Question

A body of mass 1 kg, initially at rest, explodes and breaks into three fragments of masses in the ratio $1:1:3$. The two pieces of equal mass fly off perpendiculars to each other with a speed of 30 m/s each. If the velocity of the heavier fragment is $10\sqrt{x}$, determine x.

Answer 1

Initially Body is at rest and after exploding it breaks into three pieces with different velocity. So, we can apply the momentum conservation.

Complete step by step answer:
Given, Total mass of Body $M=1\text{ kg}$
Initially at rest $v=0$
After exploding it breaks into three fragments in the ratio $=1:1:3$
So, the mass of first fragment ${{m}_{1}}=\dfrac{1}{5}\times 1=0.2\text{ kg}$
Mass of second fragment ${{m}_{2}}=\dfrac{1}{5}\times 1=0.2\text{ kg}$
Mass of third fragment ${{M}_{3}}=\dfrac{3}{5}\times 1=0.6\text{ kg}$
The momentum of the system will be conserved.
So,
Momentum of first particle $\left( {{P}_{1}} \right)={{M}_{1}}{{V}_{1}}$
$=0.2\times 30=6\text{ kg}$
Momentum of second particle $\left( {{P}_{2}} \right)={{M}_{2}}{{V}_{2}}$
$=0.2\times 30=6\text{ kg}$
Net Resultant moment of first and second Fragment when both are perpendicular
${{P}_{Net}}=\sqrt{{{P}_{1}}^{2}+{{P}_{2}}^{2}}=\sqrt{{{6}^{2}}+{{6}^{2}}}=6\sqrt{2}\text{ kg}\dfrac{M}{\text{sec}}$
Momentum of third fragment ${{P}_{3}}={{M}_{3}}\times {{V}_{3}}$
$=10\sqrt{x}\times 0.6$
$=6\sqrt{x}\text{ }\dfrac{\text{kg m}}{\text{S}}$
Thus ${{P}_{3}}$ (momentum of third Fragment) = ${{P}_{Net}}$
$\Rightarrow 6\sqrt{x}=6\sqrt{2}$
$\Rightarrow x=2$

Therefore, the value of x is 2.

Note:
When two particle of same mass fly perpendicular to each other than the Net momentum calculated by
${{P}_{Net}}=\sqrt{{{P}_{1}}^{2}+{{P}_{2}}^{2}+2{{P}_{1}}{{P}_{2}}\cos \theta }$ where $\theta =90{}^\circ$