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Change in Momentum and Impulse

AP Physics 1· difficulty 3/5

A 0.10 kg0.10~\text{kg} ball hits a wall horizontally at 20 m/s20~\text{m/s} and rebounds at 15 m/s15~\text{m/s} in the opposite direction. The magnitude of the impulse is

  • A

    3.5 N⋅s3.5~\text{N·s}

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  • B

    0.5 N⋅s0.5~\text{N·s}

  • C

    2.0 N⋅s2.0~\text{N·s}

  • D

    1.5 N⋅s1.5~\text{N·s}

Explanation

Δp=m(vf−vi)=0.10(−15−20)=−3.5 kg⋅m/s\Delta p = m(v_f - v_i) = 0.10(-15 - 20) = -3.5~\text{kg·m/s}. Magnitude 3.5 N⋅s3.5~\text{N·s}.

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