Question

A bomb of mass 4 m at rest, explodes into three pieces of masses in the ratio $1:1:2$. The two identical pieces fly in mutually perpendicular directions, each with a velocity v. The third piece is making an angle $\alpha$ with the line of motion of each piece. Find $\alpha$.

• A. ${45}^0$
• B. ${90}^0$
• C. ${135}^0$
• D. ${180}^0$

Answer 1

Answer: (C)

${135}^0$

Let the two identical pieces fly along $\hat{i}$ and $\hat{j}$ directions,
Using momentum conservation
Initial momentum $=0$, therefore,
$mv\hat{i}+mv\hat{j}+2m\overrightarrow{x}=0$

$\overrightarrow{x}=-\frac{v}{2}(\hat{i}+\hat{j})$

Let $\alpha$ be the angle between $\overrightarrow{x}$ and $\hat{i}$.
By taking dot product of $\overrightarrow{x}$ and $\hat{i}$, we get,

$\overrightarrow{x}.\hat{i}=|\overrightarrow{x}||\hat{i}|cos\alpha$

$\Rightarrow -\frac{v}{2}=\sqrt{{\left(\frac{v}{2}\right)}^2+{\left(\frac{v}{2}\right)}^2}cos\alpha$

$\Rightarrow -\frac{v}{2}=\frac{v}{\sqrt{2}}cos\alpha$

$\Rightarrow cos\alpha =-\frac{1}{\sqrt{2}}$

$\Rightarrow \alpha ={135}^0$