A particle of mass 3m is projected from the g | vedclass

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(C) The center of mass of the system follows the original parabolic trajectory because the explosion is caused by internal forces.The horizontal range

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$\frac{5R}{4}$

Vedclass · Answer

(C) The center of mass of the system follows the original parabolic trajectory because the explosion is caused by internal forces.The horizontal range of the center of mass is $R$,so it lands at $x_{com} = R$.The highest point of the trajectory is at a horizontal distance of $\frac{R}{2}$ from the point of projection.Let the smaller mass $m$ come to rest,meaning its horizontal position is $x_1 = \frac{R}{2}$.Let the larger mass $2m$ fall at a distance $x_2 = x$.The center of mass position is given by $x_{com} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$.Substituting the values: $R = \frac{m \cdot \frac{R}{2} + 2m \cdot x}{m + 2m}$.$3mR = \frac{mR}{2} + 2mx$.$3mR - \frac{mR}{2} = 2mx$.$\frac{5mR}{2} = 2mx$.$x = \frac{5R}{4}$.

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