A 0.40 kg ball moving to the right at 5.0 m/s strikes a wall and rebounds to the left at 3.0 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?
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Correct answer
A. 160 N
Principle or equation
Impulse-momentum theorem: F_avg Δt = Δp = m(v_f - v_i). Take right as positive, so v_i = +5.0 m/s and v_f = -3.0 m/s.
Why this answer is correct
Δp = 0.40 × (-3.0 - 5.0) = 0.40 × (-8.0) = -3.2 kg·m/s. Magnitude of impulse is 3.2 N·s. F_avg = |Δp| / Δt = 3.2 / 0.020 = 160 N.
Example
If a 0.2 kg ball goes from 4 m/s to -2 m/s in 0.01 s, Δp = 0.2 × (-6) = -1.2, F = 1.2 / 0.01 = 120 N.
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