NEETPhysicsCenter of Mass, Momentum and Collision
A ball of mass 0.2 kg is dropped from a height of 20 m . Upon striking the ground, the ground imparts an upward impulse of 12 N s to the ball. The height to which the ball rebounds is (g = 10 m s ⁻²) :
Options
- A180 m
- B80 m
- C320 m
- D20 m
Correct answer
B. 80 m
Step-by-step solution
Velocity of the ball just before striking the ground: v₁ = 2gh₁ = 2 10 20 = 20 m s ⁻¹ (downwards) Taking the upward direction as positive, the initial momentum of the ball is: p_i = m(-v₁) = 0.2 (-20) = -4 kg m s ⁻¹ According to the impulse-momentum theorem: J = p_f - p_i 12 = p_f - (-4) 12 = p_f + 4 p_f = 8 kg m s ⁻¹ The rebound velocity v₂ is: v₂ = p_f m = 8 0.2 = 40 m s ⁻¹ (upwards) The height to which the ball rebounds is: h₂ = v₂^2 2g = 40^2 2 10 = 1600 20 = 80 m Answer: 80 m