Solve the following :

resultant velocity of 2 particles moving along $x$ and $y$-axis

$V_{R}=\sqrt{(10)^{2}+(10)^{2}+2(10)(10) \cos 90^{\circ}}$

$=10 \sqrt{2} \mathrm{~m} / \mathrm{s}$

C.O.L.M

$\Rightarrow 0=\frac{M}{3}(V+10 \sqrt{2})$

$\Rightarrow V=-10 \sqrt{2}$

So,

$V$ is along $135^{\circ}$ below x-axis.