CBSE Important Questions Class 9 Science Chapter 8 – Force and Laws of Motion
Class 9 Science Chapter 8 Force and Laws of Motion is part of the CBSE Class 9 Science syllabus. It explains balanced and unbalanced forces, inertia and mass, and the three laws of motion given by Newton. Students learn how force is related to the rate of change of momentum, how to use F = ma in numerical problems, the law of conservation of momentum, and why action and reaction forces never cancel each other.
These CBSE important questions for Class 9 Science Chapter 8 cover reading position-time and velocity-time graphs to find force, deriving the first law from the second law, balanced forces on a cart, why dust falls off a beaten carpet, the unit of force, action-reaction pairs and why a rider falls forward when the horse stops. Every question has a clear step-by-step answer in simple English with figures, so students can practise on their own, parents can check the working, and teachers can use them for class tests.
Q1. The position-time graph of a particle with a mass of 4 kg is shown in the figure. Calculate the force on the particle in the time intervals 0 < t < 2 s and 4 s < t < 8 s respectively.
(a) 4 N, 4 N
(b) 4 N, 8 N
(c) 8 N, 4 N
(d) 0 N, 0 N

Answer: (d) 0 N, 0 N
In both the given intervals the graph is a straight line. A straight line on a position-time graph means the velocity is constant, so the acceleration is zero.
Since F = ma = 4 × 0 = 0 N in both intervals.
Q2. Derive Newton's first law from Newton's second law of motion.
Answer: According to Newton's second law, F = ma, where F is the force, m the mass and a the acceleration.
If no external force acts on the body, then F = 0.
So, ma = 0. Since the mass m cannot be zero, the acceleration a must be zero.
If a = 0, then (v − u)/t = 0, which gives v = u. The final velocity is the same as the initial velocity.
This means that in the absence of an external force, a body at rest stays at rest and a body in motion keeps moving with the same velocity in a straight line, which is exactly Newton's first law of motion.
Q3. Mohan throws a ball on a smooth surface and draws the velocity-time graph of the ball as shown. If the mass of the ball is 150 g, how much force does the surface exert on the ball to bring it to rest?

Answer: From the graph: initial velocity u = 24 cm/s, final velocity v = 0, time t = 8 s.
a = (v − u)/t = (0 − 24)/8 = −3 cm/s² = −0.03 m/s²
Mass of the ball = 150 g = 0.15 kg
F = m × a = 0.15 × 0.03 = 4.5 × 10−3 N
So the surface exerts a force of 0.0045 N on the ball, opposite to its motion.
Q4. A wheel cart is moving at a constant speed. Two forces of equal magnitude are acting upon the cart. Identify the statement that best describes the motion of the cart as a result of the forces.
(a) It slows down and stops.
(b) It accelerates in the opposite direction.
(c) It continues moving at a constant speed.
(d) It changes direction and travels in a circular path.
Answer: (c) It continues moving at a constant speed.
Two equal forces acting in opposite directions cancel each other, so the net (unbalanced) force on the cart is zero. According to Newton's first law, with no net force the cart keeps moving at the same speed in a straight line.
Q5. Explain with an example why action and reaction forces do not cancel each other.
Answer: Action and reaction forces do not cancel each other because they act on two different objects, not on the same object. Forces can cancel only when they act on the same body.
Example: When a rubber ball strikes a hard floor, the ball exerts a force on the floor (action). The floor exerts an equal and opposite force on the ball (reaction). The action acts on the floor and the reaction acts on the ball, so they cannot cancel. The reaction force on the ball makes it bounce back.
Q6. Which of the following can explain why the dust particles fall off a hanging carpet when it is struck with a stick?
(a) Newton's first law of motion
(b) Newton's second law of motion
(c) Conservation of energy
(d) Conservation of momentum
Answer: (a) Newton's first law of motion
When the carpet is struck, the carpet suddenly starts moving, but the dust particles stay at rest because of their inertia of rest. Since they resist the change in their state, they get separated from the carpet and fall down.
Q7. Which of the following is the correct unit of force?
(a) kg m s−2
(b) g s−2
(c) kg s
(d) kg
Answer: (a) kg m s−2
Force = mass × acceleration. The SI unit of mass is kg and of acceleration is m s−2, so the unit of force is kg m s−2, which is called the newton (N).
Q8. The force required to accelerate a 2 kg mass at 5 m s−2 will be
(a) 12 N
(b) 10 N
(c) 8 N
(d) 5 N
Answer: (b) 10 N
F = m × a
= 2 kg × 5 m s−2 = 10 N
Q9. Which of the following is a correct representation of the direction of an action and reaction pair?

Answer: The correct figure is the one in which the action arrow points downwards and the reaction arrow points upwards.
The book presses down on the table with its weight; this is the action. The table pushes the book up with an equal force; this is the reaction. The two forces are equal in magnitude and opposite in direction, and they act on different bodies.
Q10. A horse rider falling forward when the horse suddenly stops can be explained with the help of
(a) Newton's first law of motion
(b) Newton's second law of motion
(c) Newton's third law of motion
(d) Newton's law of gravitation
Answer: (a) Newton's first law of motion
When the horse stops suddenly, the lower part of the rider's body stops with it, but the upper part keeps moving forward because of the inertia of motion. That is why the rider falls forward. This resistance to a change of state is described by Newton's first law.
Q11. According to the second law of motion, the force acting on an object is equal to the
(a) rate of change of position
(b) rate of change of velocity
(c) product of mass and velocity
(d) product of mass and acceleration
Answer: (d) product of mass and acceleration
Newton's second law says that the rate of change of momentum is proportional to the applied force. On simplifying, this gives F = m × a, that is, force equals mass multiplied by acceleration.
Q12. According to Newton's third law, both the forces, action and reaction,
(a) act on one object
(b) act on different objects
(c) act with the same magnitude and direction
(d) act in opposite directions with different magnitude
Answer: (b) act on different objects
Newton's third law says that to every action there is an equal and opposite reaction. The two forces are equal in size and opposite in direction, but they always act on two different bodies. This is why they never cancel each other.
Related Study Material on Extramarks
- CBSE Important Questions Class 9 Science Chapter 1
- CBSE Important Questions Class 9 Science Chapter 2
- CBSE Important Questions Class 9 Science Chapter 3
- CBSE Important Questions Class 9 Science Chapter 4
- CBSE Important Questions Class 9 Science Chapter 5
- CBSE Important Questions Class 9 Science Chapter 6
- CBSE Important Questions Class 9 Science Chapter 7
- CBSE Important Questions Class 9 Science Chapter 9
- CBSE Important Questions Class 9 Science Chapter 10
- CBSE Important Questions Class 9 Science Chapter 11
- CBSE Important Questions Class 9 Science Chapter 12
Q.1 A box is being pushed on a smooth frictionless ice rink. Choose the correct statement in the context of the motion of the box.
The box speeds up.
The box moves at a constant speed.
The box slows down.
The box does not move.
Marks:1
Ans
The box speeds up.
Q.2 Calculate the net force.
10 N
60 N
70 N
130 N
Marks:1
Ans
10 N
Q.3 An 8000 kg engine pulls a train of 5 wagons, each wagon of mass 2000 kg, along a horizontal track. If the engine exerts a force of 40,000 N and the track offers a friction force of 5,000 N.
Calculate
a) the net accelerating force,
b) the acceleration of the train, and
c) the force of wagon 1 on wagon 2.
Marks:5
Ans
a) Net accelerating force = 40,000 N – 5000N = 35,000 N
b) Mass of the train, m = 5—2000 = 10,000 kg
Net force acting on the train = 35000N
So, Acceleration of the train = Force / Mass
= 35000 /10000 = 3.5 m/s2.
c) Force exerted by wagon 1 on wagon 2
= Net accelerating force – Force acting on wagon 1
= 35000 N – 2000 kg x 3.5 m/s2
= 35000 N – 7000 N = 28000 N
Q.4 A swimmer swims forward , even though he pushes water backward while swimming. Why?
Marks:1
Ans
While swimming,a person pushes the water backward with his hand.
In reaction to this action , the water pushes the person in the forward direction with equal force.