A particle of mass m is moving in the xy-plan | vedclass

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(A) Before the explosion,the particle is moving along the $x$-axis,so its initial velocity component in the $y$-direction is zero $(v_{y,i} = 0)$.Sinc

Vedclass · Accepted Answer

$ -5$

Vedclass · Answer

(A) Before the explosion,the particle is moving along the $x$-axis,so its initial velocity component in the $y$-direction is zero $(v_{y,i} = 0)$.Since there are no external forces acting in the $y$-direction,the position of the center of mass $(y_{cm})$ remains constant at $0$ (assuming the initial $y$-coordinate was $0$).Using the formula for the center of mass: $y_{cm} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2}$.Given $m_1 = m/4$,$y_1 = 15 \ cm$,$m_2 = 3m/4$,and $y_{cm} = 0$.$0 = \frac{(m/4)(15) + (3m/4)(y_2)}{m/4 + 3m/4}$.$0 = \frac{m}{4} (15 + 3y_2) / m$.$15 + 3y_2 = 0$.$3y_2 = -15$.$y_2 = -5 \ cm$.

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