Important Questions and Answers for Class 9 Science Exploration Chapter 6 – How Forces Affect Motion (2026–27)

These important questions of Chapter 6 Class 9 Science cover the major concepts from How Forces Affect Motion. Each question helps students practise a different skill, from identifying the direction of net force to applying Newton's laws in unfamiliar situations.

Multiple Choice and Assertion–Reason Questions on Force and Motion

Q1. Two students push a stationary trolley in opposite directions with forces of 45 N and 30 N. What will be the net force on the trolley?

(a) 75 N towards the larger force
(b) 15 N towards the larger force
(c) 15 N towards the smaller force
(d) Zero

Answer: (b) 15 N towards the larger force.

Explanation:

When forces act in opposite directions, their magnitudes are subtracted.

Net force = 45 − 30 = 15 N

The net force acts in the direction of the 45 N force.

Important Point: A non-zero net force produces acceleration. It does not necessarily mean that an object immediately moves in the direction of that force.

Q2. A hockey puck moves along a straight, frictionless surface at a constant velocity. What can be concluded about the net force acting on it?

(a) A constant forward force is necessary.
(b) The net force is zero.
(c) A backward force must be acting.
(d) The net force increases with time.

Answer: (b) The net force is zero.

Explanation: According to Newton's first law, an object continues moving with constant velocity when the net force acting on it is zero.

A force is required to change velocity, not to maintain constant velocity.

Q3. Assertion–Reason

Assertion (A): A heavier shopping trolley generally requires a greater net force than a lighter trolley to produce the same acceleration.

Reason (R): According to Newton's second law, the net force required for a given acceleration is directly proportional to mass.

(a) Both A and R are true, and R correctly explains A.
(b) Both A and R are true, but R does not explain A.
(c) A is true, but R is false.
(d) A is false, but R is true.

Answer: (a) Both A and R are true, and R correctly explains A.

Explanation:

Newton's second law gives:

F = ma

For the same acceleration, a larger mass requires a larger net force.

Very Short Answer Questions on Forces, Friction and Newton's Laws

Q4. Why is force called a vector quantity? How is its magnitude measured?

Answer:

Force is a vector quantity because it has both magnitude and direction.

• Magnitude: Indicates the strength of the force.

• Direction: Indicates where the force acts.

• SI unit: newton (N).

• Measuring instrument: Spring balance.

For example, a force of 10 N towards the right is different from a force of 10 N towards the left.

Q5. A weightlifter holds a barbell stationary above their head. Are the forces acting on the barbell balanced? Explain.

Answer:

Yes. The forces are balanced because the barbell remains stationary.

Two main forces act on the barbell:

1. The gravitational force acts vertically downwards.

2. The upward force applied by the weightlifter supports the barbell.

The two forces are equal in magnitude and opposite in direction.

Therefore, the net force is zero.

Important Point: Balanced forces can act on an object even when large forces are involved.

Q6. Why does a bicycle slow down when the rider stops pedalling? Would the same thing happen on a perfectly frictionless horizontal surface?

Answer:

A bicycle slows down because friction and air resistance oppose its motion.

When the rider stops pedalling, these opposing forces reduce the bicycle's speed.

On a perfectly frictionless horizontal surface, with no air resistance or other opposing force, the bicycle would continue moving with constant velocity.

This follows Newton's first law of motion.

Numerical Questions on Newton's Second Law and Acceleration

Q7. A box of mass 12 kg is pushed with a horizontal force of 50 N. Friction opposes its motion with a force of 14 N. Calculate its acceleration.

Solution:

Given:

Mass = 12 kg
Applied force = 50 N
Frictional force = 14 N

Step 1: Calculate the net force.

Net force = 50 − 14 = 36 N

Step 2: Apply Newton's second law.

$F=ma$

$a=\dfrac{F}{m}$

$a=\dfrac{36}{12}=3,\text{m/s}^2$

Answer: The box accelerates at 3 m/s² in the direction of the applied force.

Q8. A student pushes a 20 kg trolley from rest. The net horizontal force is 40 N. Calculate its acceleration and the distance travelled in 4 seconds, assuming constant acceleration.

Solution:

Given:

Mass = 20 kg
Net force = 40 N
Initial velocity = 0 m/s
Time = 4 s

Step 1: Calculate acceleration.

$a=\dfrac{F}{m}$

$a=\dfrac{40}{20}=2,\text{m/s}^2$

Step 2: Calculate distance.

$s=ut+\dfrac{1}{2}at^2$

$s=0+\dfrac{1}{2}\times2\times4^2$

$s=16,\text{m}$

Answer:

• Acceleration = 2 m/s²

• Distance travelled = 16 m

Exam Tip: Use the net force, not just the applied force, when calculating acceleration.

Q9. A student has a mass of 45 kg. Calculate the gravitational force acting on the student near the Earth's surface. Take g = 9.8 m/s².

Solution:

Given:

Mass = 45 kg
Acceleration due to gravity = 9.8 m/s²

Using:

$F=mg$

$F=45\times9.8$

$F=441,\text{N}$

Answer: The gravitational force acting on the student is 441 N downwards.

Important Point: Mass is measured in kilograms, while gravitational force or weight is measured in newtons.

Experiment-Based and Graph-Based Questions on Motion

Q10. A student launches identical stacks of coins using a rubber band stretched by the same amount on three different horizontal surfaces. The results are shown below.

(a) Which surface offers the greatest friction?
(b) Which surface offers the least friction?
(c) Why must the rubber band be stretched by the same amount each time?

Answer:

(a) The rough wooden surface offers the greatest friction because the coins travel the shortest distance.

(b) The smooth tiled surface offers the least friction because the coins travel the greatest distance.

(c) Keeping the rubber band's stretch the same helps ensure similar launching conditions. This makes the comparison between surfaces fairer.

Conclusion: Greater opposing friction generally brings a moving object to rest over a shorter distance.

Q11. A 4 kg toy vehicle moves along a straight track. Its velocity changes as shown in the table. Calculate the net force during each interval.

Solution:

Interval 1: 0–4 seconds

Initial velocity = 0 m/s
Final velocity = 8 m/s

$a=\dfrac{8-0}{4}=2,\text{m/s}^2$

$F=4\times2=8,\text{N}$

Interval 2: 4–6 seconds

Velocity remains constant at 8 m/s.

Acceleration = 0

Net force = 0 N

Interval 3: 6–10 seconds

Initial velocity = 8 m/s
Final velocity = 0 m/s

$a=\dfrac{0-8}{4}=-2,\text{m/s}^2$

$F=4\times(-2)=-8,\text{N}$

Answer:

Conclusion: A change in the slope of a velocity–time graph indicates a change in acceleration and hence in net force for a constant-mass object.

Q12. In a classroom experiment, a cart is pulled using a string passing over a pulley. Predict what happens to the cart's acceleration in the following cases.

(a) The net force is doubled while the cart's mass remains constant.
(b) The mass is doubled while the net force remains constant.
(c) Both the mass and net force are doubled.

Answer:

According to Newton's second law:

$a=\dfrac{F}{m}$

Explanation: Acceleration is directly proportional to net force and inversely proportional to mass.

Experimental Note: Actual measurements may differ slightly because of friction and measurement errors.

Application-Based Questions on Newton's Three Laws

Q13. Why does a cricket fielder move their hands backwards while catching a fast-moving ball? Explain using Newton's second law.

Answer:

A cricket fielder moves their hands backwards to increase the time taken to stop the ball.

• The ball's velocity decreases to zero.

• Increasing the stopping time reduces the magnitude of deceleration.

• A smaller deceleration means a smaller average stopping force for the same ball.

This reduces the force experienced by the hands and lowers the chance of injury.

Related Application: Vehicle airbags and high-jump landing mats also increase stopping time to reduce the force during impact.

Q14. A student sitting on a wheeled chair pushes a heavy table forwards. The chair moves backwards. Explain why this happens.

Answer:

This happens because of Newton's third law of motion.

When the student pushes the table forwards, the table simultaneously pushes the student backwards with an equal and opposite force.

Since the chair can move on its wheels, the student and chair move backwards.

Important Point: Action and reaction forces act on different objects. Therefore, they do not cancel each other on the same object.

Q15. Why can we walk forwards even though our feet push the ground backwards? What would happen on a very slippery surface?

Answer:

Walking is possible because of the interaction between our feet and the ground.

1. Our feet push the ground backwards.

2. The ground exerts a forward frictional force on our feet.

3. This forward force helps us move.

On a very slippery surface, friction is reduced. Our feet may slide backwards, making walking difficult.

Conclusion: Friction does not always oppose useful motion. It also provides the grip needed for walking and running.

Q16. A sailor jumps forwards from a small boat towards the shore. Why does the boat move backwards? Will a heavier boat move backwards with the same acceleration?

Answer:

When the sailor jumps, they push the boat backwards.

According to Newton's third law, the boat pushes the sailor forwards with an equal and opposite force.

As a result, the sailor moves towards the shore and the boat moves backwards.

A heavier boat generally experiences a smaller acceleration for the same horizontal force because:

$a=\dfrac{F}{m}$

Conclusion: Equal and opposite interaction forces can produce different accelerations because the objects have different masses.

Higher-Order Thinking and Connected-Object Questions

Q17. Two boxes of masses 3 kg and 5 kg are connected by a light string on a frictionless horizontal surface. A horizontal force of 24 N pulls the 3 kg box. Calculate the acceleration of the system and the tension in the string.

Solution:

Given:

Mass of first box = 3 kg
Mass of second box = 5 kg
Applied force = 24 N

Step 1: Calculate the total mass.

Total mass = 3 + 5 = 8 kg

Step 2: Calculate the system's acceleration.

$a=\dfrac{F}{m_1+m_2}$

$a=\dfrac{24}{8}=3,\text{m/s}^2$

Step 3: Calculate the tension.

The 5 kg box is pulled horizontally by the string.

$T=m_2a$

$T=5\times3=15,\text{N}$

Answer:

• Acceleration of the system = 3 m/s²

• Tension in the string = 15 N

Important Point: The string tension is an internal force when both boxes are considered together as one system.

Q18. Two carts of different masses collide. Cart A has a mass of 2 kg and Cart B has a mass of 6 kg. During the collision, each cart experiences a force of magnitude 18 N. Compare the forces and accelerations acting on the carts.

Solution:

According to Newton's third law, the forces exerted by the carts on each other are equal in magnitude and opposite in direction.

Acceleration of Cart A:

$a_A=\dfrac{18}{2}=9,\text{m/s}^2$

Acceleration of Cart B:

$a_B=\dfrac{18}{6}=3,\text{m/s}^2$

Answer:

The lighter cart experiences three times the acceleration of the heavier cart.

Conclusion: Equal action and reaction forces do not necessarily produce equal accelerations.

Important Concepts, Formulas and Preparation Tips for Class 9 Science Chapter 6

The science Class 9 Chapter 6 important questions students practise should include conceptual explanations, force diagrams, numerical calculations and practical applications. Understanding how net force affects acceleration is especially useful for solving unfamiliar problems.

Important Concepts in How Forces Affect Motion

Important Formulas for Class 9 Science Chapter 6

Important Note: Use SI units in numerical questions: mass in kg, force in N, time in s and acceleration in m/s².

How to Prepare Class 9 Science Chapter 6 Important Questions

• Learn the difference between balanced and unbalanced forces with examples.

• Practise drawing arrows to show the magnitude and direction of forces.

• Understand why friction sometimes slows motion and sometimes helps it.

• Revise all three of Newton's laws through everyday situations.

• Solve numerical problems involving net force, mass, acceleration and weight.

• Practise interpreting velocity–time graphs.

• Study the effect of force and mass on the acceleration of connected objects.

• Check the units and direction of the final answer.

The Class 9 Science Chapter 6 Important Questions PDF, once available, can help students revise the solved questions, numerical calculations and diagrams offline.

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Students can use these chapter-specific resources to strengthen their understanding of forces, motion and Newton's laws.

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