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

Hint: 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 \[\theta =90{}^\circ \]