Laws of Motion — Practice Quiz
Tamil Nadu Board · Class 10 · Science
Try a 4-question quiz on Laws of Motion for Tamil Nadu Board Class 10 Science: tap an answer to check it and see why.
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Quick Quiz: Laws of Motion
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A body of mass 4 kg is moving with a velocity of 10 m/s. A force acts on it for 2 seconds and its velocity becomes 20 m/s. What is the magnitude of the force acting on the body?
Two bodies A and B collide. Before collision, A (mass 3 kg) moves at 4 m/s and B (mass 2 kg) is at rest. After collision, A moves at 1 m/s. What is the velocity of B after collision?
A person stands on a weighing machine in a lift. The reading is MORE than his actual weight. Which situation correctly describes this?
The universal gravitational constant G has the value 6.674 × 10⁻¹¹ Nm² kg⁻². If the distance between two masses is doubled, what happens to the gravitational force between them?
Sample Questions
A fielder in cricket pulls his hands back while catching a ball. The scientific reason behind this action is best explained by which of the following?
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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.
A rocket of total mass 1000 kg (including 200 kg of fuel) is fired. The fuel burns and ejects gas at a rate that produces a thrust of 12000 N. What is the initial acceleration of the rocket? (g = 10 m/s²)
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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.
At what distance from the centre of the Earth will the acceleration due to gravity be 1/9th of its value on the Earth's surface? (R = radius of Earth)
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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.
Which of the following correctly explains why astronauts feel weightless in an orbiting space station?
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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 —
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