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Elastic and Inelastic Collisions

AP Physics 1· difficulty 3/5

A 0.020 kg0.020~\text{kg} bullet at 200 m/s200~\text{m/s} embeds in a 1.98 kg1.98~\text{kg} block hanging by a string. Using g≈10 m/s2g \approx 10~\text{m/s}^2, the block-bullet system rises to a height of

  • A

    0.20 m0.20~\text{m}

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  • B

    0.10 m0.10~\text{m}

  • C

    0.50 m0.50~\text{m}

  • D

    0.30 m0.30~\text{m}

Explanation

Right after collision: v′=(0.020⋅200)/2.0=2.0 m/sv' = (0.020 \cdot 200)/2.0 = 2.0~\text{m/s}. Energy: 12v2=gh⇒h=2/20=0.20 m\tfrac{1}{2}v^2 = g h \Rightarrow h = 2/20 = 0.20~\text{m}.

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