Ch 6: How Forces Affect Motion Class 9 Question Answer

Welcome to “Ch 6: How Forces Affect Motion Class 9 Question Answer? “

This comprehensive guide provides accurate, step-by-step solutions for all textbook exercises.

Fully aligned with the latest CBSE syllabus and NCERT curriculum, these answers are designed to clear your physics concepts and help you score higher in your school exams.

We have also provided the source of almost all the answers in the official NCERT textbook(Exploration).

Before attempting the chapter-end questions of Revise, Reflect, Refine, learn deeply our short notes on chapter 6, How Forces Affect Motion

Q1

Using 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?

Answer:

The frictional force = F (equal in magnitude to applied force, opposite in direction).

The table moves at constant velocity. By Newton’s 1st Law, constant velocity means zero net force.

Two horizontal forces act on the table — applied force F (forward) and friction (backwards). For the net force to be zero, these must be exactly equal and opposite. Therefore, friction = F.

Constant velocity always means zero net force.

Q2

For a ball moving on a smooth frictionless surface, choose the appropriate option that will make the following statements physically correct.

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

(ii) If a net force is applied on the ball in the direction of its motion,
The magnitude of the velocity of the ball will remain the same/ increase/decrease.

(iii) If a net force is applied on the ball in a direction opposite to the
direction of its motion, the magnitude of the velocity of the ball will remain the same/increase/decrease.

Answers:

Q3

Two blocks P and Q on a smooth horizontal surface are shown in Fig. 6.36a and Fig. 6.36b. Two forces of magnitudes 4 N and 5 N are acting in opposite directions on block P, while block Q is moving with a constant velocity.

Which of the following statements is correct?

(i) P experiences a net force, and Q does not experience a net force.
(ii) P does not experience a net force, and Q experiences a net force.
(iii) Both P and Q experience a net force.
(iv) Neither P nor Q experiences a net force.

Answer:

Option (i) — P experiences a net force; Q does not.

Q4

While practising for the snake boat race (Vallum kalli in Kerala),100 oarsmen are rowing a boat together. Out of these, 95 row backwards to propel the boat forward. But by mistake, 5 oarsmen row in the opposite direction. If each oarsman applies a horizontal force of 200 N, what is the net force on the snake boat? (Ignore drag forces, air friction, etc.)

Answer:

Step-by-step calculation:

1. Forward force (95 oarsmen): 95 × 200 = 19,000 N
2. Backward force (5 oarsmen): 5 × 200 = 1,000 N

Net Force = 19,000 − 1,000 = 18,000 N in the forward direction. Ans

Method: “When forces act in opposite directions, net force = difference between them.”

Q5

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

(i) opposite to the direction of force, with acceleration proportional
to the force acting on the object.
(ii) opposite to the direction of force, with acceleration proportional
to the mass of the object.
(iii) in the direction of force, with acceleration inversely proportional
to the force acting on the object.
(iv) in the direction of force, with acceleration proportional to the
force acting on the object.

Answers:

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

Newton’s 2nd Law: “The magnitude of the acceleration is proportional to the magnitude of the net force.”

Q6

The position-time graph for four objects A, B, C and D moving along a straight line are given in Fig. 6.37.

A net force acts on:

(i) Object A
(ii) Object B
(iii) Object C
(iv) Object D

Answers:

Answer: Net force acts on Object C and Object D.

The key principle: if net force = 0, the position-time graph is either flat (at rest) or a straight inclined line (constant velocity).

Quick rule:

1. Straight line (flat or inclined) on position-time graph → No net force
2. Curve (upward or downward) on the position-time graph → Net force is present

Q7

A 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?

Answers:

Answer: Yes — the boat moves in the direction opposite to the sailor’s jump (away from shore).

Newton’s 3rd Law: when the sailor pushes off the boat to jump forward, the boat simultaneously receives an equal and opposite force, pushing it backwards.

Flow:

Sailor pushes boat backwards → Boat pushes sailor forward (equal force) → Sailor goes to shore, boat moves away

The two forces are equal in magnitude, opposite in direction, but act on two different objects (sailor and boat) — so they do not cancel each other.

This works on the same principle as rowing a canoe: “When the canoeist pushes water backwards, water pushes the paddle forward with an equal force.”

Q8

During 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.

Answers:

A soft landing mat increases the time of impact, which reduces the force on the athlete and lowers the risk of injury.

Explanation:
By Newton’s 2nd Law (F = ma), to bring the athlete from high velocity to rest, if this happens over a longer time, the deceleration (acceleration) is smaller, so the force is smaller.

The same principle applies to:

1. A cricket fielder pulling hands back while catching — “the time duration is increased during which the high velocity reduces to zero.”
2. Airbags in vehicles — the passenger’s head pushes into the soft bag over a longer time, reducing force.

Q9

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

(i) The loaded cart exerts a force of larger magnitude on the empty cart.
(ii) the empty cart exerts a force of larger magnitude on the loaded cart.
(iii) neither cart exerts a force on the other.
(iv) the loaded cart and the empty cart, both exert an equal magnitude
of force on each other.

Answers:

Option (iv) — Both carts exert equal magnitude forces on each other.

Newton’s 3rd Law states: “Whenever one object is exerting a force on a second object, the second object is simultaneously exerting an equal and opposite force on the first object.”

This applies to all interactions regardless of mass or speed.

Important distinction:

“Even though the forces acting on the two interacting objects are always equal in magnitude, they do not, in general, produce equal acceleration. This is because the masses of the objects may be different.”

Q10

The 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.

Answers:

The force-mass graph is a horizontal straight line — force is constant.

Using F = ma, we can calculate F for each point on the graph (Fig. 6.40):

Force = 10 N at every mass value — it is constant throughout.

The force-mass graph is therefore a horizontal straight line at F = 10 N, parallel to the mass axis.

Q11

The 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 by using the graph

Answers:

Step 1 — Find acceleration:

a = (v − u) / t = (30 − 0) / 8 = 3.75 m/s²

Step 2 — Apply Newton’s 2nd Law:

F = ma = 10 × 3.75 = 37.5 N

The force acting on the object = 37.5 N in the direction of motion.

A straight line on a velocity-time graph means constant acceleration — use a = (v−u)/t, then F = ma.

Q12

A bullet of mass 50 g moving with a speed of 100 m s–1 enters a heavy, stationary wooden block and stops after penetrating a distance of 50 cm. Estimate the stopping force acting on the bullet (assume that the bullet undergoes constant acceleration within the block).

Answer:

Given: m = 0.05 kg; u = 100 m/s; v = 0 m/s; s = 0.5 m

Step 1 — Find deceleration using v² = u² + 2as:

0 = (100)² + 2 × a × 0.5

0 = 10,000 + a

a = −10,000 m/s²

Step 2 — Apply Newton’s 2nd Law:

F = ma = 0.05 × 10,000 = 500 N

The stopping force = 500 N (acting opposite to the bullet’s direction of motion).

Q13

An ace footballer converted a penalty shot by kicking the football with a speed of 108 km h–1. The estimated force they imparted was 800 N. The mass of the football was 0.4 kg. Calculate the time of contact between their foot and the ball.

Answer:

Step 1 — Convert speed:

108 km/h = 108 × (1000/3600) = 30 m/s

Step 2 — Find acceleration:

a = F/m = 800 / 0.4 = 2,000 m/s²

Step 3 — Find time (ball starts from rest, u = 0):

v = u + at → 30 = 0 + 2000 × t → t = 30/2000 = 0.015 s

Time of contact = 0.015 seconds (15 milliseconds) — extremely brief!

Q14

An object of mass 2 kg moving with a constant velocity of 10 m s–1 encounters a rough patch where the force of friction on the object is 7 N. At the same time, an additional constant force of 3 N opposing the motion is applied on the object. After entering the rough patch, how much distance does the object travel before coming to rest?

Answer:

Given: m = 2 kg; u = 10 m/s; v = 0; friction = 7 N; additional force = 3 N (both opposing)

Step 1 — Total opposing force:

7 + 3 = 10 N

Step 2 — Deceleration:

a = F/m = 10/2 = 5 m/s² (deceleration)

Step 3 — Distance using v² = u² + 2as:

0 = (10)² + 2 × (−5) × s

0 = 100 − 10s

s = 10 metres

The object travels 10 metres before stopping.

Q15

A tractor pulls a harrow (a ploughing tool) of mass m1 with a net force F resulting in an acceleration of a1. The same tractor pulls a trolley of mass m2 with a force F producing an acceleration of a2. If the tractor now pulls the trolley with the harrow placed on it (with the same force F ), then obtain an expression for the resulting acceleration in terms of a1 and a2. Ignore friction.

Answers:

Step 1 — Express masses using F = ma:

m₁ = F/a₁ and m₂ = F/a₂

Step 2 — Total mass:

m₁ + m₂ = F/a₁ + F/a₂ = F(a₁ + a₂) / (a₁ × a₂)

Step 3 — New acceleration:

a = F / (m₁ + m₂) = F / [F(a₁ + a₂)/(a₁ × a₂)]

a = (a₁ × a₂) / (a₁ + a₂)

The resulting acceleration = (a₁ × a₂) / (a₁ + a₂)

Note: the combined acceleration is always less than either a₁ or a₂ alone — the same force now moves a greater combined mass.

Q16

When the pole of a bar magnet is brought close to a magnetic compass, the bar magnet and the compass needle (which is also a magnet) exert a magnetic force on each other. As per Newton’s third law of motion, both the forces are equal in magnitude and opposite in direction. However, the compass needle moves, whereas the bar magnet does not move (Fig. 6.42). Explain why.

Answers:

The forces are equal — but the masses are very different. The compass needle has far less mass, so it accelerates much more.

Newton’s 2nd Law: a = F/m. Same force, different masses → different accelerations.

This is the same reasoning as why the Earth does not visibly move toward a falling fruit: “The mass of the Earth is so large that the acceleration caused by the force is extremely small… too small to be noticed.”

Page 104, Eq. 6.1
Page 110, Example 6.7

End of solutions: Revise, Reflect, Refine Chapter 6 “How Forces Affect Motion”

Common Mistakes to Avoid

All Important Formulas

FAQs Ch 6: How Forces Affect Motion Class 9 Question Answer

Quick formula:

1. Forces in same direction → Net force = F₁ + F₂
2. Forces in opposite directions → Net force = F₁ − F₂ (direction follows the larger)

The underlying formula in all cases:

F = m × (v − u) / t

Keep m and (v − u) fixed. Increase t. Force F drops.

The critical rule to remember:

Same object + equal and opposite forces = balanced forces (net force zero, no acceleration).

Different objects + equal and opposite forces = Newton’s 3rd Law pair (each object accelerates according to its own mass).