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A 821-kg car stopped at an intersection is rear-ended by a 1810-kg truck moving with a speed of 11.5 m/s. If the car was in neutral and its brakes were off, so that the collision is approximately elastic, find the final speed of both vehicles after the collision.

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2 answers
  1. Use the eqn:
    v1'=v1*(m1-m2)/(m1+m2)
    M2=neutral car, its v2=0 since its at rest.
    m1=truck, v1=11.5.
    Then, once you have v1' you can sub it into m1v1+m2v2=m1v1'+m2v2' and find v2'.
    The first eqn is a special eqn only to be used when v2=0 and the collision is completely elastic. NEVER under anyother circumstances. They are specialy derived for solely this case.

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  2. Vf2=(m1+m2)(Vo1)/m1
    m1=truck; m2=car; Vo1=initial speed of truck.

    mo1Vo1=mf1Vf1+mf2Vf2; substitute Vf2 to find Vf1

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