Question

Two particle $A$ and $B$ of equal mass are projected simulateously with speed $10m/s$ each as shown in figure. Find the co-ordinates of $COM$ at $t=10s$. (Take $g=10m/s^2$)

Answer 1

Velocity of A and B

$V_A=10\cos {30}^o\hat{i}+10\sin {30}^o\hat{j}and{V}_B=-10\cos {30}^o\hat{i}+10\sin {30}^o\hat{j}$

From conservation of momentum

$M{V}_A+M{V}_B=2M{V}_{cm}$

$V_{cm}=\frac{V_A+V_B}{2}=\frac{(10\cos {30}^o\hat{i}+10\sin {30}^o\hat{j})+(-10\cos {30}^o\hat{i}+10\sin {30}^o\hat{j})}{2}$

$V_{cm}=10\sin {30}^o\hat{j}=5\hat{j}m{s}^{-1}$

$initialvelocityofCOM,\overrightarrow{U}={\overrightarrow{V}}_{cm}=5\hat{j}m{s}^{-1}$

Apply equation of kinematic

$\overrightarrow{S}=\overrightarrow{U}t+\frac{1}{2}\overrightarrow{a}{t}^2$

$\overrightarrow{S}=\left(5\hat{j}\right).10-\frac{1}{2}\left(10\hat{j}\right){.10}^2$

$\overrightarrow{S}=-450\hat{j}$

Position of center of mass at time $(t=10\sec )$ is $(x,y)=(0,-450)meter$