Question

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

• A. $(0,-450m)$
• B. $(-450,0)$
• C. $(-450m,-450m)$
• D. $(+450,-450m)$

Answer 1

The correct option is A $(0,-450m)$

As we can see that the $horizontal$ components will get cancelled viz. $10Cos{30}^0-10Cos{30}^0=0$

and the vertical component of the velocity center of mass will be $u=\frac{m(10Sin{30}^0+10Sin{30}^0)}{m+m}=5m/s$

using $h=ut-\frac{1}{2}g{t}^2$ and putting $g=10m/s^2$ we get $h=50-5\times 100=-450meter$

so coordinate $\left(0\right(rest),-450m)$