Question

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 horizontal 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

Answer 1

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 $x =\frac{-R}{2}$                          

Answer 2

The shell move from its initial contact point with the surface is,

$x =\dfrac{-R}{2}$

Given-

Radius of the ball is $R$.

the mass of the ball is $m$.

The mass of the thin shell is $m$.

The radius of the thin shell is $2 R$.

For the two bodies with mass $m$, the center of mass can be given as,

$x_{cm}=\dfrac{m_{1} x_{1}+m_{2} x_{2} }{m_{1} +m_{2}}$

In the given question, the mass of both the bodies are equal and the given distance of center of mass for both bodies are also equal. Therefore,

$x_{cm}=\dfrac{m x_+mx }{m +m}$

$x_{cm}=\dfrac{2mx }{2m}$

$x_{cm} =x$

Distance for center of mass can also be given as,

$x_{cm} =m\times o$

$x_{cm} =m\times \dfrac{-R}{2m}$

$x_{cm} =\dfrac{-R}{2}$

Comparing both the values of the distance of center of mass we get,

$x =\dfrac{-R}{2}$

Hence, The shell move from its initial contact point with the surface is,

$x =\dfrac{-R}{2}$

For more about the center of mass follow the link below,

https://brainly.com/question/8662931