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Conservation of Linear Momentum

AP Physics 1· difficulty 4/5

Two carts of masses m1=1.0 kgm_1 = 1.0\,\text{kg} and m2=3.0 kgm_2 = 3.0\,\text{kg} are initially at rest with a compressed spring between them. The spring is released. If m2m_2 moves at 1.0 m/s1.0\,\text{m/s}, what is the speed of m1m_1 and what is the total kinetic energy released?

  • A

    v1=3.0 m/sv_1 = 3.0\,\text{m/s}, KE=6.0 JKE = 6.0\,\text{J}

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  • B

    v1=1.0 m/sv_1 = 1.0\,\text{m/s}, KE=2.0 JKE = 2.0\,\text{J}

  • C

    v1=3.0 m/sv_1 = 3.0\,\text{m/s}, KE=4.5 JKE = 4.5\,\text{J}

  • D

    v1=0.33 m/sv_1 = 0.33\,\text{m/s}, KE=0.5 JKE = 0.5\,\text{J}

Explanation

Momentum: 0=m1v1−m2v20 = m_1 v_1 - m_2 v_2, so v1=(3.0)(1.0)/1.0=3.0 m/sv_1 = (3.0)(1.0)/1.0 = 3.0\,\text{m/s}. Then KE=12(1)(3)2+12(3)(1)2=4.5+1.5=6.0 JKE = \tfrac{1}{2}(1)(3)^2 + \tfrac{1}{2}(3)(1)^2 = 4.5 + 1.5 = 6.0\,\text{J}.

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