Two identical balls `A` and `B`. each of mass `2 kg` and radius `R`, are suspended vertically from inextensible strings as shown in Fig. The third ball `C` of mass `1 kg` and radius `r=(sqrt(2)-1)R` falls and hits `A` and `B` symmetrically with `10 m//s`. Speed of both `A` and `B` just after the collision is `3 m//s`.

speed of `C` just asfter collision is

Options

A. `2m//s`
B. `2sqrt(2)m//s`
C. `5m//s`
D. `(sqrt(2)-1)m//s`

Answer

A

Explanation

First calculate `theta=45^(@)`. As the balls `A` and `B` are constained to move horizontally (immediately after collision) if `'I'` be the impulse impated by ball `C` to each of `A` and `B`, the mpulse received by ball `C` from them would be `2Isintheta`

Now, each of ball `B` and `C` received impulse `'I'` as shown in figure. but moves impulse imparted to ball `B` and `C` by the respective strings and hence,
`Icostheta=M_(A)V_(A)=M_(B)V_(B)`
Now, for ball `C` if its final velocity is `V_(C)^(')` downwards we have
`M_(C)V_(C)^(')=M_(C)V_(C)-2Isintheta`
`impliesV_(C)^(')=V_(C)-2(MA)/(MC)V_(A)` (`:'theta=45^(@))`
`=-2m//s` (`-ve` sign indcates that it is directed upwards)