Question

Three particles $A$,$B$ and $C$ of masses $10\text{g}$, $20\text{g}$ and $30\text{g}$ are initially moving with the velocity of $20\text{cm/s}$ each along the positive direction of the three coordinate axes $x,y$ and $z$ respectively. Due to their mutual interaction, particle $A$ comes to rest. If the particle $B$ has a velocity of $10\hat{i}+20\hat{k}\text{cm/s}$ then find the velocity of particle $C$.

• A. $\frac{40}{3}\hat{j}-\frac{20}{3}\hat{k}$
• B. $\frac{40}{3}\hat{j}+\frac{20}{3}\hat{k}$
• C. $\frac{20}{3}\hat{i}-\frac{40}{3}\hat{j}+\frac{20}{3}\hat{k}$
• D. $\frac{40}{3}\hat{i}-\frac{40}{3}\hat{j}+\frac{20}{3}\hat{k}$

Answer 1

The correct option is B $\frac{40}{3}\hat{j}+\frac{20}{3}\hat{k}$

Given,
Mass of particle $A$ ($m_1$)$=10\text{g}$
Mass of particle $B$ ($m_2$)$=20\text{g}$
Mass of particle $C$ ($m_3$)$=30\text{g}$
Final velocity of particle $A$ ($v_1$)$=0\text{cm/s}$
Final velocity of particle $B$ $\overrightarrow{v_2}=(10\hat{i}+20\hat{k})\text{cm/s}$
Let the speed of the particle $C$ be $v$
Applying $PCLM$,
(Since, only internal forces are acting)
$m_1{u}_1+m_2{u}_2+m_3{u}_3=m_1{v}_1+m_2{v}_2+m_3{v}_3$
$10\left(20\right)\hat{i}+20\left(20\right)\hat{j}+30\left(20\right)\hat{k}$
$=10\left(0\right)+20(10\hat{i}+20\hat{k})+30\left(\overrightarrow{v}\right)$
$200\hat{i}+400\hat{j}+600\hat{k}=200\hat{i}+400\hat{k}+30\overrightarrow{v}$
$\Rightarrow \overrightarrow{v}=\frac{400\hat{j}+200\hat{k}}{30}=\frac{40}{3}\hat{j}+\frac{20}{3}\hat{k}$