An isolated particle of mass $m$ is moving in a horizontal $x-y$ plane 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 $y = \dots \ cm$.

The position vectors of three particles of masses $1\, kg, 2\, kg$,and $3\, kg$ are $\vec{r_1} = (\hat{i} + 4\hat{j} + \hat{k})\,m$,$\vec{r_2} = (\hat{i} + \hat{j} + \hat{k})\,m$,and $\vec{r_3} = (2\hat{i} - \hat{j} - 2\hat{k})\,m$ respectively. The position vector of their centre of mass is:

Two particles of mass $5\, kg$ and $10\, kg$ respectively are attached to the two ends of a rigid rod of length $1\, m$ with negligible mass. The centre of mass of the system from the $5\, kg$ particle is nearly at a distance of $..........\, cm$.

$A$ uniform square plate $abcd$ has a mass of $1 \, kg$. If two point masses,each of $20 \, g$,are placed at the corners $b$ and $c$ as shown,then the centre of mass shifts on the line

Look at the figure drawn with ink of uniform linear density. $A$ mass $m$ of ink is used to draw each of the two inner circles and each of the two line segments. $A$ mass $6m$ of ink is used to draw the outer circle. The coordinates of the centers of the different parts are: outer circle $(0, 0)$,left inner circle $(-a, a)$,right inner circle $(a, a)$,and the horizontal line segment $(0, -a)$. Find the $y$-coordinate of the center of mass of the ink in the figure.

Four particles of masses $m_1 = 2m$,$m_2 = 4m$,$m_3 = m$,and $m_4$ are placed at the four corners of a square. What should be the value of $m_4$ so that the centre of mass of all the four particles is exactly at the centre of the square?