Physics is cool Part-1

A symmetric block of mass M M with a hemispherical groove of radius r r rests on a smooth horizontal surface in contact with the wall as shown in the figure. A small block of mass m slides without friction from the initial position.Find the maximum velocity of M M .

Details and Assumptions

M = 1 , m = 9 , r = 5 M=1, m=9,r=5 .


The answer is 18.

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1 solution

Aditya Kumar
Mar 21, 2016

This is a very nice problem. First, we have to analyse the motion of the blocks.

  • When the small block covers the first half of semi-circle for the first time, the big block is at rest.

  • When the small block moves to the second half of the semi-circle, the big block moves to the right.

  • Subsequently, the big block continuously move to the right.

Now, velocity of small block at the bottom most part of the semi-circle is v i 2 g r v_{i}\sqrt{2gr} . This can be obtained by conserving energy.

When the block moves to the second half of the semi-circle, the big block attains velocity. Since, there is no friction anywhere, we can conserve momentum and energy using basic equations.

Conservation of Momentum:

m v i = m v 1 + M v 2 m{v}_{i}=m{v}_{1}+M{v}_{2}

Conservation of Energy:

m g r = m v 1 2 2 + M v 2 2 2 mgr=\frac{m{{v}_{1}}^{2}}{2}+\frac{M{{v}_{2}}^{2}}{2}

Solving these solutions, we get 2 pairs of solutions.

( 1 ) (1) v 1 = 2 g r {v}_{1}=\sqrt{2gr} and v 2 = 0 {v}_{2}=0

( 2 ) (2) v 1 = m M m + M 2 g r {v}_{1}=\frac { m-M }{ m+M } \sqrt { 2gr } and v 2 = 2 m m + M 2 g r {v}_{2}=\frac { 2m }{ m+M } \sqrt { 2gr }

Hence, v 2 m a x = 2 m m + M 2 g r {v}_{2max}=\frac { 2m }{ m+M } \sqrt { 2gr }

On inserting the values the answer comes out to be 18.

i did it by using the concept of relative velocity !

A Former Brilliant Member - 4 years, 6 months ago

Why is big block at rest when small block covers first half of semicircle?

Harsh Shrivastava - 5 years, 2 months ago

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That's because the normal force acting on the big block is in the direction of the wall. Hence, it remains stationary for first half of semicircle.

Aditya Kumar - 5 years, 2 months ago

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Oh yes! Thanks!

Harsh Shrivastava - 5 years, 2 months ago

Same solution!!

Aakash Khandelwal - 5 years, 2 months ago

simply awesome sol

rakshith lokesh - 3 years, 2 months ago

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