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Change in Momentum and Impulse

AP Physics 1· difficulty 4/5

F t F_max=20 N 0.30 s

A triangular force pulse acts on a 2.0 kg2.0\,\text{kg} object initially at rest. The pulse rises linearly from 00 at t=0t=0 to 20 N20\,\text{N} at t=0.15 st=0.15\,\text{s}, then falls linearly to 00 at t=0.30 st=0.30\,\text{s}. What is the object's final speed?

  • A

    v=1.5 m/sv=1.5\,\text{m/s}

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  • B

    v=6.0 m/sv=6.0\,\text{m/s}

  • C

    v=0.75 m/sv=0.75\,\text{m/s}

  • D

    v=3.0 m/sv=3.0\,\text{m/s}

Explanation

Impulse == area =12(0.30)(20)=3.0 N⋅s=\tfrac{1}{2}(0.30)(20)=3.0\,\text{N·s}. Δv=J/m=3.0/2.0=1.5 m/s\Delta v=J/m=3.0/2.0=1.5\,\text{m/s}.

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