Question Detail
Question
A player stops a football weighing 0.5 kg which comes flying towards him with a velocity of 10 m/s. If the impact lasts for 1/50th sec. and the ball bounces back with a velocity of 15 m/s, then the average force involved is:
Correct Answer
Option D is the correct answer.
Detailed Explanation
Impulse = change in momentum. Taking direction of initial velocity as positive, initial momentum = 0.5 × 10 = 5 kg m/s. Final momentum = 0.5 × (-15) = -7.5 kg m/s (since it bounces back). Change = -7.5 - 5 = -12.5 kg m/s. Magnitude of impulse = 12.5 N s. Average force = Impulse / time = 12.5 / (1/50) = 12.5 × 50 = 625 N.
Hint
Use the impulse-momentum theorem: F_avg = Δp / Δt.