A 0.50 kg ball moving at 8.0 m/s to the right strikes a wall and rebounds to the left at 6.0 m/s. The ball is in contact with the wall for 0.040 s. What is the magnitude of the average force exerted by the wall on the ball?
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Correct answer
C. 175 N
Principle or equation
Impulse-momentum theorem: the impulse on an object equals its change in momentum. For a collision with a wall, the change in momentum is m(v_final - v_initial), taking direction into account.
Why this answer is correct
Take right as positive. Initial velocity u = +8.0 m/s, final velocity v = -6.0 m/s. Change in momentum Δp = m(v - u) = 0.50 × (-6.0 - 8.0) = 0.50 × (-14.0) = -7.0 kg·m/s. Magnitude of impulse = 7.0 N·s. Average force magnitude = |Δp| / Δt = 7.0 / 0.040 = 175 N.
Example
A 0.2 kg ball hits a wall at 5 m/s and rebounds at 3 m/s in the opposite direction. The change in momentum is 0.2 × ( -3 - 5 ) = -1.6 kg·m/s. If contact time is 0.02 s, average force = 1.6 / 0.02 = 80 N.
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