A spherical body of mass 100 g is dropped from a height of 10 m from the ground. After hitting the ground, the body rebounds to a height of 5 m. The impulse of force imparted by the ground to the body is given by : (given, g=9.8 m / s²)

Solution

Physics sir · 4 steps

The velocity reverses direction, so add the two speeds, then multiply by the mass in kilograms.

1. Speed before impact. Falling from 10 m, the body hits the ground with v₁ = √2gh₁ = √(2 × 9.8 × 10) = √196 = 14 m s⁻¹, moving downward.
2. Speed after impact. To rise to 5 m, it must leave the ground with v₂ = √(2 × 9.8 × 5) = √98 ≈ 9.9 m s⁻¹, moving upward.
3. Impulse equals momentum change. Take upward as positive. The velocity goes from -v₁ to +v₂, so the change is v₂ - (-v₁) = v₁ + v₂. Use m = 100 g = 0.1 kg.

   J = m(v₁ + v₂) = 0.1 × (14 + 9.9) = 0.1 × 23.9 ≈ 2.39 kg m s⁻¹

4. So. The ground gives the body an upward impulse of about 2.39 kg m s⁻¹, which is option (b).

The idea to keep. Momentum is a vector. When a body bounces, its velocity changes sign, so the speeds before and after add up. Always convert grams to kilograms first.

Answer from the NTA final answer key. Solution by PrepWiser.