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: 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)`

A bob of mass `10m` is suspended through an inextensible string of len

1.	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: 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» Correct Answer - A 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`

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