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15 April, 05:09

If a 2 kg ball traveling at 10 m/s hits a wall and stops in 0.03 seconds, then how much force will the ball experience?

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  1. 15 April, 08:02
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    Newton's second law is:

    F=m*a,

    where a=dv/dt, so

    F=m * (dv/dt)

    Rearranging gives:

    F*dt=m*dv.

    Basic integration gives:

    F*t=m (vf-v0),

    where vf and v0 are the final and initial velocities of the object respectively.

    In your case vf=0, because the ball stops completely, and v0=10m/s.

    Rearranging the last expression gives F = (m (vf-vo)) / t.

    Plug in numbers to find F = (2*10) / 0.03=666.6 N
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