Two blocks of masses `2kg` and M are at rest on an inclined plane and are separated by a distance of `6.0m` as shown. The coefficient of friction between each block and the inclined plane is `0.25`. The `2kg` block is given a velocity of `10.0m//s` up the inclined plane. It collides with M, comes back and has a velocity of `1.0m//s` when it reaches its initial position. The other block M after the collision moves `0.5m` up and comes to rest. Calculate the coefficient of restitution between the blocks and the mass of the block M.
[Take `sintheta=tantheta=0.05` and `g=10m//s^2`]

Answer

`e = 0.84, M = 15.21 kg`

Explanation

For body of mass m from A to B `u = 10 m//s` (given)
`a = -[(mg sin theta + f)/(m)] = -[(mg sin theta + mumg cos theta)/(m)]`
`= -[g sin theta + mug cos theta] = - g [ sin theta + mu cos theta]`
`= - 10 [0.05 + 0.25 xx 0.99]`
`= - 2.99 m//s^(2)`
`v^(2) - u^(2) = 2as`
`rArr v = sqrt(100 + 2 - (-2.99) xx 6) = 8 m//s`

After collision
Let `v_(1)` be the velocity of mass m after collision. Body `v_(2)` be the velocity of mass M after collision. Body of mass M moving from B to C and Coming to rest.
`u = v_(2), v = 0 " " a = - 2.99 m//s^(2)`
and `s = 0.5 " " v^(2) - u^(2) = 2as`
`rArr (0)^(2) = v_(2)^(2) = 2(-2.99) xx 0.5`
`rArr v_(2)^(2) = 1.73 m//s`
Body of mass m mocing from B to A after collision
`u = v_(1), v = +1 m//s`
`(K.E. + P.E.)_("initial") = (K.E. + P.E.)_("final") + W_("frication")`
`(1)/(2)mv_(1)^(2) + mgh = (1)/(2)mv^(2) + 0 + mugs`
`(1)/(2)v_(1)^(2) + 10 xx (6 xx 0.05) = (1)/(2) (1)^(2) + 0.25 xx 10 xx 6`
`v_(1) = -5 m//s`
`sin theta = (h)/(6)`
`h = 6 sin theta = 6 xx 0.05`

`:.` Coefficient of restituation
`e = |("Relative velocity of sepaeration")/("Relative velocity of approach")|`
`= |(-5 - 1.73)/(8 - 0)| = 0.84`