How Forces Affect Motion NCERT Solutions Class 9 Science Chapter 6

Step-by-step answers by SciNCERT Decoder by Kuldeep S. Take g = 9.8 m/s².

These How Forces Affect Motion NCERT Solutions give clear answers to every in-text question, activity, example and the Revise, Reflect, Refine questions of Class 9 Science Chapter 6. Each answer is short and shows the steps, so you can write it in the exam. To learn the theory first, read our How Forces Affect Motion Class 9 Notes. You can also open your NCERT Science textbook or find more chapters on our Class 9 study material page.

How Forces Affect Motion NCERT Solutions: Quick Answer Key

Use this table to check your numerical answers fast. The full steps are given below.

QuestionAnswer
Example 6.4 (weightlifter)294 N upward
Example 6.5 (block)(i) 0 m, (ii) 0.4 m
Example 6.6 (sports car)3000 N east, 0, 3000 N west
Example 6.8 (gun and bullet)Gun 0.4 m/s², bullet 20 m/s²
Revise Q4 (snake boat)18,000 N forward
Revise Q6 (position–time graphs)Object C
Revise Q11 (velocity–time graph)25 N
Revise Q12 (bullet)500 N
Revise Q13 (football)0.015 s
Revise Q14 (rough patch)10 m
Revise Q15 (tractor)a = a₁a₂ ÷ (a₁ + a₂)
Think It Over
Think It Over 1

Why does a canoe move forward when the canoeist pushes the water backwards with the paddle? Why does it move faster when the push is harder?

The paddle pushes the water backwards. By Newton’s third law, the water pushes the paddle and the canoe forwards with an equal force. A harder push on the water gives a bigger forward force. A bigger net force gives a bigger acceleration, so the canoe speeds up more.
Think It Over 2

The same canoeist uses the same paddle force in an empty canoe and in a canoe with a passenger. Which moves faster?

The empty canoe. The force is the same, but a = F ÷ m. The canoe with a passenger has more mass, so its acceleration is smaller.
Activities: What You Should Observe

The activities in these How Forces Affect Motion NCERT Solutions show what you should see and why.

ActivityWhat you observeConclusion
6.1 Coin stack on different surfacesThe stack slides farthest on polished marble and the shortest on the wooden tableFriction is smaller on smoother surfaces, so the velocity decreases more slowly
6.2 Spring balance and a blockThe reading is smallest on the surface where the coins went farthestThe reading gives an approximate measure of friction
6.3 Cart and cup (double the cup’s mass)The cart takes less time to cross the tableFor the same mass, acceleration increases as force increases (a₁ ÷ a₂ = T₂² ÷ T₁²)
6.4 Cart and cup (double the cart’s mass)The cart takes more timeFor the same force, acceleration decreases as mass increases
6.5 Wheeled chair and tablePushing the table away rolls you back. Pulling it towards you rolls you forwardThe table pushes you with a force equal and opposite to yours
6.6 Two spring balances hooked togetherBoth show the same reading every timeThe two forces are equal in magnitude
6.7 Balloon on a threadAir rushes out backwards and the balloon moves forward along the threadThe air pushes the balloon opposite to its own motion (third law, like a rocket)
Pause and Ponder
Pause and Ponder 1

List two forces acting on a barbell lifted by a weightlifter. Are they balanced when she holds it steady?

(1) The weight of the barbell, acting downwards. (2) The upward force applied by the weightlifter. When the barbell is held steady, the two forces are equal and opposite, so the net force is zero and they are balanced.
Pause and Ponder 2

In an arm-wrestling match, are the forces of the two players balanced at the instant the arms tilt? If not, who exerts the larger force?

No. If the joined arms are tilting, they are accelerating, so there is a net force and the forces are unbalanced. The player who pushes the arms in the direction in which they move exerts the larger force and is winning.
Pause and Ponder 3

An object moves with constant velocity. Is a net force acting on it?

No. Constant velocity means acceleration a = 0. By F = ma, the net force is zero. (This is Newton’s first law.)
Pause and Ponder 4

No net force acts on an object. Which are possible: (i) it stays at rest, (ii) it keeps moving with constant velocity, (iii) it moves with constant acceleration?

With no net force, a = 0. So (i) and (ii) are possible. (iii) is not possible, because constant acceleration needs a net force.
Pause and Ponder 5

Real objects always have forces on them. Show with an example that the net force can still be zero.

A book rests on a table. Its weight acts downwards and the normal force of the table acts upwards. They are equal and opposite, so the net force is zero. The book behaves as if no force acts on it and stays at rest.
Pause and Ponder 6

A toy car of 100 g moves with a constant velocity of 0.5 m/s. What is the net force on it?

The velocity is constant, so a = 0. m = 100 g = 0.1 kg. F = ma = 0.1 × 0 = 0 N.
Pause and Ponder 7

Two children of different masses sit on identical swings. For the same initial acceleration, which child needs the larger force?

The heavier child. F = ma, so for the same a, the force is proportional to the mass.
Pause and Ponder 8

How does bubble wrap or hay protect glass items during transport?

They make the glass take a longer time to stop in a bump. Since F = change in momentum ÷ time, a longer time gives a smaller force, so the glass does not break.
Pause and Ponder 9

Why does a firefighter sometimes struggle to hold a hose pipe?

The hose throws out water at high speed. By the third law, the water pushes the hose backwards with an equal force. This backward force on the firefighter makes the hose hard to hold.
Pause and Ponder 10

A spacecraft is in a region of space where gravity is negligible. How can it change its velocity?

By firing its engines (thrusters) to throw out gas in one direction. The gas pushes the spacecraft in the opposite direction with an equal force, so its velocity changes.
Textbook Examples 6.1 to 6.8

Here are the solutions of the examples in the How Forces Affect Motion NCERT Solutions, written in simple steps.

Example 6.1

Forces of 10 N and 6 N act on a block in three ways. Find the net force in each case.

(a) Both forces in the same direction: 10 + 6 = 16 N, to the right.
(b) 10 N to the right, 6 N to the left: 10 − 6 = 4 N, to the right.
(c) 6 N to the right, 10 N to the left: 10 − 6 = 4 N, to the left.
Example 6.2

A person pushes a moving box forward with a force equal to the friction on it. Will the box keep moving or stop?

Friction acts backwards and the push acts forwards. They are equal and opposite, so the net force is zero. By the first law, the box keeps moving with constant velocity.
Example 6.3

Draw the position–time and velocity–time graphs of an object with no net force on it.

There are two cases.
Object at rest: the position–time graph is a horizontal line (position does not change). The velocity–time graph is a horizontal line on the time axis (velocity = 0).
Object moving with constant velocity: the position–time graph is a straight, sloping line. The velocity–time graph is a horizontal line above the time axis.
Example 6.4

A weightlifter holds a barbell with 10 kg fixed on each side of a 10 kg bar. What force does she apply to keep it steady?

Total mass = 10 + 10 + 10 = 30 kg. Weight = mg = 30 × 9.8 = 294 N downwards. To keep it steady she must apply an equal force upwards: 294 N.
Example 6.5

A 25 kg block rests on a floor. The maximum friction is 50 N. Find the displacement in 2 s if it is pushed with (i) 50 N, (ii) 55 N.

(i) The push equals the friction, so the forces are balanced. The block stays at rest: displacement = 0.
(ii) Net force = 55 − 50 = 5 N. a = 5 ÷ 25 = 0.2 m/s². s = ut + ½at² = 0 + ½ × 0.2 × 2² = 0.4 m forward.
Example 6.6

A 1500 kg sports car moves east with the velocity–time graph shown. Find the force in 0–5 s, 5–10 s and 10–15 s.

Velocity-time graph of the sports car10051015Time (s)Velocity (m/s)0–5 s: F = 3000 N5–10 s: F = 010–15 s: F = −3000 N
Velocity–time graph of the sports car.
0–5 s: the graph is a sloping line. a = (10 − 0) ÷ 5 = 2 m/s². F = 1500 × 2 = 3000 N towards the east.
5–10 s: the graph is horizontal, so the velocity is constant. a = 0, so F = 0.
10–15 s: a = (0 − 10) ÷ 5 = −2 m/s². F = 1500 × (−2) = −3000 N, that is 3000 N towards the west (opposite to the motion).
Example 6.7

The Earth and a fruit pull each other with equal forces. Why does only the fruit seem to move?

The forces are equal. But a = F ÷ m, and the Earth’s mass is huge compared with the fruit’s. So the Earth’s acceleration is extremely small and cannot be noticed.
Example 6.8

A 0.1 kg bullet is fired from a 5 kg gun with a force of 2 N. Find the initial accelerations of the bullet and the gun.

By the third law, the recoil force on the gun is also 2 N.
Gun: a = 2 ÷ 5 = 0.4 m/s².
Bullet: a = 2 ÷ 0.1 = 20 m/s².
The forces are equal, but the accelerations differ because the masses differ.
Revise, Reflect, Refine

These are the end-of-chapter questions of the How Forces Affect Motion NCERT Solutions, with full steps.

Q1

A table is moved across the floor at constant velocity by a horizontal force F. What is the friction from the floor?

Constant velocity means zero net force, so friction is equal and opposite to the applied force. Friction = F, acting backwards.
Q2

Choose the right word for a ball on a frictionless surface.

(i) With no net force, the velocity stays the same.
(ii) With a net force along the motion, the speed increases.
(iii) With a net force opposite to the motion, the speed decreases.
Q3

Block P has forces of 5 N and 4 N in opposite directions. Block Q moves with constant velocity. Which statement is correct?

Net force on P = 5 − 4 = 1 N, so P has a net force. Q has constant velocity, so its net force is zero. The answer is (i): P experiences a net force and Q does not.
Q4

In a snake boat, 95 oarsmen row forward and 5 row the wrong way. Each applies 200 N. Find the net force.

Forward force = 95 × 200 = 19,000 N. Backward force = 5 × 200 = 1000 N. Net force = 19,000 − 1000 = 18,000 N forward.
Q5

When a net force acts on an object, it accelerates…

The acceleration is in the direction of the force and proportional to it. The answer is (iv).
Q6

Position–time graphs of four objects A, B, C and D are given. On which does a net force act?

How Forces Affect Motion NCERT Solutions: position-time graphs of objects A, B, C and D
Position–time graphs of A, B, C and D.
A net force acts only when the velocity changes. A and D are straight lines (constant velocity), and B is horizontal (at rest). Only C is a curve, so its velocity changes. The answer is (iii) Object C.
Q7

A sailor jumps from a small boat to the shore. Does the boat move?

Yes, backwards. The sailor pushes the boat backwards. By the third law, the boat pushes the sailor forwards with an equal force, and the sailor jumps ahead while the boat moves back.
Q8

Why is a landing mat or sand bed used in high jump?

It makes the athlete take a longer time to stop. F = change in momentum ÷ time, so a longer time means a smaller force on the body and no injury.
Q9

A loaded hand cart hits an identical empty hand cart. During the collision…

By the third law, both carts exert equal forces on each other. The answer is (iv).
Q10

Plot the force–mass graph from the acceleration–mass graph given in the book.

Force-mass graph from the acceleration-mass graphAcceleration–mass graph (given)12345105Mass (kg)a (m/s²)Force–mass graph (answer)1234510Mass (kg)F (N)F = 10 N (constant)
From the acceleration–mass graph to the force–mass graph.
Read two points: m = 1 kg gives a = 10 m/s², and m = 5 kg gives a = 2 m/s². F = ma = 1 × 10 = 10 N, and F = 5 × 2 = 10 N. The force is the same for every mass, so the force–mass graph is a horizontal straight line at F = 10 N.
Q11

Find the force on a 10 kg object from its velocity–time graph.

Velocity-time graph of a 10 kg object102030480Time (s)Velocity (m/s)slope = a = 20 ÷ 8 = 2.5 m/s²F = ma = 10 × 2.5 = 25 N
Velocity–time graph of the 10 kg object.
From the graph, the velocity rises from 10 m/s at t = 0 to 30 m/s at t = 8 s. a = slope = (30 − 10) ÷ 8 = 2.5 m/s². F = ma = 10 × 2.5 = 25 N.
Q12

A 50 g bullet moving at 100 m/s stops after going 50 cm into a wooden block. Find the stopping force.

m = 0.05 kg, u = 100 m/s, v = 0, s = 0.5 m.
v² = u² + 2as gives 0 = 10,000 + 2 × a × 0.5, so a = −10,000 m/s².
F = ma = 0.05 × (−10,000) = −500 N. The stopping force is 500 N, opposite to the motion.
Q13

A footballer kicks a 0.4 kg ball to 108 km/h with a force of 800 N. Find the contact time.

v = 108 km/h = 108 ÷ 3.6 = 30 m/s, u = 0. F = m(v − u) ÷ t, so t = m(v − u) ÷ F = (0.4 × 30) ÷ 800 = 12 ÷ 800 = 0.015 s.
Q14

A 2 kg object moving at 10 m/s meets a rough patch with friction 7 N, and an extra 3 N force also opposes the motion. How far does it go before it stops?

Total opposing force = 7 + 3 = 10 N. a = −10 ÷ 2 = −5 m/s². From v² = u² + 2as: 0 = 100 + 2 × (−5) × s, so s = 10 m.
Q15

A tractor with force F gives acceleration a₁ to a harrow (m₁) and a₂ to a trolley (m₂). Find the acceleration if the harrow is placed on the trolley.

m₁ = F ÷ a₁ and m₂ = F ÷ a₂. Total mass = F(1 ÷ a₁ + 1 ÷ a₂).
a = F ÷ (m₁ + m₂) = 1 ÷ (1 ÷ a₁ + 1 ÷ a₂) = a₁a₂ ÷ (a₁ + a₂).
Q16

A bar magnet and a compass needle exert equal forces on each other. Why does only the needle move?

The forces are equal, but a = F ÷ m. The needle is very light, so it gets a large acceleration and moves. The bar magnet is much heavier (and is usually held fixed), so its acceleration is too small to notice.

Summary

You have now seen all the How Forces Affect Motion NCERT Solutions for Class 9 Science Chapter 6. Remember three ideas. Only a net force changes motion. F = ma links force, mass and acceleration. Action and reaction act on different objects. To revise the theory with diagrams, examples and MCQs, open our How Forces Affect Motion Class 9 Notes, and then continue with Work, Energy and Simple Machines.

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