a lighter body collides with a much more massive body at rest. Prove that the direction of lighter body is reversed and massive body remains at rest.

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for one dimensional elastic collision velocities after the collsiion is given by:V1=m1-m2u1m1+m2+2m2u2m1+m2V2=m2-m1u2m1+m2+2m1u1m1+m2since u2=0and m1<<m2V1=m1m2-1u1m1m2+1+0=-u1v2=0+2m1u1m2m1m2+12m1m2u110

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  • 22

Consider a ball of mass m colliding elastically with a stationary object of larger mass M. Draw the picture before and after the collision. The conservation equations are:

[1]

 

mv1 = mv2 + MV2

    [2]

    (1/2)mv12 = (1/2)mv22 + (1/2)MV2

      where v1 is the initial velocity of the smaller ball, v2 is its final velocity after collision, and V2 is the velocity of the larger mass after the collision.

      Multiply the energy equation by 2 to eliminate the (1/2) factors.

      [3]

      mv12 = mv22 + MV2

        Divide this by m on both sides.

        [4]

        v12 = v22 + (M/m)V22

          Rearrange.

          [5]

          v12 - v22 = (M/m)V22

            Divide the momentum equation by m on both sides.

            [6]

            v1 = v2 + (M/m)V2

              Rearrange and square both sides.

              [7]

              (v1 - v2)2 = (M/m)2V22

                Multiply Eq. [5] by (M/m) and combine with [7] to eliminate V22.

                [8]

                (M/m)(v12 - v22) = (v1 - v2)2

                  Multiply both sides by (m/M)

                  [9]

                  (v12 - v22) = (m/M)(v1 - v2)2

                    Take the limit as (m/M) goes to zero.

                    [10]

                    (v12 - v22) = 0

                      So one solution of this is v1 = -v2. Another solution is v1 = v2, corresponding to the case where the two objects do not collide at all. therefore v1=vwill be rejected.

                      • 8
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