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Uniform Circular Motion

A uniform bar...

Question

A uniform bar of negligible cross sectional area of length 6a and mass 8m lies on a horizontal frictionless table. Two point masses m and 2m moving in opposite directions but in the same horizontal plane with speeds 2v and v respectively strike the bar at a distance a and 2a from one end and stick to the bar after the collision. Then, after the collision,

The velocity of centre of mass is

4 m s^{- 1}

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The angular speed of the bar with the masses stuck to it is

\frac{v}{15 a}

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The moment of inertia of the bar with masses stuck to it about the axis passing through the centre of the bar and perpendicular to its plane is $30 m a^{2}$

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The total energy of the bar is

\frac{3}{25} m v^{2}

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Solution

The correct option is C
The moment of inertia of the bar with masses stuck to it about the axis passing through the centre of the bar and perpendicular to its plane is $30 m a^{2}$

Since no external force is applied, the linear momentum is conserved. Hence, $V_{C M} = 0$
Moment of inertia of the system is,
$I = 2 m a^{2} + m (2 a)^{2} + [\frac{1}{12} \times 8 m \times (6 a)^{2}] = 30 m a^{2}$
Since no external torque is applied, the angular momentum of the system is conserved.
$\Rightarrow (2 m v \times a) + (m \times 2 v \times 2 a) = I w$
$\Rightarrow 6 m v a = 30 m a^{2} w \Rightarrow w = \frac{v}{5 a}$
$= \frac{1}{2} I w^{2}$
$= \frac{1}{2} (30 a m a^{)} {(\frac{V}{5 a})}^{2} = \frac{3}{5} m v^{2}$

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