The Second Law of Motion – Long Answer Questions (Class 9 Science)

Medium Level (Application & Explanation)

Q1. Explain Newton’s Second Law of Motion (F = ma) in simple words and relate it to the examples of a shopping cart, a baseball, and a car.

Answer:

Newton’s Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This is written as F = ma, where F is net force, m is mass, and a is acceleration.
In simple words: if you push harder (more force) an object speeds up more (greater acceleration). If the object is heavier (more mass), the same push produces less change in speed.
For the shopping cart: when you push an empty cart with a small force, it accelerates quickly. If you add many groceries (increasing mass), the same push produces smaller acceleration. To get the same acceleration you must push harder (increase force).
For the baseball: the ball has small mass, so when you apply a strong force by throwing, it gets a large acceleration and travels fast. A stronger throw (greater force) gives more speed.
For the car: cars have large mass, so the same force that quickly moves a ball will cause only a small acceleration in a car. That is why a large force (engine power or many people pushing) is needed to move a heavy car.
Thus F = ma helps predict how objects change motion when forces and masses change.

Q2. A student applies a constant force of 20 N on a trolley of mass 5 kg. Calculate the acceleration and explain what the result tells about the relationship between force, mass and acceleration.

Answer:

Using F = ma, acceleration a = F/m. Here F = 20 N and m = 5 kg. So a = 20 / 5 = 4 m/s².
This means the trolley’s speed increases by 4 metres per second every second while the force acts.
The calculation shows two important points: (1) For a fixed mass, increasing force increases acceleration proportionally. If the student applied 40 N, acceleration would be 8 m/s². (2) For a fixed force, increasing mass decreases acceleration inversely. If the trolley mass were 10 kg with same 20 N, acceleration would be 2 m/s².
So the result directly demonstrates the idea that more force → more acceleration, and more mass → less acceleration for the same force. This is the core of Newton’s Second Law.

Q3. Why does an empty shopping cart accelerate more easily than a fully loaded cart? Explain using the second law and real-life reasoning.

Answer:

An empty cart has much smaller mass compared to a fully loaded cart. According to F = ma, for the same applied force (F), acceleration a = F/m. Therefore, with smaller m, a is larger.
In real life, when you push both carts with the same effort, the empty cart gains speed quickly because its inertia (resistance to change of motion) is small. The loaded cart has greater inertia because of its larger mass, so it resists changes in motion more.
Also, a loaded cart may have higher rolling friction and distribution of weight may increase contact resistance; both effects require more force to achieve the same acceleration.
To move the loaded cart at the same acceleration as the empty cart, you must apply a larger force. This shows that both mass and net force determine how easily an object speeds up, as expressed by F = ma.

Q4. Two forces act on a 6 kg box: 30 N to the right and 10 N to the left. Find the net force and acceleration. Explain why only the net force matters for acceleration.

Answer:

First compute the net force by vector addition: rightward 30 N and leftward 10 N give net force F_net = 30 − 10 = 20 N to the right.
Using F = ma, acceleration a = F_net / m = 20 / 6 ≈ 3.33 m/s² to the right.
Only the net force matters because forces in opposite directions partly cancel each other. The box responds to the combined effect (resultant) of all forces, not to each force separately. If their magnitudes balance exactly, net force is zero and there is no acceleration (object moves at constant velocity or stays at rest).
In this case, the smaller leftward force reduces the effect of the larger rightward force, so the box accelerates less than it would under 30 N alone. This demonstrates the importance of calculating net force before using F = ma.

Q5. Two objects, A (2 kg) and B (6 kg), are pushed with the same force of 12 N. Calculate accelerations and explain how mass affects acceleration using these numbers.

Answer:

For object A (mass 2 kg): a = F/m = 12 / 2 = 6 m/s².
For object B (mass 6 kg): a = 12 / 6 = 2 m/s².
These calculations show that with the same force, the lighter object A gains three times the acceleration of the heavier object B (because 6 m/s² vs 2 m/s²). This clearly illustrates the inverse relationship between mass and acceleration in F = ma.
Mass represents inertia, the property that resists change in motion. Object B, being heavier, has more inertia, so the same push changes its velocity more slowly. Object A, having less inertia, changes velocity faster.
Therefore, to achieve equal accelerations for A and B, the heavier object would need a proportionally larger force. This is why vehicles with larger mass require more engine power to reach the same acceleration.

High Complexity (Analytical & Scenario-Based)

Q6. A 70 kg passenger is in a car moving at 20 m/s (about 72 km/h). If the car stops suddenly in 0.1 s (without a seatbelt) versus 0.5 s (with a seatbelt stretching), calculate and compare the average decelerations and the forces experienced by the passenger. Explain why seatbelts reduce injury using the second law.

Answer:

Change in velocity Δv = 20 − 0 = 20 m/s.
Case 1 (stop in 0.1 s): average deceleration a1 = Δv / Δt = 20 / 0.1 = 200 m/s² (direction negative). Force on passenger F1 = m a1 = 70 × 200 = 14,000 N forward relative to the car’s frame.
Case 2 (stop in 0.5 s): average deceleration a2 = 20 / 0.5 = 40 m/s². Force F2 = 70 × 40 = 2,800 N.
The numbers show that with a shorter stopping time, deceleration is much higher and the force on the passenger is much larger. Seatbelts increase the stopping time slightly by stretching or by distributing the stopping force across the body, which lowers the average deceleration and therefore the force experienced. According to F = ma, reducing a (by increasing Δt for fixed Δv) reduces F on the person.
Less force means less risk of severe injury: internal organs, bones, and the body experience smaller sudden loads. Hence, seatbelts are life-saving because they reduce the acceleration (deceleration) and so the forces acting on the passenger.

Q7. A 10 kg block rests on a rough inclined plane inclined at 30°. The coefficient of kinetic friction is 0.2 and the block is sliding down. Calculate the net force along the incline and the acceleration of the block. Explain each step and physical meaning. (Use g = 9.8 m/s².)

Answer:

Weight component along the incline: W_parallel = m g sinθ = 10 × 9.8 × sin30° = 98 × 0.5 = 49 N down the slope.
Normal reaction: N = m g cosθ = 10 × 9.8 × cos30° = 98 × 0.866 ≈ 84.87 N.
Friction force (opposing motion up the plane): f = μ_k N = 0.2 × 84.87 ≈ 16.97 N up the slope.
Net force down the slope: F_net = W_parallel − f = 49 − 16.97 ≈ 32.03 N.
Acceleration a = F_net / m = 32.03 / 10 ≈ 3.203 m/s² down the incline.
Physically: the component of gravity along the incline tries to pull the block down, while friction resists motion. The difference (net force) causes the block to accelerate. Smaller friction or larger slope increases W_parallel and thus larger acceleration. This calculation shows how to combine gravitational components and friction to use F = ma on an inclined surface.

Q8. Two people push a stalled car of mass 800 kg along a flat road. Each person pushes with 300 N in the same direction. The friction and resistance opposing the motion total 250 N. Determine the car’s acceleration and explain the role of net force and mass in this situation.

Answer:

Total applied forward force = 300 + 300 = 600 N. Opposing resistance = 250 N. So net force F_net = 600 − 250 = 350 N forward.
Acceleration a = F_net / m = 350 / 800 = 0.4375 m/s² (approximately 0.44 m/s²) forward.
The net force is the resultant of all pushing and resisting forces. Despite a fairly large total push, the heavy mass (800 kg) makes the resulting acceleration small. This shows the two-fold effect: you need to overcome resistances first, and even after overcoming them, the leftover net force must accelerate a large mass, producing only small changes in speed.
If the people could push more (increase applied force) or if the car’s mass were reduced, acceleration would increase. This illustrates how force and mass together determine the motion, as stated in F = ma.

Q9. Explain qualitatively how rockets accelerate in space using the second law. Address the fact that mass of a rocket changes as fuel is expelled and why rockets still follow Newton’s laws.

Answer:

Rockets accelerate by expelling mass (exhaust gases) backwards at high speed. By conservation of momentum, the backward momentum of the exhaust yields forward momentum for the rocket, causing it to accelerate.
Newton’s Second Law in the simple form F = ma applies directly when the mass is constant. For a rocket, the mass changes as fuel is consumed, so we must consider the rate of change of momentum: force equals the rate at which momentum of the system changes. In practice, thrust equals mass flow rate of exhaust times exhaust velocity relative to the rocket.
Even thou...