`PQR` is a wedge fixed to the floor. Two blocks `A` & `B` of equal mass `m=1kg` is placed on either side of wedge connected through a string passing over the pulley `P`. Initially `PB=PA`. Another block `C` of mass `m=1kg` moving with a velocity `10m//s` collides inelasically with `A`. All the surfaces are smooth and pulley & string are massless.
(`a`) What should be the minimum length of the string `AB` so that block `A` does not hit the pulley ?
(`b`) What is the loss in meechanical energy from the instant `C` hits `A` to the instant when `B` starts moving upward ? (`g=10m//s^(2)`)

Explanation

(`a`) Conservation of momentum,
`m_(c )v_(c )=(m_(A)+m_(C ))xxv_(1)`, `v_(1)=` common velocity of (`A+C`) mass.
`:. V_(1)=(m_(c )v_(c ))/(m_(A)+m_(C ))=(1xx10)/(2)=5m//s`.
String slacks and `B` starts moving downward under `ac c=g sin 45^(@)`.
`:. (A+C)` moves under retardation of `g sin45^(@)` and come to rest after time
`5=10xx(1)/(sqrt(2))t rArr t=0.707 sec`.
Distance moved in this time
`S_(1)=(v_(1)^(2))/(2gsin45^(@))=(5xx5xxsqrt(2))/(2xx10)=(5)/(4)sqrt(2)=1.76m`.
Now, string will again become taut and `(A+C)` will experience a jerk.
Velocity of `B` at this time `=5m//s`
`:.` Conservation of momentum
for `B : 1xx5-J=1xxv_(2)`, `J=`impulse generated
for `(A+C) : J=2xxv_(2)`, `v_(2)=` common velocity of `(A+C)` after string has become taut
`:. v_(2)=(5)/(3)m//s`
Now `(A+C)` will start moving upward with `v_(2)` velocity under retardation
`a=((m_((A+C))-m_(B))/(m_(A+C)+m_(B)))gsin45^(@)=((2-1)/(3))xx10xx(1)/(sqrt(2))=(10)/(3sqrt(2))m//s^(2)`
`:. (A+C)` will come to rest after moving distance
`S_(2)=((5//3)xx(5//3))/(2xx(10//3sqrt(2)))=(5)/(6sqrt(2))m=0.59m`
`:. ` Net length risen by `(A+C)=S_(1)+S_(2)=2.35m`.
`:. ` Length of the string must be at least `l=2xx2.35=4.7m`
(`b`) Energy loss in collision `=(1)/(2)m_(C )v_(c)^(2)-(1)/(2)(m_(A)+m_(C ))v_(1)^(2)`
`=(1)/(2)xx1xx100-(1)/(2)xx2xx25=25J`
Loss in impulsive jerk
`=(1)/(2)m_(B)v_(1)^(2)-(1)/(2)(m_(a)+m_(A)+m_(c ))v_(2)^(2)=(1)/(2)xx1xx25-(1)/(2)xx3xx(25)/(9)`
`=(25)/(2)[1-(1)/(3)]=(25)/(3)J`
`:. ` Net loss `=25+(25)/(3)=(4xx25)/(3)=(100)/(3)=33.33J`