An isolated particle of mass m is moving in a | vedclass

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(A) Since the particle is isolated and moving along the $x-$axis,there is no external force acting on it in the $y-$direction. Therefore,the center of

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$-5$

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(A) Since the particle is isolated and moving along the $x-$axis,there is no external force acting on it in the $y-$direction. Therefore,the center of mass of the system remains at $y = 0$ (the $x-$axis).Let $y_1$ and $y_2$ be the $y-$coordinates of the fragments of masses $m_1 = m/4$ and $m_2 = 3m/4$ respectively.The position of the center of mass in the $y-$direction is given by $Y_{cm} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2} = 0$.Substituting the given values: $\frac{(m/4)(15) + (3m/4)(y_2)}{m} = 0$.This simplifies to: $\frac{15}{4} + \frac{3}{4}y_2 = 0$.Multiplying by $4$: $15 + 3y_2 = 0$.Solving for $y_2$: $3y_2 = -15$,which gives $y_2 = -5 \ cm$.

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