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1.

`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)`)

Answer» (`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`
Discussion
2.

A particle of mass `m` is released from height `h` on smooth quarter circular fixed wedge. The horizontal surface `AB` following the circular path at bottom of wedge is rough with coefficient of friction `mu` between surface and `m`. Find the distance from bottom of wedge where the mass will stop. A. `(2h)/(mu)`B. `(h)/(mu)`C. `(h)/(mug)`D. none of these

Answer» By work energy theorem,
`mgh-mu mgs=0`
`rArrs=(h)/(mu)`
Discussion
3.

An iron ball of mass `m`, suspended by a light inextensible string of length `l` from a fixed point `O`, is shifted by `theta_(0)` as shown so as to strike the vertical wail perpendicularly.The maximum angle made by the string with vertical after the first collision, if `e` is the coefficient of restitution, is A. `sin^(-1){1-e^(2)(1-costheta_(0))}`B. `cos^(-1){1-e^(2)(1-costheta_(0))}`C. `tan^(-1){1-e^(2)(1-costheta_(0))}`D. zero

Answer» Speed just before collision `=sqrt(2gl(1-costheta_(0)))`
Speed just after collision`=esqrt(2gl(1-costheta_(0)))`
From `COE`,
`(1)/(2)me^(2)[2gl(1-costheta_(0))]=mgl(1-costheta)`
`:. theta=cos^(-1){1-e^(2)(1-costheta_(0))}`
Discussion
4.

A ping-pong ball of mass `m` is floating in air by a jet of water emerging out of a nozzle. If the water strikes the ping-pong ball with a speed `v` and just after collision water falls dead, the mass flow rate of water in the nozzle is equal to :A. `(2mg)/(v)`B. `(mv)/(g)`C. `(mg)/(v)`D. None of these

Answer» The impact force `F=(Deltap//Deltat)`, where `Deltap=` change of momentum of water of mass `Deltam` striking the ball with a speed `v` during time `Delta t`. Since water falls dead after collision with the ping-pong ball
`Deltap=DeltamvrArrF=v(Deltam)/(Deltat)`
where `(Deltam)/(Deltat)=` rate of flow of water in the nozzle.
Since the ball is in equilibrium
`F-mg=0rArrF=mg`
`rArr(vDeltam)/(Deltat)=mg`
`rArr(Deltam)/(Deltat)=(mg)/(v)`
Discussion