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

A, B, D

Explanation

Let `v_1=` velocity of block `2kg` just before collision
`v_2=` velocity of block `2kg` just after collision
and `v_3=` velocity of block M just after collision.
Applying work energy theorem
(change in kinetic energy=work done by all the forces) at different stages as shown in figure.

`[1/2m{v_1^2-(10)^2}]=-6mumgcostheta-mgh_1`
`(m=2kg)`
or `v_1^2-100=2[6mugcostheta+gh_1]`
`costheta=sqrt(1-sin^2theta)=sqrt(1-(0.05)^2)=0.99`
`:. v_1^2=100-2[(6)(0.25)(10)(0.99)+(10)(0.3)]`
`impliesv_1=8m//s`

`DeltaKE=W_(f riction)+W_(gravity)`
`1/2m[(1)^2-(v_2^2)]=-6mumgcostheta+mgh_1`
or `1-v_1^2=2[-6mucostheta+gh_1]`
`=2[(-6)(0.25)(10)(0.99)+(10)(0.3)]`
`=-23.7`
`:. v_2^2=24.7` or `v_2=5m//s`

`DeltaKE=W_(f riction)+W_(gravity)`
`1/2M[0-v_3^2]=-(0.5)(mu)(M)gcostheta-Mgh_2`
or `-v_3^2=-mucostheta-2gh_2`
or `v_3^2=(0.25)(10)(0.99)+2(10)(0.025)`
or `v_3^2=2.975`
`:. v_3~~1.72m//s`
Now,
(i) Coefficient of restitution
`=("Relative velocity of separation")/("Relative velocity of approach")`
`=(v_2+v_3)/(v_1)=(5+1.72)/(8)=(6.72)/(8)`
or `e~~0.84`
(ii) Applying conservation of linear momentum before and after collision
`2v_1=Mv_3-2v_2`
`:. M=(2(v_1+v_2))/(v_3)=(2(8+5))/(1.72)=(26)/(1.72)`
`M=15.12kg`