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A small cube of mass m slides down a circular path of radius R cut into a larger block of mass M, as shown in the figure. M rests on a table, and both blocks move without friction. The blocks are initially at rest, and m starts from the top of the path. The velocity v of the cube as it leaves the block is

vineet chatterjee , 11 Years ago

Grade 12

2 Answers

Sumit Majumdar

Last Activity: 11 Years ago

Dear student,
We dont have the figure given with the question.
So if we assume that a small cube starts from the top of a semi circular path, we would have the following:
$mgR=\frac{1}{2}\left ( m+M \right )v^{2}$
Hence, we have:
$v=\sqrt{\frac{2mgR}{m+M}}$ .
This is the velocity with which both the blocks move together.
Regards
Sumit

Aayush

Last Activity: 7 Years ago

v= velocity of mass m

V= velocity of block M

Gravitational force is acting which is conservative in nature therefore mechanical energy will remain constant

MEi=MEf

0+mgR=0+1/2m(v-V)²

^{now conserve the momentum}

^Pi=pf

^0=m(v-V)+MV

^V=mv/mM

^{Now put it in mechanical conservation equation}

mgR=1/2m(v-mv/m+M)²

v² = 2gr(1-- m/m+M)

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