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Change in Momentum and Impulse

AP Physics 1· difficulty 4/5

A 0.15 kg0.15\,\text{kg} baseball traveling east at 40 m/s40\,\text{m/s} leaves a bat moving west at 50 m/s50\,\text{m/s}. Contact time is 1.0 ms1.0\,\text{ms}. What is the magnitude of the average force exerted by the bat on the ball?

  • A

    F=1.35×104 NF=1.35\times 10^4\,\text{N}

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  • B

    F=6.00×103 NF=6.00\times 10^3\,\text{N}

  • C

    F=9.00×103 NF=9.00\times 10^3\,\text{N}

  • D

    F=1.50×103 NF=1.50\times 10^3\,\text{N}

Explanation

Δp=m(vf−vi)=0.15(−50−40)=−13.5 kg⋅m/s\Delta p=m(v_f-v_i)=0.15(-50-40)=-13.5\,\text{kg·m/s}. ∣F∣=∣Δp∣/Δt=13.5/0.001=1.35×104 N|F|=|\Delta p|/\Delta t=13.5/0.001=1.35\times10^4\,\text{N}.

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