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`]

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)]

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A block of mass m slides down an inclined plane of inclination theta with uniform speed The coefficient of friction between the block and the plane is mu . The contact force between the block and the plane is .

A block of mass m slides down an inclined plane of inclination theta with uniform speed. The coefficient of friction between the block and the plane is mu . The contact force between the block and the plane is