Question

A rod of mass $6\text{m}$ and length $6a$ is kept on a horizontal smooth surface. Two point masses of masses $2\text{m}$ and $\text{m}$ respectively, moving perpendicular to the rod, collide with the rod and stick to the rod after collision as shown in the column-II. Point C represents centre of mass of the rod. $AC=2a$ and $CB=a$. $v$ is the linear speed of the centre of mass of the system just after collision and $\omega$ is the angular speed of the system just after collision.

• A. A – (q,r), B – (s), C – (p,s), D – (p,q, r)
• B. A – (r,s), B – (p,r), C – (q,s), D – ( r,q)
• C. A – (q,r), B – (p,r), C – (q,s), D – ( p,s)
• D. A – (q,r), B – (p,q), C – (r,s), D – ( p,r)

Answer 1

The correct option is A A – (q,r), B – (s), C – (p,s), D – (p,q, r)

$I=\frac{1}{12}\times 6m\times 36a^2+m\left(4a^2\right)+2m\left(a^2\right)=24m{a}^2$ for $p,q,r\text{and}s$

For $p$,
Applying conservation of linear momentum in vertical direction
$2m{v}_0-m{v}_0+6m\times 0=(2m+m+6m)v$
$\Rightarrow \overrightarrow{v}=\frac{v_0}{9}\hat{j}$
Now applying conservation of angular momentum we get
$24m{a}^2\overrightarrow{\omega }=m{v}_{0}2a\hat{k}+2m{v}_{0}a\hat{k}$
$\Rightarrow \overrightarrow{\omega }=\frac{v_0}{6a}\hat{k}$

For $q$,
$2m{v}_0-2m{v}_0+6m\times 0=(2m+m+6m)v$
$\Rightarrow \overrightarrow{v}=024m{a}^2\overrightarrow{\omega }=2m{v}_{0}2a\hat{k}+2m{v}_{0}a\hat{k}\Rightarrow \overrightarrow{\omega }=\frac{v_0}{4a}\hat{k}$

For $r$,
$2m{v}_0-2m{v}_0+6m\times 0=(2m+m+6m)v$
$\Rightarrow \overrightarrow{v}=024m{a}^2\overrightarrow{\omega }=4m{v}_{0}a\hat{k}+2m{v}_{0}a\hat{k}$
$\Rightarrow \overrightarrow{\omega }=\frac{v_0}{4a}\hat{k}$

For $s$,
$-m{v}_0-2m{v}_0+6m\times 0=(2m+m+6m)v$
$\Rightarrow \overrightarrow{v}=\frac{v_0}{3}(-\hat{j}),24m{a}^2\overrightarrow{\omega }=m{v}_{0}2a\hat{k}+2m{v}_{0}a\widehat{(-k)}$
$\Rightarrow \overrightarrow{\omega }=\overrightarrow{0}$