Question

Three masses $m,2m$ and $3m$ are moving in x-y plane with speed $3u,2u$ and $u$ respectively as shown in figure. The three masses collide at the same point at $P$ and stick together. The velocity of resulting mass will be :

• A. $\frac{u}{12}\left(\hat{i}+\sqrt{3}\hat{j}\right)$
• B. $\frac{u}{12}\left(\hat{i}-\sqrt{3}\hat{j}\right)$
• C. $\frac{u}{12}\left(-\hat{i}+\sqrt{3}\hat{j}\right)$
• D. $\frac{u}{12}\left(-\hat{i}-\sqrt{3}\hat{j}\right)$

Answer 1

Answer: (D)

From the law of conservation of momentum we know that,
$m_1{u}_1+m_2{u}_2+....=m_1{v}_1+m_2{v}_2+....$
Given $m_1=m,m_2=2m$ and $m_3=3m$ and $U_1=3u,u_2=2u$ and $u_3=u$
Let the velocity when they stick $=\vec{v}$
Then, according to question.
$m\times 3u\left(\hat{i}\right)+2m\times 2u\left(-\hat{i}cos60^{°}-\hat{j}sin60^{°}\right)$
$+3m\times u\left(-\hat{i}cos60^{°}+\hat{j}sin60^{°}\right)$
$=\left(m+2m+3m\right)\vec{v}$
$\Rightarrow 3mu\hat{i}-4mu\frac{\hat{i}}{2}-4mu\left(\frac{\sqrt{3}}{2}\hat{j}\right)-3mu\frac{\hat{i}}{2}+3mu\left(\frac{\sqrt{3}}{2}\hat{j}\right)=6m\vec{v}$
$\Rightarrow mu\hat{i}-\frac{3}{2}mu\hat{i}-\frac{\sqrt{3}}{2}mu\hat{j}=6m\vec{v}$
$\Rightarrow \vec{v}$$\frac{u}{12}\left(-\hat{i}-\sqrt{3\hat{j}}\right)$From the law of conservation of momentum we know that,
$m_1{u}_1+m_2{u}_2+....=m_1{v}_1+m_2{v}_2+....$
Given $m_1=m,m_2=2m$ and $m_3=3m$ and $U_1=3u,u_2=2u$ and $u_3=u$
Let the velocity when they stick $=\vec{v}$
Then, according to question.
$m\times 3u\left(\hat{i}\right)+2m\times 2u\left(-\hat{i}cos60^{°}-\hat{j}sin60^{°}\right)$
$+3m\times u\left(-\hat{i}cos60^{°}+\hat{j}sin60^{°}\right)$
$=\left(m+2m+3m\right)\vec{v}$
$\Rightarrow 3mu\hat{i}-4mu\frac{\hat{i}}{2}-4mu\left(\frac{\sqrt{3}}{2}\hat{j}\right)-3mu\frac{\hat{i}}{2}+3mu\left(\frac{\sqrt{3}}{2}\hat{j}\right)=6m\vec{v}$
$\Rightarrow mu\hat{i}-\frac{3}{2}mu\hat{i}-\frac{\sqrt{3}}{2}mu\hat{j}=6m\vec{v}$
$\Rightarrow \vec{v}$$\frac{u}{12}\left(-\hat{i}-\sqrt{3\hat{j}}\right)$