A uniform bar of length $6 a$ and mass $8 m$ lies on a smooth horizontal table. Two point masses $m$ and $2 m$ , moving in the same horizontal plane with speeds $2 v$ and $v$ respectively strike the bar (as shown in figure) and stick to the bar after collision. By denoting angular velocity (about centre of mass), total energy and velocity of centre of mass after collision as $ω, E$ and $v_{o}$ respectively, we can say that

v_{o} = 0

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ω = \frac{6 v}{5 a}

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ω = \frac{3 v}{5 a}

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E = \frac{3}{5} m v^{2}

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Solution

The correct options are
A $v_{o} = 0$
C $ω = \frac{3 v}{5 a}$
D $E = \frac{3}{5} m v^{2}$
From conservation of momentum
$P_{i} = P_{f}$
$2 m v - 2 m v = 11 m v_{c m}$
$v_{c m} = 0$
Position of COM from O
$C . M \equiv \frac{2 a \times 2 m + 8 m \times 3 a + m \times 5 a}{11 m} = 3 a$
Conserving angular momentum about COM
$L_{n e t} = I ω$
$\Rightarrow 2 m v a + m (2 v) (2 a) = I ω$
$∴ I ω = 6 m v a$
$\Rightarrow (2 m a^{2} + m \times 4 a^{2} + \frac{1}{12} 8 m \times 36 a^{2}) \times ω = 6 m v a$
$\Rightarrow (6 a^{2} + 24 a^{2}) ω = 6 v a ∴ ω = \frac{v}{5 a}$
Energy of the system:
$E = \frac{1}{2} I ω^{2} = \frac{1}{2} \times 30 m a^{2} \times \frac{v^{2}}{25 a^{2}}$
$= \frac{3}{5} v^{2} \times m$
$∴ E = \frac{3}{5} m v^{2}$

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Similar questions

Q. A uniform bar of length

6 a

and mass

8 m

lies on a smooth horizontal table. Two point masses

m

and

2 m

moving in the same horizontal plane with speed

2 v

and

v

respectively, strike the bar (as shown in figure) and stick to the bar after collision. C represents centre of mass of bar. Denoting angular velocity (about the centre of mass), total energy and centre of mass by

ω

, E and

v_{c}

respectively, then after collision we have

A uniform bar of length 6a and mass 8 m lies on a smooth horizontal table. Two point masses m and 2m moving in the same horizontal plane with speed 2v and v respectively, strike the bar and stick to the bar after collision. Denoting angular velocity (about the centre of mass), total energy and centre of mass velocity by $ω$ , $E$ and $v_{c}$ respectively, we have after collision