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

v_{c} = 0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

ω = \frac{3 v}{5 a}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

ω = \frac{v}{5 a}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct options are
A $v_{c} = 0$
C $ω = \frac{v}{5 a}$
D $E = \frac{3 m v^{2}}{5}$
As $F_{e x t} = 0$ , so linear momentum of the system is conserved, Thus
$- 2 m v + m \times 2 v + 0 = (2 m + m + 8 m) v_{c}$
or
$v_{c} = 0$
Now as $τ_{e x t} = 0$ , angular momentum of the system is also conserved, Thus
$m_{1} v_{1} r_{1} + m_{2} v_{2} r_{2} = (I_{1} + I_{2} + I_{3}) ω$
or
$2 m v a + m (2 v) (2 a) = (2 m a^{2} + m (2 a)^{2} + \frac{8 m \times (6 a)^{2}}{12}) ω$
or
$6 m v a = 30 m ω a^{2}$
or
$ω = \frac{v}{5 a}$
Now as the system has no translatory motion but only rotatory motion we have
$E = \frac{1}{2} I ω^{2} = \frac{1}{2} (30 m a^{2}) (\frac{v}{2 a})^{2} = \frac{3}{5} m v^{2}$

Suggest Corrections

Similar questions

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