Rotation Ques 88
- 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 the figure] and stick to the bar after collision. Denoting
angular velocity (about the centre of mass), total energy and centre of mass velocity by $\omega, E$ and $v _c$ respectively, we have after collision
(1991)
(a) $v _c=0$
(b) $\omega=\frac{3 v}{5 a}$
(c) $\omega=\frac{v}{5 a}$
(d) $E=\frac{3}{5} m v^{2}$
Analytical Answer Type Questions
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Answer:
Correct Answer: 88.(a, c, d)
Solution:
$$ \begin{aligned} \therefore \quad P _f & =0 \text { or } v _c=0 \\ L _i & =L _f \text { or }(2 m v) a+(2 m v)(2 a)=I \omega \end{aligned} $$
Here, $I=\frac{(8 m)(6 a)^{2}}{12}+m(2 a)^{2}+(2 m)\left(a^{2}\right)=30 m a^{2}$
Substituting in Eq. (i), we get
$$ \omega=\frac{v}{5 a} $$
Further, $E=\frac{1}{2} I \omega^{2}=\frac{1}{2} \times\left(30 m a^{2}\right) \frac{v}{5 a}^{2}=\frac{3 m v^{2}}{5}$
BETA
