Two balls `A` and `B` of masses 0.3 kg and 0.2 kg respectively are moving along positive X-axis and negaive X-axis with velocities 2.0 m `//` s. They collide and thereafter move in the directions opposite to their original directions. Find the velocities of `A` and `B` after collision. Also, calculate total K.E. of the ball before and after collision.

Explanation

Here, `m_(1)=0.3kg, m_(2)=0.2kg`
`u_(1)=2.9m//s, u_(2)=-2.0m//s`
Let `upsilon_(1)=` vel. of `A` after collision , along negative X-axis, figure,
`upsilon_(2)=` vel. of `B` after collision, along positive X-axis
According to the law of conservation of linear momentum,
`m_(1)upsilon_(1)+m_(2)upsilon_(2)=m_(1)u_(1)+m_(2)u_(2)`
`0.3upsilon_(1)+0.2upsilon_(2)=0.3xx2.0+0.2(-2.0)`
`0.3upsilon_(1)+0.2upsilon_(2)=0.2` ...(i)
Relative velocity of separation after collision `=` Relative velocity of approach before collision
`upsilon_(2)-upsilon_(1)=u_(1)-u_(2)=2.0-(-2.0)=4.0`
`upsilon_(2)=4+upsilon_(1)` ...(ii)
Put in (i), `0.3upsilon_(1)= o .2(4+upsilon_(1))=0.2`
`0.5upsilon_(1)=0.2-0.8=-0.6`
`upsilon_(1)=(-0.6)/(0.5)=-1.2ms^(-1) ` `:.` Ball `A` moves with velocity `1.2m//s` along negative X-axis.
From (ii), `upsilon_(2)=4-1.2=2.8m//s` `:.` Ball `B` moves with velocity `2.8m//s` along `+` X-axis.
Total KE of balls before collision `=(1)/(2)m_(1)u_(1)^(2)=(1)/(2)m_(2)u_(2)^(2)=(1)/(2)[0.3(2.0)^(2)+0.2(-2.0)^(2)]=1.0J`
Total KE of balls after collision `=(1)/(2)m_(1)upsilon_(1)^(2)+(1)/(2)m_(2)upsilon_(2)^(2)=(1)/(2)[0.3(-1.2)^(2)+0.2(2.8)^(2)]=1.0J`
Thus KE of balla is conserved during the elastic collision.