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A particle moving with kinetic energy `= 3J` makes an elastic head-on collision with a stationary particle which has twice its mass. During the impact :-

A

the minimum kinetic energy of the system is `1 J`

B

the change in elastic potential energy of the system is `2 J`

C

momentum and total energy are conserved at every instant.

D

the ratio of kinetic energy to potential energy of the system first decreases and then increases.

Text Solution

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The correct Answer is:
A, B, C, D

At the time of collision both particles have common velocity and hence the system has minimum kinetic energy.
`COLM : m u + 0 = 3mv rArr v = u//3`
`KE_("Initial") = (1)/(2) m u^(2) = 3 J`
`KE_("collision") = (1)/(2)(3m)v^(2) = (1)/(2)(3m)(U^(2))/(9) = 1 J`
`PE_("collision") = (3 - 1) = 2J`
Total energy remains constant and hence `KE` of system First decreases & then increases.
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