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Conservation of Linear Momentum

AP Physics 1· difficulty 3/5

Two identical balls collide on a frictionless table. Ball 1 was moving east at 3 m/s3~\text{m/s} and ball 2 was at rest. After collision, ball 1 moves north at 3 m/s\sqrt{3}~\text{m/s}. By momentum conservation, ball 2 moves at angle and speed

  • A

    South at 3 m/s\sqrt{3}~\text{m/s}

  • B

    South-east, 12 m/s\sqrt{12}~\text{m/s}

  • C

    Combination — east 3 m/s3~\text{m/s} and south 3 m/s\sqrt{3}~\text{m/s} vector summed

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  • D

    East at 3 m/s3~\text{m/s}

Explanation

East: 3=v2x⇒v2x=33 = v_{2x} \Rightarrow v_{2x} = 3. North: 0=3+v2y⇒v2y=−30 = \sqrt 3 + v_{2y} \Rightarrow v_{2y} = -\sqrt 3. So ball 2 has east-component 33 and south-component 3\sqrt 3. Magnitude =9+3=12= \sqrt{9 + 3} = \sqrt{12}.

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