Question

A bomb moving with velocity $\left(40\hat{i}+50\hat{j}-25\hat{k}\right)m{s}^{-1}$ explodes into two pieces of mass ratio $1:4$. After explosion the smaller piece moves away with velocity $\left(200\hat{i}+70\hat{j}-15\hat{k}\right)m{s}^{-1}$. The velocity of larger piece after explosion is

• A. $45\hat{j}-35\hat{k}$
• B. $45\hat{i}-35\hat{j}$
• C. $45\hat{k}-35\hat{j}$
• D. $-35\hat{i}+45\hat{k}$

Answer 1

Answer: (A)

Let the mass of the unexploded bomb be 5m. It explodes into the two pieces of masses m and 4m respectively.
Initial momentum of the unexploded bomb
$=5m\left(40\hat{i}+50\hat{j}-25\hat{k}\right)$
After explosion, momentum of the smaller piece $=m\overrightarrow{v_1}=m\left(200\hat{i}+70\hat{j}+15\hat{k}\right)$
and momentum of the larger piece = $4m\overrightarrow{v_2}$ where $\overrightarrow{v_1}$ and $\overrightarrow{v_2}$ are the velocities of the two pieces respectively.
According to the law of conservation of momentum, we get
$5m\left(40\hat{i}+50\hat{j}-25\hat{k}\right)=m\left(200\hat{i}+70\hat{j}+15\hat{k}\right)+4m\overrightarrow{v_2}$
$4m\overrightarrow{v_2}=5m\left(40\hat{i}+50\hat{j}-25\hat{k}\right)-m\left(200\hat{i}+70\hat{j}+15\hat{k}\right)$
$\overrightarrow{v_2}=\frac{1}{4}\left(180\hat{j}-140\hat{k}\right)=45\hat{j}-35\hat{k}$