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Elastic and Inelastic Collisions

AP Physics 1· difficulty 4/5

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A bullet of mass mm moving at speed vv embeds in a block of mass MM at the end of a string of length LL. To what maximum height hh does the combined mass swing?

  • A

    h=v22gh=\dfrac{v^2}{2g}

  • B

    h=m2v22g(m+M)2h=\dfrac{m^2 v^2}{2g(m+M)^2}

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  • C

    h=mv22g(m+M)h=\dfrac{mv^2}{2g(m+M)}

  • D

    h=m2v22gM2h=\dfrac{m^2 v^2}{2gM^2}

Explanation

Inelastic: V=mvm+MV=\dfrac{mv}{m+M}. Energy: 12(m+M)V2=(m+M)gh⇒h=V22g=m2v22g(m+M)2\tfrac{1}{2}(m+M)V^2=(m+M)gh\Rightarrow h=\dfrac{V^2}{2g}=\dfrac{m^2v^2}{2g(m+M)^2}.

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