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

0.84, 15 kg

Explanation

Considering the motion of block of mass m from A to B. Acceleration of the block.
`a = -[(m g sin theta + f)/(m)] = - [(m g sin theta + mu m g cos theta)/(m)]`
Here
`cos theta = sqrt(1 - sin^2 theta) = sqrt(1 - (0.05)^2) = 0.99 =1`
`a = - [g sin theta + mu g cos theta] = -[sin theta + mu cos theta)g`
`= -10[0.05 + 0.25 xx 1] -3 m//s^2`
Let velocity of block A just before collision is `v_1`.
Using `v^2 - u^2 = 2as`
`v_1^2 = (10)^2 + 2(-3) xx 6 implies v_1 = 8m//s`
Now considering situation just after collision

Let `v_2` be the velocity of mass m after collision and `v_3` be the velocity of mass M after collision.
The block of mass M moves from B to C (distance 0.5 m) and comes to rest. Now considering the motion of M from B to C. Again using `v^2 - u^2 = 2 as`
Here `u = v_3, v = 0` and acceleration `a = –3 m//s^2`
(here acceleration of M is same as a previous case because all other parameters are the same)
Hence `(0)^2 - v_3^2 = 2(-3) xx 0.5 implies v_3 = sqrt(3) = 1.73 m//s`
The block of mass m moves from B to A after collision
Using W.E. theorem
`W_("total") = Delta K = W_("friction") + W_("gravity") = K_A - K_B`
`=(mu m g cos theta)s + mgh = 1/2 mv^2 - 1/2 mv_2^2`
Here s = 6m, h = 0.3 m and v = 1 m/s

Substituting the value in above equation
We get `v_2 = ¬5 m//s" ":." "` Coefficient of restitution
`e = |("Relative velocity of seperation")/("Relative velocity of approach")| = |(5 + 1.73)/(8 - 0)| = 0.84`
Now for calculating mass apply conservation of linear momentum before and after collision
`m v_1 = M v_3 - mv_2 , M = (m(v_1 + v_2))/(v_3) = (2(8+5))/(1.73) = 15.02kg`.