Laws of Motion — Important Questions

It increases the time of action of force, reducing the force experienced

Step 1: The concept of impulse states J = F × t = change in momentum (Δp). Step 2: The momentum change (Δp) of the ball is fixed — it must be brought to rest from its initial velocity. Step 3: By pulling hands back, the cricketer increases the time 't' over which the force acts. Step 4: Since Δp = F × t is constant, if t increases, then F decreases. This means less force acts on the hands, reducing injury. Step 5: The impulse received (= Δp) stays the same — it is the force that decreases, not the impulse. Option C is wrong as the fielder doesn't add momentum to the ball.

2 m/s²

Step 1: Total mass of rocket = 1000 kg; Thrust (upward force) = 12000 N. Step 2: Weight of rocket (downward) = mg = 1000 × 10 = 10000 N. Step 3: Net upward force = Thrust - Weight = 12000 - 10000 = 2000 N. Step 4: Using F = ma: a = F/m = 2000/1000 = 2 m/s². Step 5: 12 m/s² ignores gravity; 10 m/s² is just g; 22 m/s² incorrectly adds instead of subtracting gravity. The key concept is that the net force drives acceleration, not just the thrust.

3R from the centre

Step 1: Use the relation g = GM/r² where r is the distance from the centre. Step 2: Let the required distance be r. Then g'/g = R²/r². Step 3: Given g'/g = 1/9, so R²/r² = 1/9. Step 4: Therefore r² = 9R², which gives r = 3R. This means the object must be at 3 times the Earth's radius from the centre. Step 5: R/3 would make g nine times larger (not smaller); 9R would make g 1/81 of surface value; 2R would give g = g/4. Only 3R satisfies the 1/9 condition.

The space station and astronauts are both in free fall with the same acceleration around Earth

Step 1: Weightlessness occurs when apparent weight R = 0. This happens in free fall when a = g. Step 2: An orbiting space station is constantly falling towards Earth due to gravity, but its high tangential velocity keeps it in orbit — this is free fall around Earth. Step 3: Since both the space station and astronauts fall with the same acceleration (a = g), the normal reaction force R = m(g - g) = 0. Step 4: Gravity does act at the space station's altitude — it's what keeps it in orbit! So option A is completely wrong. Step 5: Speed doesn't cancel gravity; mass is unaffected by air pressure —