Three spheres, each of mass `m`, can slide freely on a frictionless, horizontal surface. Spheres `A` and `B` are attached to an inextensible, inelastic cord of length `l` and are at rest in the position shown where sphere `B` is struck by sphere `C` which is moving to the right with a velocity `v_(0)`. Knowing that the cord is taut where sphere `B` is struck by sphere `C` and assuming 'head on' inelastic impact between `B` and `C`, we cannot conserve kinetic energy of the entire system.
The velocity of `B` immediately after collision is along unit vector

Options

A. `hati`
B. `(hati+hatj)/sqrt(2)`
C. `(sqrt(3))/2hati+1/2hatj`
D. none of these

Answer

D

Explanation

Let after collision the velocities are as shown i the figure,
Here `theta=60^(@)`. Let the impulse `C` and `B` during collision be `J`.
For the system of `A` and `B`
`Jcostheta=mv_(1)`…………i
`Jsintheta=mv_(2)`…………..ii

`impliesv_(1)=J/m (costheta)/2, v_(2)=J/msitheta`
velocity of `B`:
`vecv_(B)=v_(1)costhetahati+vecvsinthetahatj+vecv_(2)sinthetahati-v_(2)costhetahatj`
`=(v_(1)costheta+v_(2)sintheta)hati+(v_(1)sintheta-v_(2)costheta)hatj`
`=J/m[((cos^(2)theta)/2+sin^(2))hati+((costhetasintheta)/2-costhetasintheta)hatj]`
`=J/m[7/8hati-(sqrt(3))/8hatj]`
From here we can find directioin of unit vector in the direction of `vecv_(B)`.