Q1. Explain Newton’s Second Law of Motion with simple numbers. How does doubling the force or mass change acceleration?
Answer:
Newton’s Second Law says F = m a.
Acceleration is directly proportional to force and inversely proportional to mass.
Example: Apply 10 N on a 5 kg cart. Then a = 10/5 = 2 m/s².
If you double the force to 20 N, the same cart gets a = 4 m/s². Acceleration doubles.
If you double the mass to 10 kg, with 10 N, then a = 1 m/s². Acceleration halves.
So, more force → more acceleration. More mass → less acceleration.
This matches our daily life: light objects respond faster to the same push.
Q2. A shopping cart moves only when you push it hard enough. Explain using net force and friction.
Answer:
The cart moves due to net force, not just your push.
Net force = your push − friction (opposite direction).
If your push equals friction, net force = 0, so a = 0. No speeding up.
If you push harder than friction, net force > 0, so the cart accelerates.
Heavier cart means more normal force, often more friction, so harder to move.
Adding wheels with less friction or oiling them helps. Net force increases.
Thus, motion depends on the resultant force, not individual forces.
Q3. Two people push a box in opposite directions: 50 N to the right and 30 N to the left. The box has a mass of 10 kg. Find the acceleration and its direction.
Answer:
Forces in opposite directions subtract.
Net force = 50 N − 30 N = 20 N to the right.
Use F = m a. So, a = F/m = 20/10 = 2 m/s².
The direction of acceleration is the direction of the net force: to the right.
If both forces were equal, net force would be zero. Then acceleration would be zero.
This shows why the resultant matters, not each force alone.
Balanced forces mean no acceleration, even if forces are large.
Q4. Why is pushing a stalled car much harder than pushing a bicycle? Explain using F = ma and practical tips to increase acceleration.
Answer:
A car has a large mass. A bicycle has a small mass.
For the same push (force), a = F/m is small for the car.
So the car’s acceleration is low, and it speeds up very slowly.
To increase acceleration, you can increase force (more people push).
Or you can reduce mass (remove heavy items in the car).
You can also reduce friction (inflate tires, push on smoother ground).
All these increase net force or reduce m, so a becomes larger.
Q5. What is the SI unit of force? Derive it from F = ma and explain what 1 Newton means.
Answer:
From F = m a, force depends on mass and acceleration.
SI unit of mass is kilogram (kg).
SI unit of acceleration is meter per second squared (m/s²).
So, the SI unit of force is kg·m/s², called a Newton (N).
1 N is the force needed to give a 1 kg mass an acceleration of 1 m/s².
So, if you push a 1 kg object and it gains 1 m/s², you applied 1 Newton.
This unit connects force, mass, and acceleration directly.
High Complexity (Analysis & Scenario-Based)
Q6. You push a 10 kg suitcase with 60 N on a floor with 20 N friction. Find the acceleration. If the mass doubles but your push stays the same, what happens?
Answer:
Opposing friction is 20 N; your push is 60 N.
Net force = 60 − 20 = 40 N in the push direction.
a = F/m = 40/10 = 4 m/s².
If mass doubles to 20 kg, friction stays 20 N (assume unchanged for simplicity).
Net force is still 40 N, so a = 40/20 = 2 m/s².
Acceleration becomes half when mass doubles.
This shows the inversely proportional link between a and m.
Q7. In a tug-of-war, Team A pulls with 800 N to the right, Team B pulls with 800 N to the left. The flag in the middle has mass 2 kg. Will it accelerate? What if Team A increases to 900 N?
Answer:
With equal pulls, net force = 800 − 800 = 0 N.
With net force zero, the acceleration is zero. The flag does not speed up.
It may still move at constant velocity if it was already moving.
If Team A increases to 900 N, net force = 900 − 800 = 100 N to the right.
For the 2 kg flag, a = 100/2 = 50 m/s² to the right (idealized case).
In real life, the ground reactions on teams matter, but the idea is the same.
Direction of acceleration is the direction of net force.
Q8. You push a 15 kg box up a ramp with 100 N. Down-slope resistance (gravity component + friction) totals 70 N. Find the acceleration and predict what happens if the ramp gets steeper.
Answer:
Upward push is 100 N. Down-slope resistance is 70 N.
Net force = 100 − 70 = 30 N up the ramp.
a = F/m = 30/15 = 2 m/s² up the ramp.
If the ramp gets steeper, the down-slope component of weight increases.
Then the opposing force grows, so net force becomes smaller.
As a result, acceleration decreases. It could even become negative if resistance > push.
This shows how net force depends on all forces, not just your effort.
Q9. Two carts, A (2 kg) and B (5 kg), are pulled by the same 12 N motor. Compare their accelerations and time to reach 6 m/s.
Answer:
For cart A: a = F/m = 12/2 = 6 m/s².
For cart B: a = 12/5 = 2.4 m/s².
To reach 6 m/s, time t = v/a.
Cart A: t = 6/6 = 1 s.
Cart B: t = 6/2.4 = 2.5 s.
So, lighter cart accelerates faster and reaches the speed sooner.
This proves a ∝ 1/m for the same force.
Q10. A car’s engine provides a nearly constant 3000 N driving force. The car’s mass is 1000 kg. When towing a 500 kg trailer, how does acceleration change? What does this mean for overtaking?
Answer:
Without trailer: a = F/m = 3000/1000 = 3 m/s².
With trailer: total mass = 1000 + 500 = 1500 kg.
Then a = 3000/1500 = 2 m/s².
Acceleration drops when mass increases while force is fixed.
Overtaking needs quick acceleration. With the trailer, the car accelerates slower.
So the driver must allow more distance and more time to overtake safely.
This is a direct use of F = m a in driving decisions.
