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`]

Explanation

For the block of mass 2 kg (=m) in going from the A to B
`(KE+PE)_("initial")=(KE+PE)_("final")+W_("friction")`
`(1)/(2)mv_(1)^(2)+0=(1)/(2)mv_(f)^(2)+mgh_(1)+F_(f)s`
of `v_(f)^(2)=2[(1)/(2)v_(l)^(2)-gh_(1)-mu gs]`
Substituting the given data, we get
`v_(f)^(2)=2[(1)/(2)(10.0 m//s^(2))-(10m//s^(2))(0.30m)-(0.25)(10m//s^(2))(6.0m)]`
Or, `v_(f)=8 m//s`
For the block of mass 2 kg (=m) coming back from the position B to A
`(KE+PE)_("initial")=(KE+PE)_("final")+W_("friction")`
`(1)/(2)mu_(l)^(2)+mgh_(1)=(1)/(2)mu_(f)^(2)+0+F_(f)s`
`u_(l)^(2)=2[(1)/(2)u_(f)^(2)+ mu gs-gh_(l)]`
Substituting the given data, we get
`u_(l)^(2)=2[(1)/(2)(-1.0m//s^(2))+(0.25)(10 m//s^(2))(6.0m)-(10m//s^(2))(0.30m)]`
Or `u_(l)=-5.0 m//s`

For the block of Mass M in going from the position B to C
`(KE+PE)_("initial")=(KE+PE)_("final")+W_("friction")`
`(1)/(2)Mc_(1)^(2)+Mgh_(1)=(1)/(2)Mc_(r )^(2)+Mg(h_(1)+h_(2))+F_(f)S_(2)`
or `(1)/(2)c_(1)^(2)=(1)/(2)c_(1)^(2)+gh_(2)+mu gs_(2)`
`(1)/(2)c_(i)^(2)=(1)/(2)(0)+(10 m//s^(2))(0.25 m)+(0.25)(10 m//s^(2))(0.5m)`
`c_(i)=sqrt(3)m//s`
`e=(u_(i)-c_(i))/(v_(f)-c_(i))=((-5.0m//s)-(-sqrt(3)m//s))/((8.0 m//s)-0)=(5+sqrt(3))/(8)=0.84`