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Anni [7]
2 years ago
10

A ball of radius R and mass m is magically put inside a thin shell of the same mass and radius 2R. The system is at rest on a ho

rizontal frictionless surface initially. When the ball is, again magically, released inside the shell, it sloshes around in the shell and eventually stops at the bottom of the shell. How far does the shell move from its initial contact point with the surface
Physics
1 answer:
dlinn [17]2 years ago
7 0

Answer:

x =\frac{-R}{2}

Explanation:

From the question we are told that mass

Thin layer radius = 2R

Generally the expression for ths solution is given as

Xcm =(m*0 =m(-2R))/2m =-mR/(2m)=-R/2

the center of mass will not move at initial state  

Considering the center of mass of both bodies

xcm=\frac{m*x+m*x)}{2m} =x

x =\frac{-R}{2}

Therefore the enclosing layer moves x =\frac{-R}{2}                          

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Answer:

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Explanation:

FOR CONSTANT KINETIC ENERGY:

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FOR CONSTANT POTENTIAL ENERGY:

The potential energy of a body depends upon its height according to its formula:

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So, for Δh = 0 m/s

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So, for keeping potential energy constant, the books must be moved at constant height.

So, the correct option is:

<u>a. by moving the book without acceleration and keeping the height of the book constant</u>

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3 years ago
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Answer:

I think it is better if you read and shortly write my explanation

Explanation:

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