Question

A block of mass in is pushed towards
a movable wedge of mass 2m and
height h with a velocity u. All surfaces
are smooth. The minimum value
of a for which the block will reach
the top of the wedge is:​

Answer 1

Given:

The mass of the m= 2m

The height is h

The velocity is u

To find:

The minimum value of the block reach the top.

Solution:

The center of mass is the combination of two that is Block+ wedge

v=2mu/(2m+2mn)

v=u/(1+n)

Now the kinetic energy to potential energy

(2m+2mn)v²/2=μ²/2 - 2mgh

m(1+n)u²/(1+n)²=μ²/2-2mgh

u=√(1+n)(μ²/2 - 2mgh)/m

That is the minimum value of the block  to reach the top