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Change in Momentum and Impulse

AP Physics 1· difficulty 4/5

F t F_max 0 τ/2 τ

A force on a 0.50 kg0.50\,\text{kg} object varies as a triangle: it rises linearly from 00 to Fmax⁡=12 NF_{\max} = 12\,\text{N} over 0.10 s0.10\,\text{s}, then falls linearly back to 00 over the next 0.10 s0.10\,\text{s}. The object starts at rest. What is its final speed?

  • A

    v=0.6 m/sv = 0.6\,\text{m/s}

  • B

    v=4.8 m/sv = 4.8\,\text{m/s}

  • C

    v=2.4 m/sv = 2.4\,\text{m/s}

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  • D

    v=1.2 m/sv = 1.2\,\text{m/s}

Explanation

Impulse equals area under F(t)F(t): J=12(0.20)(12)=1.2 N⋅sJ = \tfrac{1}{2}(0.20)(12) = 1.2\,\text{N·s}. Then v=J/m=1.2/0.5=2.4 m/sv = J/m = 1.2/0.5 = 2.4\,\text{m/s}.

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