Question

A uniform bar of length $6\text{m}$ and mass $16\text{kg}$ lies on a smooth horizontal table. Two point masses $2\text{kg}$ and $4\text{kg}$ moving in the square horizontal plane with $6\text{m/s}$ and $3\text{m/s}$ respectively strike the bar as shown in figure and stick to the bar after collision. Denote angular velocity (about the centre of mass), total energy and centre of mass velocity by $\omega ,E$ and $v_c$ respectively. Then after the collision:

• A. $v_c=0\text{m/s}$
• B. $\omega =\frac{9}{5}\text{rad/s}$
• C. $\omega =\frac{3}{5}\text{rad/s}$
• D. $E=\frac{54}{5}\text{J}$

Answer 1

Answer: (A), (C), (D)

$E=\frac{54}{5}\text{J}$

Given that,
Length of the bar $L=6\text{m}$
Mass of the bar $m=16\text{kg}$
Mass of the ball $m_1=4\text{kg}$; $m_2=2\text{kg}$


No external force is acting on the system, so we can apply conservation of linear momentum before and after collision.
$P_i=P_f$
$\Rightarrow [4\times (-3\left)\right]+(2\times 6)+(16\times 0)=(4+2+16)\times v_c$
$\Rightarrow 22v_c=0\Rightarrow v_c=0$
So, the velocity of centre of mass of the system after collision is zero.

Now, apply angular momentum conservation about the centre of bar.
The centre of the bar $O$, will lie at $3\text{m}$ from one end.
$L_{i,O}=L_{f,O}$
$\Rightarrow (12\times 1)+(12\times 2)=I_{total}\omega$
[taking anticlockwise sense +ve]
$\Rightarrow 12+24=\omega [(\frac{1}{12}\times 16\times 6^2)+(4\times 1^2)+(2\times 2^2\left)\right]$
$[\because I_{total}=\frac{1}{2}m{L}^2+m_1{a}_1^2+m_2{a}_2^2]$
$\Rightarrow 36=\omega [48+4+8]$
$\Rightarrow \omega =\frac{36}{60}=\frac{3}{5}\text{rad/s}$

Now, energy after collision
$E=\frac{1}{2}{I}_{total}{\omega }^2=\frac{1}{2}\times 60\times \frac{3}{5}\times \frac{3}{5}$
$\Rightarrow E=\frac{54}{5}\text{J}$

Hencem options (a), (c) and (d) are correct.