# Momentum

A small block of mass m is placed on the top of a right triangular block of mass M, which in turn is placed on a horizontal surface. All the surfaces are frictionless. Angle of inclination is α. Height of the triangular block is h. Find the velocity of the triangular block when the smaller block reaches the bottom

146
Amartya Kumar Mistry ·

m√2gh/M

15
Anomitro Kundu ·

Mαh2m2

486
Niraj kumar Jha ·

v=(√2gh*m)/M

145

The answer is actually quite a complicated expression

1163
Akash Anand ·

For now ..just take a hint.
You have to do just two things here.
1) Since initially system was at rest and only force acting on the block is gravity so momentum of the system along x-axis will be zero.

2) Since the body go down by a height h so loss in potential energy will be gain in K.E of both the blocks will be equal to that.

Just apply this two concepts, one more thing which is constraint here is ..since the block is itself on the wedge so whatever be the speed of the block w.r.t ground, you have to take the speed of the wedge into consideration as well.

• Ranadeep Roy In case of initial energy, we have to consider the PE of only small block and ignore the dimension of wedge??
• Akash Anand Which initial energy you are talking about?