Q1. 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.

Answer:
Yes, it is possible for an object to travel with a non-zero velocity when the net external force is zero. According to Newton’s first law, an object in motion continues to move with constant velocity unless acted upon by an unbalanced force. The conditions are that both the magnitude and direction of the velocity must remain constant throughout the motion.

Q2. When a carpet is beaten with a stick, dust comes out of it. Explain.

Answer:
When a carpet is beaten with a stick, the carpet suddenly moves but the dust particles tend to remain at rest due to their inertia. This creates relative motion between the carpet and dust particles, causing the dust to separate from the carpet fibers and fall out.

Q3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?

Answer:
When a bus suddenly starts, stops, or changes direction, the luggage on the roof tends to remain in its previous state of motion due to inertia. This can cause the luggage to slide off the roof. Tying it with a rope provides the necessary force to make the luggage move along with the bus, preventing it from falling.

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.

Answer:
(c) there is a force on the ball opposing the motion.
The ball stops due to friction between the ball and the ground, which acts opposite to the direction of 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 tonnes (Hint: 1 tonne = 1000 kg.)

Answer:
Given: u = 0, s = 400 m, t = 20 s, m = 7 tonnes = 7000 kg
Using s = ut + ½at²
400 = 0 + ½a(20)²
400 = 200a
a = 2 m/s²
Using F = ma
F = 7000 × 2 = 14,000 N
The acceleration is 2 m/s² and the force acting on the truck is 14,000 N.

Q6. 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?

Answer:
Given: m = 1 kg, u = 20 m/s, v = 0, s = 50 m
Using v² = u² + 2as
0² = 20² + 2a(50)
0 = 400 + 100a
a = -4 m/s²
Using F = ma
F = 1 × (-4) = -4 N
The magnitude of friction force is 4 N acting opposite to the direction of motion.

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 and
(b) the acceleration of the train.

Answer:
Given: Mass of engine = 8000 kg, Mass of each wagon = 2000 kg, Number of wagons = 5, Applied force = 40000 N, Friction force = 5000 N
Total mass = 8000 + (5 × 2000) = 18,000 kg
(a) Net accelerating force = Applied force – Friction force = 40000 – 5000 = 35,000 N
(b) Using F = ma: a = F/m = 35000/18000 = 1.94 m/s²

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 s⁻²?

Answer:
Given: m = 1500 kg, a = -1.7 m/s²
Using F = ma
F = 1500 × (-1.7) = -2550 N
The force between the vehicle and road must be 2550 N in the direction opposite to motion.

Q9. What is the momentum of an object of mass m, moving with a velocity v?
(a) (mv)²
(b) mv²
(c) ½ mv²
(d) mv

Answer:
(d) mv
Momentum is defined as the product of mass and velocity.

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?

Answer:
Since the cabinet moves with constant velocity, the acceleration is zero. Therefore, the net force is zero. This means the friction force must be equal and opposite to the applied force.
Friction force = 200 N (opposite to the direction of applied force).

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 s⁻¹ before the collision during which they stick together. What will be the velocity of the combined object after collision?

Answer:
Given: m₁ = m₂ = 1.5 kg, u₁ = +2.5 m/s, u₂ = -2.5 m/s
Using conservation of momentum:
m₁u₁ + m₂u₂ = (m₁ + m₂)v
(1.5)(2.5) + (1.5)(-2.5) = (1.5 + 1.5)v
3.75 – 3.75 = 3v
v = 0 m/s
The combined object will be at rest after the collision.

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.

Answer:
The student’s logic is incorrect. The action and reaction forces act on different objects, so they cannot cancel each other. When we push the truck, we apply a force on the truck (action) and the truck applies an equal and opposite force on us (reaction). The truck doesn’t move because the force we apply is not sufficient to overcome the static friction between the truck’s tires and the road surface.

Q13. A hockey ball of mass 200 g 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 magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

Answer:
Given: m = 200 g = 0.2 kg, u = 10 m/s, v = -5 m/s (opposite direction)
Initial momentum = mu = 0.2 × 10 = 2 kg⋅m/s
Final momentum = mv = 0.2 × (-5) = -1 kg⋅m/s
Change in momentum = mv – mu = -1 – 2 = -3 kg⋅m/s
Magnitude of change in momentum = 3 kg⋅m/s

Q14. A bullet of mass 10 g travelling horizontally with a velocity of 150 m/s 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:
Given:
Mass of bullet,

$m = 10$

g = 0.01 kg
Initial velocity,

$u = 150$

m/s
Final velocity,

$v = 0$

m/s
Time taken to stop,

$t = 0.03$

s
Distance penetrated,

$s = ?$

Force exerted by block on bullet,

$F = ?$

Formula:
Acceleration,

$a = \frac{v – u}{t}$

Using equation of motion,

$v^2 = u^2 + 2 a s$

Force,

$F = m \times a$

Substitution:

$a = \frac{0 – 150}{0.03} = -5000$

m/s²
Using

$v^2 = u^2 + 2 a s$

$0 = (150)^2 + 2 \times (-5000) \times s$

Final calculation:

$0 = 22500 – 10000 s$

$10000 s = 22500$

$s = \frac{22500}{10000} = 2.25$

m

Force:

$F = 0.01 \times (-5000) = -50$

N

Magnitude of force

$F = 50$

N

Distance of penetration is 2.25 meters and the magnitude of the force exerted by the wooden block on the bullet is 50 N.

Q15. 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.

Answer:
Given: m₁ = 1 kg, u₁ = 10 m/s, m₂ = 5 kg, u₂ = 0
Total momentum before impact = m₁u₁ + m₂u₂ = (1)(10) + (5)(0) = 10 kg⋅m/s
Using conservation of momentum: (m₁ + m₂)v = 10
(1 + 5)v = 10, v = 10/6 = 1.67 m/s
Total momentum after impact = (1 + 5)(1.67) = 10 kg⋅m/s
Velocity of combined object = 1.67 m/s

Q16. An object of mass 100 kg is accelerated uniformly from a velocity of 5 m/s to 8 m/s in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
Answer:
Given:
Mass,

$m = 100$

kg
Initial velocity,

$u = 5$

m/s
Final velocity,

$v = 8$

m/s
Time,

$t = 6$

s
Initial momentum,

$p_i = ?$

Final momentum,

$p_f = ?$

Force,

$F = ?$

Formula:
Momentum,

$p = m \times v$

Acceleration,

$a = \frac{v – u}{t}$

Force,

$F = m \times a$

Substitution:
Initial momentum,

$p_i = 100 \times 5 = 500$

kg·m/s
Final momentum,

$p_f = 100 \times 8 = 800$

kg·m/s
Acceleration,

$a = \frac{8 – 5}{6} = \frac{3}{6} = 0.5$

m/s²
Force,

$F = 100 \times 0.5 = 50$

N

Initial momentum = 500 kg·m/s, Final momentum = 800 kg·m/s, and magnitude of force exerted = 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.

Answer:
Rahul’s explanation is correct. According to Newton’s third law, both the motorcar and the insect experience equal and opposite forces during the collision. However, due to the vast difference in their masses, the insect experiences a much larger acceleration and change in velocity than the motorcar. The change in momentum is equal and opposite for both objects, but the insect suffers greater consequences due to its much smaller mass.

Q18. How much momentum will a dumb-bell 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:
Given: m = 10 kg, h = 80 cm = 0.8 m, g = 10 m/s²
Using v² = u² + 2gh, where u = 0
v² = 0 + 2(10)(0.8) = 16
v = 4 m/s
Momentum transferred = mv = 10 × 4 = 40 kg⋅m/s

A1. The following is the distance-time table of an object in motion:

(a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?
(b) What do you infer about the forces acting on the object?

Answer:
(a) Looking at the distances, we see they follow the pattern: 1³, 2³, 3³, 4³, 5³, 6³, 7³. This means distance = t³. Taking derivatives: velocity = 3t², acceleration = 6t. The acceleration is increasing with time.
(b) Since acceleration is increasing, the net force on the object is increasing with time (F = ma, and mass is constant).

A2. Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 m s⁻². With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.)

Answer:
Given: m = 1200 kg, acceleration with 3 persons = 0.2 m/s²
When 2 persons push at uniform velocity, net force = 0
Force by 2 persons = Friction force
When 3 persons push with acceleration 0.2 m/s²:
Net force = ma = 1200 × 0.2 = 240 N
Force by 3 persons = Friction force + 240 N
Let F = force by each person
2F = Friction force and 3F = Friction force + 240
Therefore: 3F = 2F + 240, F = 240 N
Each person pushes with a force of 240 N.

A3. A hammer of mass 500 g, moving at 50 m s⁻¹, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?

Answer:
Given: m = 500 g = 0.5 kg, u = 50 m/s, v = 0, t = 0.01 s
Using a = (v – u)/t = (0 – 50)/0.01 = -5000 m/s²
Using F = ma = 0.5 × (-5000) = -2500 N
The force of the nail on the hammer is 2500 N opposite to the direction of motion.

A4. A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.
Answer:
Given:
Mass,

$m = 1200$

kg
Initial velocity,

$u = 90$

km/h =

$\frac{90 \times 1000}{3600} = 25$

m/s
Final velocity,

$v = 18$

km/h =

$\frac{18 \times 1000}{3600} = 5$

m/s
Time,

$t = 4$

s
Acceleration,

$a = ?$

Change in momentum,

$\Delta p = ?$

Force,

$F = ?$

Formula:
Acceleration,

$a = \frac{v – u}{t}$

Change in momentum,

$\Delta p = m (v – u)$

Force,

$F = m \times a$

Substitution:

$a = \frac{5 – 25}{4} = \frac{-20}{4} = -5$

m/s²

$\Delta p = 1200 \times (5 – 25) = 1200 \times (-20) = -24000$

kg·m/s

$F = 1200 \times (-5) = -6000$

N

Magnitude of acceleration = 5 m/s²
Magnitude of change in momentum = 24000 kg·m/s
Magnitude of force required = 6000 N

Acceleration = -5 m/s², Change in momentum = -24000 kg·m/s, and Force required = 6000 N (negative sign indicates deceleration).