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Change in Momentum and Impulse

AP Physics 1· difficulty 2/5

A baseball (0.15 kg0.15~\text{kg}) approaches at 40 m/s40~\text{m/s} and leaves at 50 m/s50~\text{m/s} in the opposite direction. Contact time 1.0 ms1.0~\text{ms}. Average force on the ball?

  • A

    1500 N1500~\text{N}

  • B

    13,500 N13{,}500~\text{N}

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  • C

    45,000 N45{,}000~\text{N}

  • D

    7500 N7500~\text{N}

Explanation

Δp=0.15(50−(−40))=13.5 kg⋅m/s\Delta p = 0.15(50 - (-40)) = 13.5~\text{kg·m/s}. F=Δp/Δt=13.5/0.001=13,500 NF = \Delta p/\Delta t = 13.5/0.001 = 13{,}500~\text{N}.

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