(A) Before the explosion,the particle was moving along the $x$-axis,which means it had no $y$-component of velocity. Consequently,the center of mass o

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(A) Before the explosion,the particle was moving along the $x$-axis,which means it had no $y$-component of velocity. Consequently,the center of mass of the system does not move in the $y$-direction,so $y_{CM} = 0$. Using the formula for the center of mass: $y_{CM} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2}$. Substituting the given values: $0 = \frac{(m/4)(+15) + (3m/4)(y)}{m}$. This simplifies to: $0 = \frac{15}{4} + \frac{3y}{4}$. Solving for $y$: $\frac{3y}{4} = -\frac{15}{4}$,which gives $y = -5 \ cm$.

Class 11 Physics System of Particles and Rotational Motion Centre of mass (Point Mass)

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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 two fragments of masses $m/4$ and $3m/4$. An instant later,the smaller fragment is at $y = +15 \ cm$. The larger fragment at this instant is at

IIT 1997 All IIT

A
$-5 \ cm$
B
$+20 \ cm$
C
$+5 \ cm$
D
$-20 \ cm$

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Centre of mass (Point Mass)

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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 two fragments of masses $m/4$ and $3m/4$. An instant later,the smaller fragment is at $y = +15 \ cm$. The larger fragment at this instant is at

Similar Questions

The coordinates of the center of mass of the triangular lamina shown in the figure are ....... .

Four identical spheres, each of radius $10 \,cm$ and equal mass $1 \,kg$, are placed on a horizontal surface touching each other such that their centers are located at the vertices of a square of side $20 \,cm$. What is the distance of their center of mass from the center of any sphere?

The position vectors of two particles of masses $1 \ kg$ and $3 \ kg$ are $\hat{i} + 2\hat{j} + \hat{k}$ and $-3\hat{i} - 2\hat{j} + \hat{k}$ respectively. The position vector of the center of mass of this system is:

The center of mass of three particles of masses $10 \ kg, 20 \ kg$,and $30 \ kg$ is at $(0, 0, 0)$. Where should a fourth particle of mass $40 \ kg$ be placed so that the center of mass of the system shifts to $(3, 3, 3)$?

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