Two balls with masses `m_(1)=3` and `m_(2)=5`kg have initial velocities `v_(1)=v_(2)=5m//s` in the directions shown in figure. They collide at the origin.
a. find the velocioty of the CM `3s` before the collision.
b. Find the position of the CM `2s` after the collision.

Explanation

a. Given time is of no consequence since `v_(CM)` is fixed for all times
`v_(CM)(x)=(m_(1)v_(1x)+m_(2)v_(2x))/(m_(1)+m_(2))`
`=((3)(-5cos37^(@))+(5)(0))/(8kg)=-1.5m//s`
`v_(CM)(y)=(m_(1)v_(1)y+m_(2)y_(2))/(m_(1)+m_(2))`
`=(3)(-5sin37^(@))+(5xx5)/(8kg)=+2m//s`
`:. vecv_(CM)=1.5hati+2hatjm//s`
b. Since the collision occurs at the origin `(vecr_(1)=0)`, the position of the centre of mass `2s` later is `vecr_(CM)=vecr_(1)+vecv_(CM^(t))`
`=vecv_(CM^(t))=-3hati+4hatjm`