Chapter: 6 – How Forces Affect Motion
Maximum Marks: 50
Suggested Time: 1 Hour 30 Minutes
This Class 9 Science Exploration Chapter 6 Sample Paper 1 is designed to help students practise important concepts from “How Forces Affect Motion.” The paper covers force and its effects, contact and non-contact forces, balanced and unbalanced forces, inertia, mass, momentum, Newton’s laws of motion, action and reaction, and simple numerical applications. The paper includes objective, short-answer, application-based, and numerical questions, followed by complete solutions.
Sample Paper 1
General Instructions
- All questions are compulsory.
- Read each question carefully before answering.
- Show all steps in numerical questions.
- Use SI units wherever applicable.
- Draw neat diagrams wherever required.
- Take the direction of motion as positive unless otherwise stated.
Section A — Objective Questions
10 × 1 = 10 Marks
Q1. A force can change the:
a) Colour of an object only
b) Shape or motion of an object
c) Mass of an object only
d) Temperature of an object only
Q2. Which of the following is a contact force?
a) Gravitational force
b) Magnetic force
c) Frictional force
d) Electrostatic force
Q3. When the net force acting on an object is zero, the forces are said to be:
a) Magnetic
b) Unbalanced
c) Balanced
d) Frictional
Q4. Which property of an object resists a change in its state of motion?
a) Momentum
b) Inertia
c) Acceleration
d) Velocity
Q5. The SI unit of force is:
a) Joule
b) Newton
c) Pascal
d) Watt
Q6. Newton’s first law of motion is closely associated with:
a) Energy
b) Inertia
c) Pressure
d) Density
Q7. Momentum of an object is equal to:
a) Mass × acceleration
b) Mass × velocity
c) Force × distance
d) Mass ÷ velocity
Q8. The SI unit of momentum is:
a) kg m/s
b) kg/m/s
c) N/m
d) m/s²
Q9. According to Newton’s third law, forces of action and reaction:
a) Act on the same object
b) Are always unequal
c) Act on different interacting objects
d) Always cancel each other
Q10. If the mass of an object is doubled while its velocity remains unchanged, its momentum becomes:
a) Half
b) Unchanged
c) Double
d) Four times
Section B — Very Short Answer Questions
5 × 2 = 10 Marks
Q11. What is force? State any two effects that a force can produce.
Q12. Distinguish between contact force and non-contact force with one example of each.
Q13. What are balanced forces? Do balanced forces always mean that an object is at rest?
Q14. Define inertia. Why does a heavy object generally have greater inertia than a light object?
Q15. State Newton’s third law of motion and give one everyday example.
Section C — Short Answer and Numerical Questions
4 × 3 = 12 Marks
Q16. A force of 20 N acts on an object towards the east, while another force of 12 N acts towards the west. Find the magnitude and direction of the net force.
Q17. Explain Newton’s first law of motion with the help of two everyday examples.
Q18. A ball of mass 2 kg moves with a velocity of 6 m/s. Calculate its momentum.
Q19. Explain why passengers tend to fall forward when a moving bus suddenly stops.
Section D — Application-Based and Numerical Questions
2 × 4 = 8 Marks
Q20. A trolley of mass 5 kg is initially at rest. A horizontal net force of 15 N acts on it.
Answer the following:
a) What is the initial velocity of the trolley?
b) Calculate its acceleration.
c) What will happen to its motion if the net force is removed after it has started moving, assuming friction is negligible?
d) Which law of motion explains the situation in part (c)?
Q21. A cricket player catches a fast-moving ball by moving his hands backwards while catching it.
Answer the following:
a) Why does the player move his hands backwards?
b) How does this affect the time over which the ball’s momentum changes?
c) What happens to the average force on the player’s hands?
d) Which concept of force and motion is illustrated by this example?
Section E — Long Answer and Numerical Questions
2 × 5 = 10 Marks
Q22. Explain Newton’s three laws of motion in your own words. Give one suitable everyday example for each law.
Q23. A car of mass 1000 kg is moving at 10 m/s. The driver applies the brakes and the car comes to rest in 5 seconds.
Calculate:
a) Initial momentum of the car.
b) Final momentum of the car.
c) Change in momentum.
d) Average rate of change of momentum.
e) State the relation between rate of change of momentum and force.
SOLUTIONS
Section A — Answers
Q1. b) Shape or motion of an object
Q2. c) Frictional force
Q3. c) Balanced
Q4. b) Inertia
Q5. b) Newton
Q6. b) Inertia
Q7. b) Mass × velocity
Q8. a) kg m/s
Q9. c) Act on different interacting objects
Q10. c) Double
Section B — Solutions
Q11. Force and Its Effects
A force is a push or pull that can change the state of motion or shape of an object.
A force can:
- Change the speed or direction of a moving object.
- Change the shape or size of an object.
For example, kicking a football changes its motion, while squeezing a sponge changes its shape.
Q12. Contact and Non-Contact Forces
Contact force: A force that acts only when two objects are in physical contact.
Example: Frictional force or muscular force.
Non-contact force: A force that can act without physical contact between objects.
Example: Gravitational or magnetic force.
| Contact Force | Non-Contact Force |
|---|---|
| Requires physical contact | Does not require physical contact |
| Example: Friction | Example: Gravity |
Q13. Balanced Forces
Balanced forces are forces acting on an object whose vector sum, or net force, is zero.
Balanced forces do not necessarily mean that the object is at rest.
An object already moving with constant velocity can continue moving at constant velocity when the net force is zero.
Thus, zero net force means no acceleration, not necessarily zero velocity.
Q14. Inertia
Inertia is the tendency of an object to resist any change in its state of rest or motion.
Mass is a measure of an object’s inertia. Therefore, an object with greater mass generally has greater inertia.
For example, it is harder to change the motion of a loaded truck than that of a bicycle.
Q15. Newton’s Third Law
Newton’s third law states:
When two objects interact, they exert forces on each other that are equal in magnitude and opposite in direction.
For example, when a person pushes the ground backwards while walking, the ground exerts a forward force on the person, helping the person move forward.
The two forces act on different objects.
Section C — Solutions
Q16. Net Force
Take east as the positive direction.
Force towards east:F1=20 N
Force towards west:F2=12 N
Since the forces act in opposite directions:Fnet=20−12Fnet=8 N
The larger force is towards the east.
Answer: The net force is 8 N towards the east.
Q17. Newton’s First Law
Newton’s first law states that an object remains at rest or continues moving with uniform velocity in a straight line unless acted upon by a non-zero net external force.
This law describes the concept of inertia.
Example 1:
When a stationary bus suddenly starts, passengers may lean backwards because their bodies tend to remain in the state of rest.
Example 2:
When a moving bus suddenly stops, passengers tend to move forward because their bodies tend to continue in motion.
Q18. Momentum
Given:
Mass:m=2 kg
Velocity:v=6 m/s
Momentum is:p=mv
Therefore:p=2×6p=12 kg m/s
Answer: The momentum of the ball is 12 kg m/s.
Q19. Passenger Falling Forward
When a moving bus suddenly stops, the lower part of a passenger’s body comes to rest with the bus because of contact with the floor.
However, the upper part of the body tends to continue moving forward due to inertia of motion.
As a result, the passenger tends to fall forward.
This is an example of Newton’s first law of motion.
Section D — Solutions
Q20. Trolley and Net Force
Given:
Mass:m=5 kg
Net force:F=15 N
Initially, the trolley is at rest.
a) Initial velocity
Since the trolley is initially at rest:u=0 m/s
b) Acceleration
Using Newton’s second law:F=ma
Therefore:a=mFa=515a=3 m/s2
c) Motion after the net force is removed
If friction is negligible and the net force becomes zero after the trolley starts moving, the trolley will continue moving with constant velocity in a straight line.
d) Law involved
This situation is explained by Newton’s first law of motion.
Q21. Catching a Fast-Moving Ball
a) The player moves his hands backwards to increase the time taken to bring the ball’s velocity down to zero.
b) Moving the hands backwards increases the time over which the ball’s momentum changes.
c) Since the change in momentum occurs over a longer time, the average force on the hands is reduced.
This follows from:Faverage=ΔtΔp
For the same change in momentum, increasing Δt reduces the average force.
d) This example illustrates the relationship between force, change in momentum and time, based on Newton’s second law.
Section E — Solutions
Q22. Newton’s Three Laws of Motion
Newton’s First Law
An object remains at rest or continues moving with uniform velocity in a straight line unless acted upon by a non-zero net external force.
It is also called the law of inertia.
Example: A passenger moves forward when a moving bus suddenly stops.
Newton’s Second Law
The net force acting on an object is related to the rate of change of its momentum.
For constant mass:F=ma
This means that a greater force produces a greater acceleration for the same mass.
Example: An empty shopping trolley accelerates more easily than a heavily loaded trolley when the same push is applied.
Newton’s Third Law
When two objects interact, they exert forces on each other that are equal in magnitude and opposite in direction.
Example: A swimmer pushes water backwards, and the water exerts a forward force on the swimmer.
Summary
| Law | Main Idea | Example |
|---|---|---|
| First Law | Objects resist changes in motion | Passenger moves forward when bus stops |
| Second Law | Force produces acceleration | Pushing a trolley |
| Third Law | Interaction forces occur in pairs | Swimming or walking |
Q23. Car Braking Problem
Given:
Mass of car:m=1000 kg
Initial velocity:u=10 m/s
Final velocity:v=0 m/s
Time:t=5 s
a) Initial Momentum
Momentum is:p=mv
Therefore:pi=1000×10pi=10,000 kg m/s
b) Final Momentum
pf=mvpf=1000×0pf=0 kg m/s
c) Change in Momentum
Δp=pf−piΔp=0−10,000Δp=−10,000 kg m/s
The negative sign indicates that the change in momentum is opposite to the car’s initial direction of motion.
The magnitude of the change is:10,000 kg m/s
d) Average Rate of Change of Momentum
ΔtΔp=5−10,000−2,000 N
Thus, the average force is:−2,000 N
The negative sign indicates that the braking force acts opposite to the car’s direction of motion.
e) Relation Between Force and Momentum
Newton’s second law states that the net force acting on an object is equal to the rate of change of its momentum:F=ΔtΔp
For constant mass, this becomes:F=ma
Final Answers
- Initial momentum = 10,000 kg m/s
- Final momentum = 0 kg m/s
- Change in momentum = −10,000 kg m/s
- Average force = −2,000 N
- The negative sign indicates that the force acts opposite to the initial motion.