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Perfectly Inelastic Collision

A sphere of r...

Question

A sphere of radius $R$ and mass $M$ collides elastically with a cubical block of mass $M$ and side $2 R$ . The entire system is on a smooth horizontal ground. Given that the sphere was rolling without slipping with an angular velocity $ω$ at the time of collision. The velocities of the sphere and the block after the collision are :

ω_{s p h e r e} = 0, v_{s p h e r e} = 0, v_{b l o c k} = v

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ω_{s p h e r e} = ω, v_{s p h e r e} = 0, v_{b l o c k} = v

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ω_{s p h e r e} = 0, v_{s p h e r e} = 0, v_{b l o c k} = 0

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ω_{s p h e r e} = \frac{ω}{2}, v_{s p h e r e} = \frac{v}{2}, v_{b l o c k} = \frac{v}{2}

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Solution

The correct option is B $ω_{s p h e r e} = ω, v_{s p h e r e} = 0, v_{b l o c k} = v$
Conservation of Angular momentum
$I w + m . V_{1 i} R = I w^{'} + m V_{1 f} R + m V_{2 f} R$
where $V_{1 f}$ final velocity of sphere
$V_{2 f}$ final velocity of block-----(1)
For elastic collision, we know
$V_{1 f} = (\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) V_{1 i} + \frac{2 m_{1}}{m_{1} + m_{2}} V_{2 i}$
$m_{1} = m_{2}$ and $V_{2 i} = 0$
$V_{2 f} = (\frac{2 m_{1}}{m_{1} + m_{2}}) V_{1 i} + (\frac{m_{2} - m_{1}}{m_{1} + m_{2}}) V_{2 i}$
$m_{1} = m_{2}$ and $V_{2 i} = 0 \Rightarrow V_{2 f} = 2 v i$
Using in $(1)$ gives $w = w^{'}$
$∴ w_{s p h e r e} = w, V_{s p h e r e} = 0, V_{b l o c k} = V$

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