A ball moving with velocity 2m/s collides head on with another stationar...

Q. A ball moving with velocity 2m/s collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5, then their velocities (inm/s) after collision will be

AIPMT/NEET 2010 | W.P.E Q 1 | In case of collision The two bodies of masses m1 and m2 are moving along a line with velocities u1 and u2 respectively, u1 should be greater then u2 for two bodies to collide.

As the velocity of m1 (u1) is greater than that of m2 (u2), there will be a deformation in the bodies. This deformation continues till they acquire same
velocity. Now a question arises, that why will they acquire same velocity. The velocity of the front body m2 (which was slower
initially) will gradually increase, and the velocity of the rear body m1 (which was faster initially) will gradually decrease. By conservation momentum.
m1u1 + m2u2 = m1 v1+ m2 v2
During the whole process, the momentum of the two -block system remains constant.
Before and after collision kinetic energy may be same but not during the whole process.

Newton's Experimental Law.
This is an experimental law which relates the velocity of approach of the bodies before collision and the velocity of separation after the impact.
When the collision is neither perfectly elastic nor perfectly inelastic, for such cases, Newton did certain experiments and gave another law which is now called as Newton's Experimental Law.

Relative velocity of impact = e [Relative velocity of approach before impact]

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