486v=(√2gh*m)/M
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
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5 Answers
15Mαh2m2
Sourish Ghosh Incorrect dimension.Upvote·0· Reply ·2013-03-23 01:26:45
Shaswata Roy How can alpha come in the answer?
1161For 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?
