Two particles A and B with particle A having mass m, are moving towards each other under their mutual force of attraction. If their speeds at certain instant are v and 2v respectively, then find the K.E. of the system.

Explanation

Since the particles are moving under their mutual attraction net force acting on the system `(m_(1)+m_(2))` is equal to `vec(F)_(1)+vec(F)_(2)=0`.
Therefore the linear momentum of the system is constant
`rArr |m_(1)vec(v)_(1)+m_(2)vec(v)_(2)|=` constant.
Since initially the particles were released from rest, `P_(i)=0`
`rArr p_(f)=m_(1)vec(v)_(2)=0`
`rArr m_(2)=(m_(1)v_(1))/(v_(2))` (numerically)
`rArr m_(2)=(mv)/(2v)=(m)/(2)("where " m_(1)=m, v_(1)=v & v_(2)=2v)`
The KE of the system `= K.E.=(1)/(2)m_(1)v_(1)^(2)+(1)/(2)m_(2)v_(2)^(2)`
`=(1)/(2)mv^(2)+(1)/(2)((m)/(2))(2v)^(2)=(1)/(2)mv^(2)+mv^(2)=(3)/(2)mv^(2)`