A bob of mass `10m` is suspended through an inextensible string of length `l`. When the bob is at rest in equilibrium position, two particles, each of mass `m`, strike it as shown in Fig. The particles stick after collision. Choose the correct statement from the following:

Options

A. Impulse in the string due to tension is `2"mu"`
B. Velocity of the system just after collision is `v=(usqrt(3))/14`
C. Loss of energuy is `137/28"mu"^(2)`
D. Loss of energy is `137/56"mu"^(2)`

Answer

A

Explanation

Momentum in vertical direction
`"mu"cos60^(@)+3"mu"cos60^(@)=4"mu"cos60^(@)=2"mu"`
This momentum becomes zero due to impulse in string.
Hence, impulse in string `=2"mu"`
Conservation of linear momentum in horizontal direction
`m(3u)sin60^2-"mu"sin60^(@)=12mv`
`impliesv=(sqrt(3)u)/12`
Loss of energy `=1/2m(9u^(2))+1/2"mu"^(2)-1/2xx12mxx((sqrt(3)u)/12)^(2)`
`=(39"mu"^(2))/8`