A small block of mass M moves on a frictionless surface of an inclined plane, as shown in the figure. The…

A small block of mass M moves on a frictionless surface of an inclined plane, as shown in the figure. The angle of the incline suddenly changes from 60° to 30° at point B. The block is initially at rest at A. Assume that collisions between the block and the incline are totally inelastic (g=10 m s-2). Figure:



The speed of the block at point C, immediately before it leaves the second incline is -
  1. 120 m s-1
  2. 105 m s-1
  3. 90 m s-1
  4. 75 m s-1

Solution

The height of the point B is

3 3 m tan 3 0 = 3 3 m 1 3 = 3 m

The energy of the block at B is

KE+PE=12M45 m s-12+Mg3m

The energy at the block at C is

KE=12MV2

The principle of conservation of energy gives

12MV2=12M45 m s-12+M10 m s-23m

This gives V=105 m s-1

Therefore, the choice (105  m s-1) is correct.

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