The body of mass 10kg moving with a velocity of 5 m/s hits a body of 1gm at rest. The velocity of the second body after collision, assuming it to be perfectly elastic is (in m/s)
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Answers
For a perfectly elastic collision, coefficient of restitution is e=1.
Let u and v be the velocities of 10 kg and 1 g masses after collision.
Thus, e= rel. vel. of approach v−u v-u
_________________ =___ = __
rel. vel.of separation 5−0 5
But, as e=1, we have v−u=5.
Total momentum before collision is P=10×5=50 kg m/s
Total momentum after collision is P=10×(v−5)+0.001×v=10.001v−50
Thus, 10.001v−50=50
⇒v= 10.001
=100
≈10 m/s
For a perfectly elastic collision, coefficient of restitution is e=1.
Let u and v be the velocities of 10 kg and 1 g masses after collision.
rel. vel. of approach. v - u. v - u
Thus , e = ----------------------------- = --------- = -------
rel. vel.of separation. 5 - 0. 5
But, as e=1, we have v−u=5.
Total momentum before collision is P=10×5=50 kg m/s
Total momentum after collision is P=10×(v−5)+0.001×v=10.001v−50
Thus, 10.001v−50=50⇒v= 100 / 10.001 = 10m/s
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