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Change in Momentum and Impulse

AP Physics 1· difficulty 4/5

A 0.20 kg ball strikes a wall at 15 m/s and rebounds at 10 m/s in the opposite direction. Contact lasts 0.050 s. The average force on the ball is closest to

  • A

    200 N

  • B

    5 N

  • C

    20 N

  • D

    100 N (F=Δp/ΔtF = \Delta p / \Delta t, Δp=m(vf−vi)\Delta p = m(v_f - v_i))

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Explanation

Take initial direction as positive: Δv=−10−15=−25\Delta v = -10 - 15 = -25 m/s. Δp=0.20(−25)=−5\Delta p = 0.20(-25) = -5 N·s. ∣F∣=5/0.050=100|F| = 5/0.050 = 100 N.

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