Class 9 Science · Chapter 6

6. How Forces Affect Motion

Complete solutions to Revise, Reflect, Refine. CBSE Class 9 Science (NCERT Exploration).

Revise, Reflect, Refine

1Using a horizontal force F, a table is moved across the floor at a constant velocity. How much is the frictional force exerted by the floor on the table?

The table moves with constant velocity, so its acceleration is zero. Hence the net force on it is zero. The friction must balance the applied force.

Frictional force = F, in the direction opposite to the applied force

2For a ball moving on a smooth frictionless surface, choose the appropriate option.

(i) If no net force is applied on the ball, the velocity of the ball will remain the same.

(ii) If a net force is applied in the direction of its motion, the magnitude of the velocity will increase.

(iii) If a net force is applied opposite to the direction of its motion, the magnitude of the velocity will decrease.

3Forces of 5 N and 4 N act in opposite directions on block P, while block Q moves with a constant velocity (Fig. 6.36). Which statement is correct?

Answer: (i) P experiences a net force and Q does not experience a net force.

On P, net force = 5 N − 4 N = 1 N. Q moves with constant velocity, so its acceleration is zero and the net force on it is zero.

4In a snake boat race, 95 oarsmen row backwards to propel the boat forward, but 5 oarsmen row in the opposite direction. If each applies a horizontal force of 200 N, what is the net force on the boat?

Forward force = 95 × 200 = 19,000 N
Backward force = 5 × 200 = 1,000 N
Net force = 19,000 − 1,000 = 18,000 N in the forward direction

5When a net force acts on an object, we observe that the object accelerates:

Answer: (iv) in the direction of force, with acceleration proportional to the force acting on the object.

This is Newton’s second law of motion, F = ma.

6The position-time graphs for four objects A, B, C and D are given in Fig. 6.37. A net force acts on:

Answer: (iii) Object C

A and D have straight-line graphs, so they move with constant velocity. B is at rest. Only C has a curved graph, so its velocity is changing. It is accelerating, so a net force acts on it.

7A sailor jumps out from a small boat to the shore (Fig. 6.38). As the sailor jumps forward, will the boat move? If yes, in which direction and why?

Yes, the boat moves backward (away from the shore).

To jump forward, the sailor pushes the boat backward with his feet. By Newton’s third law, the boat pushes the sailor forward with an equal and opposite force. So the sailor moves forward and the boat moves backward.

8During a high jump event, a landing mat or sand bed is placed for the athlete to fall upon (Fig. 6.39). Explain the reason behind it.

The athlete comes down with a high velocity and must be brought to rest. A soft mat or sand bed sinks in and increases the time taken to stop. So the deceleration becomes smaller. By F = ma, the force on the body also becomes smaller. This protects the athlete from injury.

9A hand cart loaded with vegetables collides with an identical but empty hand cart. During the collision:

Answer: (iv) the loaded cart and the empty cart both exert an equal magnitude of force on each other.

By Newton’s third law, action and reaction are always equal and opposite. The empty cart has less mass, so it gets a larger acceleration.

10The acceleration-mass graph for the acceleration produced by a force on objects of different masses is plotted in Fig. 6.40. Plot the force-mass graph for this case.

From the graph: (1 kg, 10 m s⁻²), (2 kg, 5 m s⁻²), (4 kg, 2.5 m s⁻²), (5 kg, 2 m s⁻²).

F = m × a = 1 × 10 = 2 × 5 = 4 × 2.5 = 5 × 2 = 10 N

The force is the same for all masses. So the force-mass graph is a straight line parallel to the mass axis at F = 10 N.

Force-mass graph: a horizontal line at 10 N.

11The velocity-time graph of an object of mass 10 kg moving along a straight line is shown in Fig. 6.41. Calculate the force acting on the object.

From the graph, velocity changes from 10 m s⁻¹ at 0 s to 30 m s⁻¹ at 8 s.

a = slope = (30 − 10)/8 = 2.5 m s⁻²
F = ma = 10 × 2.5 = 25 N

12A bullet of mass 50 g moving with 100 m s⁻¹ enters a heavy wooden block and stops after penetrating 50 cm. Estimate the stopping force on the bullet.

m = 50 g = 0.05 kg, u = 100 m s⁻¹, v = 0, s = 0.5 m
v² = u² + 2as ⇒ 0 = 10000 + 2 × a × 0.5 ⇒ a = −10000 m s⁻²
F = ma = 0.05 × 10000 = 500 N (opposite to the motion of the bullet)

13A footballer kicked a football with a speed of 108 km h⁻¹ using a force of 800 N. The mass of the football was 0.4 kg. Calculate the time of contact between foot and ball.

v = 108 km h⁻¹ = 108 × 5/18 = 30 m s⁻¹, u = 0
a = F/m = 800/0.4 = 2000 m s⁻²
t = (v − u)/a = 30/2000 = 0.015 s

14An object of mass 2 kg moving at 10 m s⁻¹ enters a rough patch where friction is 7 N. At the same time, an extra force of 3 N opposing the motion is applied. How much distance does it travel before coming to rest?

Total opposing force = 7 + 3 = 10 N
a = −10/2 = −5 m s⁻²
v² = u² + 2as ⇒ 0 = 100 − 2 × 5 × s
s = 100/10 = 10 m

15A tractor pulls a harrow of mass m₁ with force F producing acceleration a₁, and a trolley of mass m₂ with the same force producing acceleration a₂. If it pulls the trolley with the harrow placed on it (same force F), find the acceleration in terms of a₁ and a₂.

m₁ = F/a₁ and m₂ = F/a₂
Total mass = m₁ + m₂ = F/a₁ + F/a₂ = F(a₁ + a₂)/(a₁a₂)
a = F/(m₁ + m₂) = a₁a₂ / (a₁ + a₂)

16A bar magnet and a compass needle exert equal and opposite magnetic forces on each other. However, the compass needle moves, whereas the bar magnet does not (Fig. 6.42). Explain why.

The forces are equal, but the masses are very different. The compass needle is very light and turns freely on a pivot. So the force gives it a large acceleration (a = F/m) and it moves.

The bar magnet is much heavier. It is also held by the hand or by friction with the table. So the same small force produces almost no acceleration, and it does not move.

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