(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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System of Particles and Rotational Motion

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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

$A$ uniform wire is bent into a $U$ shape with side lengths $l, 2l,$ and $l$. The coordinates of the center of mass of each segment are shown in the figure. The coordinates of the center of mass of the system are:

Particles of masses $8 \, kg, 2 \, kg, 4 \, kg$,and $2 \, kg$ are placed at the vertices $A, B, C$,and $D$ of a square respectively. The distance of the center of mass of the system from vertex $A$ is $....... \, cm$. The length of the diagonal of the square is $80 \, cm$.

Write the meaning of homogeneous bodies.

Two particles of masses $1\,kg$ and $3\,kg$ have position vectors $2\hat{i} + 3\hat{j} + 4\hat{k}$ and $-2\hat{i} + 3\hat{j} - 4\hat{k}$ respectively. The position vector of the centre of mass is:

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