Three masses m, 2 ~m and 3 ~m are moving in x- y plane with speed 3 u, 2 u and u respectively as shown in figure. The three masses collide at the…

From the law of conservation of momentum we know that, $$\mathrm{m}_1 \mathrm{u}_1+\mathrm{m}_2 \mathrm{u}_2+\ldots .=\mathrm{m}_1 \mathrm{v}_1+\mathrm{m}_2 \mathrm{v}_2+\ldots .$$m1​u1​+m2​u2​+….=m1​v1​+m2​v2​+…. Given $\mathrm{m}_1=\mathrm{m}, \mathrm{m}_2=2 \mathrm{~m}$m1​=m,m2​=2 m and $\mathrm{m}_3=3 \mathrm{~m}$m3​=3 m and $\mathrm{u}_1=3 \mathrm{u}, \mathrm{u}_2=2 \mathrm{u}$u1​=3u,u2​=2u and $\mathrm{u}_3=\mathrm{u}$u3​=u Let the velocity when they stick $=\overrightarrow{\mathrm{v}}$=v $$\text { Then, according to question, }$$ Then, according to question, 

$$\begin{aligned} &m \times 3 u(\hat{i})+2 m \times 2 u\left(-\hat{i} \cos 60^{\circ}-\hat{j} \sin 60^{\circ}\right) \\ &+3 \mathrm{~m} \times \mathrm{u}\left(-\hat{\mathrm{i}} \cos 60^{\circ}+\hat{\mathrm{j}} \sin 60^{\circ}\right) \\ &=(\mathrm{m}+2 \mathrm{~m}+3 \mathrm{~m}) \overrightarrow{\mathrm{v}} \\ &\Rightarrow 3 m u \hat{\mathrm{i}}-4 \mathrm{mu} \frac{\hat{\mathrm{i}}}{2}-4 \mathrm{mu}\left(\frac{\sqrt{3}}{2} \hat{\mathrm{j}}\right)-3 \mathrm{mu} \frac{\hat{\mathrm{i}}}{2} \\ &+3 m u\left(\frac{\sqrt{3}}{2} \hat{\mathrm{j}}\right)=6 \mathrm{~m} \mathrm{v} \\ &\Rightarrow m u \hat{i}-\frac{3}{2} m u \hat{i}-\frac{\sqrt{3}}{2} m \hat{j}=6 \mathrm{~m} \mathrm{v} \\ &\Rightarrow-\frac{1}{2} m u \hat{i}-\frac{\sqrt{3}}{2} m u \hat{j}=6 \mathrm{~m} \mathrm{v} \\ &\Rightarrow \overrightarrow{\mathrm{v}}=\frac{\mathrm{u}}{12}(-\hat{\mathrm{i}}-\sqrt{3} \hat{\mathrm{j}}) \\ & \end{aligned}$$​m×3u(i^)+2m×2u(−i^cos60∘−j^​sin60∘)+3 m×u(−i^cos60∘+j^​sin60∘)=(m+2 m+3 m)v⇒3mui^−4mu2i^​−4mu(23​​j^​)−3mu2i^​+3mu(23​​j^​)=6 mv⇒mui^−23​mui^−23​​mj^​=6 mv⇒−21​mui^−23​​muj^​=6 mv⇒v=12u​(−i^−3​j^​)​