A particle of mass `m_1` moving with velocity `v` in a positive direction collides elastically with a mass `m_(2)` moving in opposite direction also at velocity v. If `m_(2)gt gtm_(1)`, then

Options

A. the velocity of `m_(1)` immediately after collision is nearly `3v`
B. the change in momentum of `m_(1)` is nearly `4m_(1)v`
C. the change in kinetic energy of `m_(1)` is nearly `4mv_(2)`
D. all of the above

Answer

D

Explanation

`v_(1)=((m_(1)-m_(2))/(m_(1)+m_(2)))u_(1)+(2m_(2)u_(2))/(m_(1)+m_(2))`
`=-u_(1)+2u_(2)` (neglecting `m_(1)` in comparison to `m_(2)`)
Hence a is correct
change in omentum of `m_(1)=P_(f)-P_(i)=mv_(1)-mv`
`=-m3v-mv=4mv`
hence b is correct.
Change in `KE` of `m_(1)=K_(f)-K_(i)=1/2mv_(1)^(2)-1/2mv^(2)`
`=1/2m(3v)^(2)-1/2mv^(2)=4mv^(2)`
Hence c is correct