Three particles `A, B` and `C` of respective masses `m_(1), m_(2)` and `m_(3)` lie on a smooth horizontal surface, and an fastened to two light inextensible strings as shown in Fig. The particle `A` is imparted an impulse `J` along `vec(BA)` . Find the initial speed of each particle.

Explanation

`J-J_(1)=m_(1)v_(1)` ………i
`J_(1)-J_(2)(cos45^(@))=m_(2)v_(2y)`…………..ii

`J_(2)sin45^(@)=m_(2)v_(2x)`…………..iii
`J_(2)=m_(3)v_(3)` ……………iv.
Now `v_(2y)=v_(1)` and `v_(2y)cos45^(@)-2_(2x)cos45^(@)=v_(3)`
Solve to get `v_(y)=(sqrt2m_(2)J)/(m_(1)m_(3)+2m_(2)(m_(1)+m_(2)+m_(3)))`
`v_(1)=((m_(3)+2m_(2))J)/(m_(1)m_(3)+2m_(2)(m_(1)m_(2)+m_(3)))`
`v_(2x)=(m_(3)v_(3)cos45^(@))/m_(2)=(m_(3)J)/(m_(1)m_(3)+2m_(2)(m_(1)+m_(2)+m_(3)))`
Net velocity of `m_(2)`
`=sqrt(v_(1)^(2)+v_(2x)^(2))=(sqrt2(m_(2)^(2)+(m_(2)+m_(3))^(2))^(1//2))/(m_(1)m_(3)+2m_(2)(m_(1)+m_(2)+m_(3)))`