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A particle of mass m(1) = 4 kg moving at...

A particle of mass `m_(1) = 4 kg` moving at `6hatims^(-1)` collides perfectly elastically with a particle of mass `m_(2) = 2` moving at `3hati ms^(-1)`

A

`200/3J`

B

`500/3J`

C

`400/3J`

D

`800/3J`

Text Solution

Verified by Experts

The correct Answer is:
C

`v_(CM) = (m_(1)(dr_(1))/(dt)+m_(2)(dr_(2))/dt)/(m_(1) + m_(2)) = (4 xx 5hat(i) + 2 xx 10hat(j))/(4+2)`
`v_(CM) = (40hat(i))/6 = 20/3hat(i)`
The kinetic energy,
`K = 1/2mv^(2) = 1/2xx(4+2) xx(20 xx 20)/(3 xx 3)`
`=1/2 x 6 xx (20 xx 20)/(3 xx 3) = 400/3 J`
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