(From Tipler, problem 8-40 on p. 282). A wedge of mass M is placed on a frictionless horizontal surface, and a block of mass m is placed on the wedge, which also has a frictionless surface. The block’s center of mass moves downward a distance h as the block slides from its initial position to the horizontal floor. (a) What are the speeds of the block and of wedge as they separate from each other and go their own ways? (b) Check the plausibility of your answer to (a) by considering the limiting case when M is much greater than m.

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Xema8 Xema8 · Answer 1 · 2019-11-04T14:02:56+01:00

Answer:

Explanation:

The loss of potential energy of the block is converted into kinetic energy of both the block and wedge.

loss of potential energy

= mgh

If v₁ and v₂ be the velocity of block and wedge at the time of separation

applying conservation of energy in horizontal direction ( no force acting in horizontal direction )

m v₁ = M v₂

Applying conservation of mechanical energy

mgh = 1/2 mv₁² + 1/2 M v₂²

1/2 mv₁² + 1/2 M ( mv₁ / M )² = mgh

1/2 mv₁² ( 1 + m/M ) = mgh

v₁² = 2gh / ( 1 + m/M )

v₁ = [tex]\sqrt{\frac{2Mgh}{m+M} }[/tex]

v₂ =( m / M) v₁

v₂= m / M [tex]\sqrt{\frac{2Mgh}{m+M} }[/tex]

If M is much greater

M+m = M

v₁ = [tex]\sqrt{\ 2gh }[/tex]

v₂ = [tex]\frac{m}{M} \times\sqrt{\ 2gh }[/tex]