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# An isolated particle of mass $m$ is moving in a horizontal plane [x-y] along the x-axis at a certain height above the ground. It suddenly explodes into fragments of masses $\dfrac{m}{4}$ and $\dfrac{{3m}}{4}$ . An instant later the smaller fragment is at $y = 15cm$. The larger fragment at this instant is at:A) $y = - 5cm$B) $y = + 20cm$C) $y = + 5cm$D) $y = - 20cm$

Last updated date: 02nd Aug 2024
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Hint: Since no external force is acting on the system, the y coordinate for the centre of mass of the system will remain the same before and after the explosion i.e. it will be 0. You have to use the formula for centre of mass considering only the y coordinates of the two pieces and equate it to 0 (as it will remain the same) and solve the equation to get the final answer.

Formula Used:
Centre of mass for a 2 body system (Y coordinate) is given by $\dfrac{{{m_1}{y_1} + {m_2}{y_2}}}{{{m_1} + {m_2}}}$ where, ${m_1},{m_2}$ are the mass of body 1 and body 2 respectively and ${y_1},{y_2}$ are the y coordinates of body 1 and body 2 respectively.

Complete Step by Step Solution:
Initially, the particle is moving along the x axis, therefore its y coordinate is 0. Since this is a single particle system, its centre of mass will be at the same location as of the particle itself. Therefore, its centre of mass will have y coordinate equal to 0.
Then the particle explodes and splits into 2 pieces. No external force is acting on the particle here. Therefore, the centre of mass will lie at the same point as before the explosion. That is, the y coordinate for the centre of mass after the explosion is 0.
Also, we can find the centre of mass after an explosion using its formula. We have, Centre of mass for a 2 body system (Y coordinate) is given by $\dfrac{{{m_1}{y_1} + {m_2}{y_2}}}{{{m_1} + {m_2}}}$ where, ${m_1},{m_2}$ are the mass of body 1 and body 2 respectively and ${y_1},{y_2}$ are the y coordinates of body 1 and body 2 respectively.
As we have calculated that this will be equal to 0, we will equate the above formula to 0 and solve after putting the values.
$\dfrac{{{m_1}{y_1} + {m_2}{y_2}}}{{{m_1} + {m_2}}} = 0$
$\Rightarrow$ $\dfrac{{\dfrac{m}{4} \times 15 + \dfrac{{3m}}{4}{y_2}}}{{{m_1} + {m_2}}} = 0$ (position of first particle after explosion is given)
$\Rightarrow$ $\dfrac{{15m}}{4} + \dfrac{{3m}}{4}{y_2} = 0$
$\Rightarrow$ ${y_2} = - \dfrac{{15m}}{4} \times \dfrac{4}{{3m}} = - \dfrac{{15}}{4} \times \dfrac{4}{3}$ (like terms get cancelled out)
We get ${y_2} = - 5cm$
This will be the position of the second particle.

Option (A) is the correct answer.

Note: when no external force acts on a system, then the centre of mass of the system remains the same even if the system has transformed into more pieces than before, like in this case. In this question, when the explosion happened, the particles were spread in such a way, respective to their masses, that the centre of mass was the same as before. This is exactly what happens in every such case.