 
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? 
Ans: 
Since, a horizontal force of 200 N is used to move a wooden cabinet. Thus, from Newton’s third law of motion, an equal amount of force will act in the opposite direction. This opposite force is the frictional force exerted on the cabinet. Hence, a frictional force of 200 N is exerted on the cabinet. 

Q11 
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 m s1 before the collision during which they stick together. What will be the velocity of the combined object after collision? 
Ans: 
Mass of first objects, m_{1} = 1.5 kg
Mass of second object, m_{2} = 1.5 kg
Velocity of m_{1} before collision, v_{1} = 2.5 m/s
Velocity of m_{2}, moving in opposite direction before collision, v_{2} = −2.5 m/s
(Negative sign arises because mass m_{2} is moving in an opposite direction)
We know that
Total momentum before collision = Total momentum after collision
m_{1}v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v
1.5(2.5) + 1.5 (−2.5) = (1.5 + 1.5) v
3.75 − 3.75 = 3 v
v = 0 / 3 = 0
Hence, the velocity of the combined object after collision is 0 m/s. 

Q12 
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. 
Ans: 
Because of the huge mass of the truck, the force of static friction is very high. To move the car / truck, one has to apply a force more than the static friction. Therefore, when someone pushes the truck and the truck does not move, then it can be said that the applied force in one direction is cancelled out by the frictional force of equal amount acting in the opposite direction. Hence , the rationale given by the students is correct. 

Q13 
A hockey ball of mass 200 g travelling at 10 m s1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 m s1. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick. 
Ans: 
Mass of the hockey ball, m = 200 g = 200 / 1000 = 0.2 kg
Initial velocity, v_{1} = 10 m/s
Initial momentum = mv_{1}
Final velocity, v_{2} = −5 m/s
Final momentum = mv_{2}
Change in momentum = mv_{1} − mv_{2} = m(v_{1}  v_{2})
= 0.2 [10 − (−5)] = 0.2 (15) = 3.0 kg m/s
Hence, the change in momentum of the hockey ball is 3 kg m/s. 

Q14 
A bullet of mass 10 g travelling horizontally with a velocity of 150 m s1 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. 
Ans: 
Mass of bullet, m = 10 g = 10 / 1000 = 0.01 kg
Initial velocity, u = 150 m/s
Final velocity, v = 0
Time taken, t = 0.03 s
Using the first equation of motion
v = u + at
0 = 150 + (a ×0.03 s)
(Negative sign indicates that the velocity of the bullet is decreasing.)
Now, using the third equation of motion:
v_{2} = u_{2} + 2as
0 = (150)2 + 2 (−5000) s
Hence, the distance of penetration of the bullet is 2.25 m.
We know that
Force, F = Mass × Acceleration
F = ma = 0.01 × 5000 = 50 N
Hence, the magnitude of force is 50 N. 

Q15 
An object of mass 1 kg travelling in a straight line with a velocity of 10 m s1 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. 
Ans: 
Mass of the object, m1 = 1 kg
Mass of the wooden block, m2 = 5 kg
Velocity of the object before collision, v1 = 10 m/s
Velocity of the wooden block before collision, v2 = 0 m/s
Total mass of the combined system = m1 + m2 = 1 + 5 = 6 kg
∴ Total momentum before collision = m1 v1 + m2 v2
= 1 (10) + 5 (0) = 10 kg m s^{−1}
Velocity of the combined object = v
Momentum = total mass x velocity of combined object
= (m_{1} + m_{2}) v = 6v
According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v
10 = 6 v
The total momentum after collision is also 10 kg m/s.
Hence, velocity of the combined object after collision = 

Q16 
An object of mass 100 kg is accelerated uniformly from a velocity of 5 m s1 to 8 m s1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object. 
Ans: 
Mass of the object, m = 100 kg
Initial velocity of the object, u = 5 m/s
Final velocity of the object, v = 8 m/s
Time take by the object to accelerate, t = 6 s
Initial momentum = mu = 100 × 5 = 500 kg m/s
Final momentum = mv = 100 × 8 = 800 kg m/s
By using Newton’s second law of motion, F = change in momentum / time taken =
Force exerted on the object is 50 N. 

Q17 
Akhtar, Kiran and Rahul were riding in a motorcar 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 
Ans: 
According to the law of conservation of momentum, the momentum of a system before collision is equal to the momentum of the system after collision.
Hence, the change in momentum of the car and insect system is zero.
Thus, Kiran is wrong with the logic as momentum is always conserved, i.e change in momentum of insect must be equal to that of a motorcar.
Akhtar’s logic is incorrect because the mass of the car is very large as compared to the mass of the insect.
Rahul gave the correct logic as insects and motorcar experience the same force and change in momentum. 

Q18 
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 s2. 
Ans: 
Mass of the dumbbell, m = 10 kg
Distance, s = 80 cm = 80 / 100 = 0.8 m
Acceleration, a = 10 m/s^{2}
Initial velocity of the dumbbell, u = 0
Final velocity of the dumbbell = v
Using the third equation of motion:
v_{2} = u_{2} + 2as
v_{2} = 0 + 2 (10) 0.8
v = 4 m/s
Hence, the momentum with which the dumbbell hits the floor is
= mv = 10 × 4 = 40 kg m/s 
