Two blocks of mass `2` kg and `M` are at rest as an inclined plane and are sparated of by a distance of `6.0` m as shown in figure . The coefficient of friction between each of the block and the inclined plane is `0.25` . The `2` kg block is given a velocity of `10.0` m//s up the inclined plate if collied with `M` comes back and has a velocity of `1.0`m//s when if rached its inital position , The ofter block `M` ofter the collisiin its resitition . THe ofter the colition its `0.5` m up and come to rest caleslate the coefficient of resttuion between the block and the mass of the block `M` [Take sin theta +n tan theta = 0.05`and `a = 10m//s^(2)]`

Answer

A, B, C, D

Explanation

From `A` and `B`
u = 10 m//s ` (given)
a = -[(mg sin theta + f)/(m)] = -[(mg sin theta + mu mg cos theta )/(m)] `

= -[ g sin theta + mu cos theta ] = -g[sin theta + mu cos theta ]`
= -10[0.05 + 0.25 xx 0.99 m//s^(2)`
`v = ?`
`s = 6m`
`v^(2) - u^(2) = 2as rArr x^(2) = 100 + 2(-2.99) xx 6 rArr v = 8 m//s`
`rArr The velocity of mass `M` just befor collision is `8 m//s ` THe velocity of mass `M` just before collision is `0 m//s ` (given)
After Collision
Let `v_(1)` be the velocity of mass `m` after collision and `v_(2) ` be the velocity of mass `M` after collision . Body of mass `M` moving from `B` and `C` and coming to rest .
`u = v_(2) `
`u = 0`
`u = -2.99`
(same mass a is independent of mass `s = 0.5`
`v^(2) - u^(2) = 2as rArr (0)^(2) - v_(2)^(2) = 2 (-2.99) xx 0.5`
`rArr v_(2) = 1.75 m//s `
Body of mass `m` moving from `B` to `A` after collision

`sin theta = (h)/(6) `
`h = 6 sin theta = 6 xx 0.05`
`u = v_(1)`
`v = + 1 m//s`
(K.E. + P.E. )_(intial ) = (K.E. + P.E. )_(final)+ W_(friction)`
`(1)/(2) mv_(1)^(2) + mgh = (1)/(2) mv^(2) + 0 + mu mgs`
`(1)/(2) v_(1)^(2) + 10 xx (6 xx 0.05) = (1)/(2) (i) ^(2) + 0.25 xx 10 xx 6 `
`v_(1) = -5 m//s`
Coefficient of restitution
`e = [(Relative velocity of separation )/(Relative velocity of approach])`
`= (|-5-1.73)/(8 - 0)|` = 0.84`
On appling converation of linear momention and after collision we get
` 2xx 8 +M xx 0 = 2 xx (-5) + M(1.73)` ltbbrgt `:. M = (26)/(1.75) = 15.02 kg `