A 0.50 kg ball moving at 6.0 m/s to the right strikes a wall and rebounds to the left at 4.0 m/s. The ball is in contact with the wall for 0.020 s. What is the magnitude of the average force exerted by the wall on the ball?
Show the answer and explanation
Correct answer
C. 250 N
Principle or equation
Impulse-momentum theorem: F_avg Δt = Δp = m(v_f - v_i), taking right as positive.
Why this answer is correct
Initial velocity +6.0 m/s, final velocity -4.0 m/s. Δp = 0.50 × (-4.0 - 6.0) = 0.50 × (-10) = -5.0 kg·m/s. Magnitude of impulse = 5.0 N·s. F_avg = 5.0 / 0.020 = 250 N.
Example
A 0.20 kg ball at 5 m/s rebounds at 3 m/s in 0.01 s: Δp = 0.20×(-3-5)= -1.6, F = 160 N.
This published item includes a stored explanation and passed the platform’s publication workflow. It is independent preparation material, not a claim of an official or recalled examination question.