Rakhshan S

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3. A small block, mass m, slides down the inner curved surface of a larger block (mass M), which has a circular path cut into the block, as shown in the figure. The large block rests on a table. Both blocks move in the absence of friction. The blocks are initially at rest. The smaller block starts sliding from the top of the path, as shown. Find the velocity v of the cube as it leaves the block. [5 marks]

          3. A small block, mass m, slides down the inner curved surface of a larger block (mass M), which has a circular path cut into the block, as shown in the figure. The large block rests on a table. Both blocks move in the absence of friction. The blocks are initially at rest. The smaller block starts sliding from the top of the path, as shown. Find the velocity v of the cube as it leaves the block.
[5 marks]

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3. A small block, mass m, slides down the inner curved surface of a larger block (mass M), which has a circular path cut into the block, as shown in the figure. The large block rests on a table. Both blocks move in the absence of friction. The blocks are initially at rest. The smaller block starts sliding from the top of the path, as shown. Find the velocity v of the cube as it leaves the block.
[5 marks]

Added by Jason S.

University Physics with Modern Physics

Hugh D. Young 14th Edition

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Solved by Expert Rakhshan S

03/20/2022

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The potential energy it had at the top of the path will be converted into kinetic energy. Show more…

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Transcript

00:01 Hello students, today we have a question in hand and the question state as follows.

00:05 We have a small block of mass m and we have a big block of mass capital m and they are both the big block of mass m is sitting on a surface and this small block of mass m is at this point and this is the curve surface area and which is a part of a bigger circle if you can see we can imagine a big circle like this and it is a one -fourth of that big circle so the question state is when this small block of mass small m falls from here and it goes here to this point what will be the velocity at this point right here when it leaves the surface so what we have to find is we have to find the velocity v the velocity with which the block of mass small m leaves the surface and in the question it is given that mu is zero now we will start by saying using the law of conservation of energy at this point and at this point so m g h it tells us about this point and this will be equal to mj h minus r plus half mv square capital m v square plus half small mv square now this is the velocity of the big box of this block and this is the velocity of the small block and this is the potential energy at this height at this height and this will be our equation also using the conservation of momentum by using conservation of momentum what we can write is capital m velocity equals to small m velocity of the small box let me correct it and this will be our second equation moving on so now what we have to do is we have to substitute the second equation in first and then we do that we can write m g h equals to m g h minus r plus half mv square plus half mv square we can write this because from second location now m g h equals to m g h minus r plus mv square and and m is getting cancelled from this equation m, m and m, and so we can write g h equals to g h minus r plus v square.

04:21 Now we have to find the velocity and that was our initial intention...

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