A 0.25 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 0.050 s. What is the magnitude of the average force exerted by the wall on the ball?
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Correct answer
C. 50 N
Principle or equation
Impulse-momentum theorem: the impulse (F_avg Δt) equals the change in momentum (m Δv). Take right as positive; the velocity changes from +6.0 m/s to -4.0 m/s, so Δv = -10 m/s.
Why this answer is correct
Change in momentum = m(v_final - v_initial) = 0.25 kg × (-4.0 - 6.0) m/s = -2.5 kg·m/s. Magnitude = 2.5 kg·m/s. Average force magnitude = |Δp|/Δt = 2.5 / 0.050 = 50 N.
Example
If a 0.5 kg ball changes velocity from 2 m/s to -2 m/s in 0.1 s, impulse magnitude = 0.5×4 = 2 N·s, force = 20 N.
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