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Oblique Collision

If m2 again...

Question

If $m_{2}$ again comes down and collides with $m_{1}$ , then after the collision :

m_{1}

will rise the same height as risen by

m_{2}

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m_{1}

will rise the less height as risen by

m_{2}

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m_{1}

will rise the more height as risen by

m_{2}

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m_{1}

and

m_{2}

will move in opposite directions

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Solution

The correct option is A $m_{1}$ will rise the same height as risen by $m_{2}$
Let velocity of $m_{1}$ just before the collision is $v_{1}$ and velocity of $m_{2}$ just after the collision is $v_{2}$ .
Energy conservation for $m_{1}$ from highest point to the lowest point:
$\frac{m_{1} v_{1}^{2}}{2} = m_{1} g (2 r)$
Energy conservation for $m_{2}$ from lowest point to the level of the center:
$m_{2} g (r) = \frac{m_{2} v_{2}^{2}}{2}$
Conservation of momentum gives
$m_{1} v_{1} = m_{2} v_{2}$
solving the first and second equations gives:
$v_{1} = \sqrt{4 g r}$
and
$v_{2} = \sqrt{2 g r}$
Now when B returns to the lowest point, its velocity must be the same as its initial velocity and let velocity of A just after the collision is $V_{1}$ and that of B is $V_{2}$ . Conservation of momentum implies:
$m_{2} v_{2} = m_{1} V_{1} + m_{2} V_{2}$
and since coefficient of restitution is $\frac{1}{\sqrt{2}}$ , we get:
$\frac{V_{1} + V_{2}}{v_{2}} = \frac{1}{\sqrt{2}}$
Solving for $V_{1}$ and $V_{2}$ gives:
$V_{1} = V_{2} = \frac{m_{2}}{(m_{1} - m_{2})} \sqrt{4 g r}$
Since velocity of the two blocks after second collision is same, they will rise to the same height.

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If m2 again comes down and collides with m1, then after the collision :

If $m_{2}$ again comes down and collides with $m_{1}$ , then after the collision :