19. Two biods of mass 2 kg and M are at rest on an inclined pa mi me saade by a distance of 6.0 m as shown. The officient of friction between each block zad the inclined plane is 0.25. The 2 g blod is given a velocity of 10.0 m/s up the ninei place. It collides with M. comes back and has a velocity of 10 m's when it reaches its initial position. The of block that the collision moves 0.5 m up and comes to test. Calculate the efficient of restitution between the blodas mod the mass of the block M. [Take in tan =0,05 and g = 10 m/s) (1999, 10M)

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 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 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)]

A block of mass 0.5 kg has an initial velocity of 10ms^(-1) down an inclined plane of angle 30^@ , the coefficient of friction between the block and the inclined surface is 0.2. The velocity of the block after it travels a distance of 10 m is

A block of mass m=4kg is placed oner a rough inclined plane as shown in figure. The coefficient of friction between the block and the plane is mu=0.5 . A force F=10N is applied on the block at an angle of 30^(@) .

A block of mass m is placed at rest on an inclination theta to the horizontal.If the coefficient of friction between the block and the plane is mu , then the total force the inclined plane exerts on the block is

A block of mass m is placed at rest on an inclination theta to the horizontal. If the coefficient of friction between the block and the plane is mu , then the total force the inclined plane exerts on the block is

Block of mass 10 kg is moving on inclined plane with constant velocity 10 m/s. The coedfficient of kinetic friction between incline plane and block is :-