Looking for accurate SEBA Class 9 Science Chapter 9 (Force and Laws of Motion) textual solutions for the 2026–27 academic year? Access step-by-step textbook exercise answers, Newton’s laws explanations, momentum calculations, and solved numerical problems designed to help you score top marks in your annual exams.
Blue question
Q1. Which of the following has more inertia:
(a) A rubber ball and a stone of the same size?
(b) A bicycle and a train?
(c) A five-rupee coin and a one-rupee coin?
Answer:
(a) The stone has more inertia because it has a greater mass than a rubber ball of the same size.
(b) The train has more inertia because its mass is significantly larger than that of a bicycle.
(c) The five-rupee coin has more inertia because it is heavier and has a larger mass than a one-rupee coin.
(Note: Inertia is directly proportional to mass—an object with greater mass has higher inertia).
Q2. In the following example, identify the number of times the velocity of the ball changes and state the agent supplying the force in each case:
“A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team.”
Answer: The velocity of the ball changes 4 times:
- First Change: When the first player kicks the stationary ball.
- Agent supplying force: Foot of the first player.
- Second Change: When the second player kicks the moving ball toward the goal.
- Agent supplying force: Foot of the second player.
- Third Change: When the goalkeeper catches and stops the ball.
- Agent supplying force: Hands/body of the goalkeeper.
- Fourth Change: When the goalkeeper kicks the ball to a teammate.
- Agent supplying force: Foot of the goalkeeper.
Q3. Explain why some leaves may get detached from a tree if we vigorously shake its branch.
Answer: When a tree branch is vigorously shaken, the branch moves to and fro rapidly. However, due to the inertia of rest, the leaves tend to remain in their original stationary position. This creates a strong pulling force on the delicate leaf stems, causing them to detach and fall.
Q4. Why do you fall in the forward direction when a moving bus brakes to a stop, and fall backwards when it accelerates from rest?
Answer:
- When a moving bus brakes: Your lower body comes to rest along with the bus, but your upper body continues to move forward due to the inertia of motion, causing you to fall forward.
- When a bus accelerates from rest: Your lower body in contact with the floor moves forward with the bus, while your upper body tends to stay at rest due to the inertia of rest, causing you to fall backward.
SEBA Class 9 Science Chapter 9 Numericals
Text Exercise
Q1. An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on its motion.
Answer: Yes, it is possible. According to Newton’s First Law of Motion, if the net external force acting on an object is zero, it can continue moving with a non-zero velocity provided:
- The motion is in a straight line (constant direction).
- The speed remains constant (zero acceleration).
Q2. When a carpet is beaten with a stick, dust comes out of it. Explain why.
Answer: Beating the carpet with a stick sets the fabric of the carpet into sudden motion. Due to the inertia of rest, the dust particles present on the carpet tend to remain at rest. As the carpet fabric moves backward, the dust particles separate from it and fall off due to gravity.
Q3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer: When a bus accelerates, stops suddenly, or takes sharp turns, the luggage on the roof tends to maintain its previous state of rest or uniform motion due to inertia. This relative motion can cause the luggage to slide or fall off the roof. Tying it securely with a rope provides the required force to keep it in place.
Q4. A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because:
(a) The batsman did not hit the ball hard enough.
(b) Velocity is proportional to the force exerted on the ball.
(c) There is a force on the ball opposing the motion.
(d) There is no unbalanced force on the ball, so the ball would want to come to rest.
Answer: (c) There is a force on the ball opposing the motion.
Explanation: Opposing frictional forces between the ball and the ground, along with air resistance, act against the direction of motion to gradually bring the ball to rest.
Q5. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes. (Hint: 1 tonne = 1000 kg)
Answer:
Given:
- Initial velocity (u) = 0m/s
- Distance (s) = 400m
- Time (t) = 20s
- Mass (m) = 7tonnes=7000kg
Step 1: Calculate acceleration (a)
Using the second equation of motion:
s=ut+12at2
400=(020)+12a(20)2
400=200a
a=400200=2m/s2
Step 2: Calculate force (F)
Using Newton’s Second Law of Motion:
F=ma
F=7000kg2m/s2=14000N
- Acceleration: 2m/s2
- Force: 14,000N
Force and Laws of Motion Class 9 SEBA Textual Exercise Solutions
Q6. A stone of 1 kg is thrown with a velocity of 20 m s⁻¹ across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
Answer:
Given:
- Mass (m) = 1 kg
- Initial velocity (u) = 20 m/s
- Final velocity (v) = 0 m/s
- Distance (s) = 50 m
Step 1: Calculate acceleration (a)
Using the third equation of motion:
v2=u2+2as
02=(20)2+2×a×50
0=400+100a
100a=−400
a=100−400=−4 m/s2
Step 2: Calculate force of friction (F)
Using Newton’s Second Law of Motion:
F=m×a
F=1 kg×(−4 m/s2)=−4 N
The negative sign indicates that the force opposes the direction of motion.
Magnitude of frictional force: 4 N
Q7. An 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, calculate:
(a) The net accelerating force
(b) The acceleration of the train
Answer:
Given:
- Mass of engine (mengine) = 8000 kg
- Number of wagons = 5
- Mass of each wagon (mwagon) = 2000 kg
- Engine force (Fengine) = 40000 N
- Frictional force (Ffriction) = 5000 N
(a) Net accelerating force (Fnet):
Fnet=Fengine−Ffriction
Fnet=40000 N−5000 N=35000 N
(b) Acceleration of the train (a):
Total mass of 5 wagons=5×2000 kg=10000 kg
Total mass of system (m)=8000 kg+10000 kg=18000 kg
Using a=mFnet:
a=1800035000≈1.94 m/s2
Q8. An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 m s⁻²?
Answer:
Given:
- Mass (m) = 1500 kg
- Acceleration (a) = −1.7 m/s2
Calculation:
Using Newton’s Second Law of Motion:
F=m×a
F=1500 kg×(−1.7 m/s2)=−2550 N
The negative sign indicates that the force acts opposite to the direction of motion to bring the vehicle to rest.
Magnitude of force: 2550 N
Q9. What is the momentum of an object of mass m, moving with a velocity v?
(a) (mv)2
(b) mv2
(c) 21mv2
(d) mv
Answer: (d) mv
Explanation: Momentum (p) is defined as the product of the mass (m) and velocity (v) of an object (p=mv).
Q10. Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
Answer:
Since the wooden cabinet moves at a constant velocity, its acceleration is zero (a=0), which means the net external force acting on it is zero.
Net Force (Fnet)=Applied Force−Friction Force=0
Friction Force=Applied Force=200 N
Therefore, the magnitude of the frictional force exerted by the floor on the cabinet is 200 N, acting in the direction opposite to motion.
Class 9 Science Chapter 9 Question Answer SEBA 2026-27
Q11. According to the third law of motion, when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
Answer:
Comment on Logic:
The student’s reasoning is incorrect. Action and reaction forces never cancel each other because they act simultaneously on two different bodies, not on the same body.
Explanation:
- When a person pushes the truck, the action force acts on the truck, while the reaction force acts back on the person’s hands.
- The reason the massive truck remains at rest is due to static friction between the truck’s tires and the road.
- The small force applied by a person is easily balanced by this static friction acting on the truck. Since the applied force is less than the maximum static friction, the net force on the truck remains zero and it does not accelerate.
Q12. A hockey ball of mass 200 g travelling at 10 m s⁻¹ is struck by a hockey stick so as to return it along its original path with a velocity of 5 m s⁻¹. Calculate the magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Given: Mass (m) = 200 g = 0.2 kg Initial velocity (u) = +10 m/s (taking initial direction as positive) Final velocity (v) = -5 m/s (since it returns in the opposite direction)
Step 1: Calculate change in momentum (p)
p = m(v – u) p = 0.2(-5 – 10) p = 0.2 * (-15) = -3 kg m/s
The negative sign indicates that the change in momentum is in the direction opposite to the initial motion.
Magnitude of change in momentum: 3 kg m/s
Q13. A bullet of mass 10 g travelling horizontally with a velocity of 150 m s⁻¹ strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.
Answer:
Given:
- Mass of bullet (m) = 10g=0.01kg
- Initial velocity (u) = 150m/s
- Final velocity (v) = 0m/s
- Time (t) = 0.03s
Step 1: Calculate acceleration (a)
Using v=u+at:
0=150+a(0.03)
a=-1500.03=-5000m/s2
Step 2: Calculate distance of penetration (s)
Using s=ut+12at2:
s=(1500.03)+12(-5000)(0.03)2
s=4.5-25000.0009
s=4.5-2.25=2.25m
Step 3: Calculate force exerted by the block (F)
Using F=ma:
F=0.01kg(-5000m/s2)=-50N
- Distance of penetration: 2.25m
- Magnitude of force: 50N (acting opposite to the motion)
Q14. An object of mass 1 kg travelling in a straight line with a velocity of 10 m s⁻¹ collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.
Answer:
Given: Mass of moving object (m1) = 1 kg Initial velocity of moving object (u1) = 10 m/s Mass of block (m2) = 5 kg Initial velocity of block (u2) = 0 m/s
Step 1: Total momentum before impact
p_initial = m1 * u1 + m2 * u2 p_initial = (1 * 10) + (5 * 0) = 10 kg m/s
Step 2: Calculate combined velocity (V) By the Law of Conservation of Momentum:
Total momentum before impact = Total momentum after impact
10 = (m1 + m2) * V 10 = (1 + 5) * V 6V = 10 => V = 10 / 6 = 5 / 3 ≈ 1.67 m/s
Step 3: Total momentum after impact p_final = (m1 + m2) * V = 6 * 1.67 ≈ 10 kg m/s
Momentum before impact: 10 kg m/s Momentum after impact: 10 kg m/s Velocity of combined object: 5/3 m/s (or 1.67 m/s)
Q15. An object of mass 100 kg is accelerated uniformly from a velocity of 5 m s⁻¹ to 8 m s⁻¹ in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
Answer:
Given: Mass (m) = 100 kg Initial velocity (u) = 5 m/s Final velocity (v) = 8 m/s Time (t) = 6 s
Step 1: Calculate initial momentum (p_i) p_i = m * u = 100 * 5 = 500 kg m/s
Step 2: Calculate final momentum (p_f) p_f = m * v = 100 * 8 = 800 kg m/s
Step 3: Calculate magnitude of force (F) Using F = (p_f – p_i) / t:
F = (800 – 500) / 6 = 300 / 6 = 50 N
Initial momentum: 500 kg m/s Final momentum: 800 kg m/s Magnitude of force: 50 N
SEBA Class 9 Science Chapter 9 Solutions
Q16. A motorcar moving with high velocity hits an insect which gets stuck on the windscreen. Kiran says the insect suffered a greater change in momentum than the car. Akhtar says the motorcar exerted a larger force on the insect because the car had a larger velocity. Who is correct? Explain briefly.
Answer:
Neither Kiran nor Akhtar is correct.
Explanation:
- Regarding Kiran’s statement: By the Law of Conservation of Momentum, the magnitude of change in momentum (p) experienced by the insect is equal and opposite to the change in momentum experienced by the motorcar. Therefore, both undergo the exact same magnitude of momentum change.
- Regarding Akhtar’s statement: According to Newton’s Third Law of Motion, the force exerted by the motorcar on the insect is equal in magnitude and opposite in direction to the force exerted by the insect on the motorcar.
Reason for the insect’s death:
Although both experience equal forces and equal changes in momentum, the insect has an extremely small mass (m). According to a=Fm, the small mass of the insect results in a huge deceleration/acceleration, causing severe damage and fatal injury. Conversely, the motorcar’s large mass results in an imperceptible change in its motion.
🔬 Updated Solutions Notice: This page features complete, step-by-step SEBA Class 9 Science Chapter 9 (Force and Laws of Motion) Textual Exercise Solutions updated for the 2026–27 academic session. All derivations, numerical solutions ($F = ma$, momentum conservation), and Newton’s laws examples strictly follow the latest revised Assam Board (SEBA) textbook.
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