# What is the velocity of the wedge after the block reaches the horizontal surface and what is the height of the wedge in the following case:A small block of mass 80 g is released from rest at the...

What is the velocity of the wedge after the block reaches the horizontal surface and what is the height of the wedge in the following case:

A small block of mass 80 g is released from rest at the top of a curved frictionless wedge of mass 396 g which sits on a frictionless horizontal surface. The acceleration due to gravity is 9.81 m/s^2. Right is the positive direction. When the block leaves the wedge, its velocity is 4.95 m/s to the right.

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A small block of mass 80 g is released from rest at the top of a curved frictionless wedge of mass 396 g which sits on a frictionless horizontal surface.

The velocity of the block when it leaves the wedge is 4.95 m/s to the right. Acceleration due to gravity is 9.81 m/s^2. Let the height that the block has to slide down before its velocity increases from 0 to 4.95 m/s be h. Use the relation 3.95^2 - 0^2 = 2*9.81*h

=> h = 0.7952 m/s

When the block leaves the wedge its momentum is 80*3.95 g*m/s. Following the law of conservation of momentum the wedge moves to the left with a speed v such that 396*v = 80*3.95

=> v = 0.7979 m/s

The wedge moves to the left at a velocity 0.7979 m/s and the height of the wedge is 0.7952 m.