Three masses $\mathrm{m}, 2 \mathrm{~m}$ and $3 \mathrm{~m}$ are moving in $\mathrm{x}-\mathrm{y}$ plane with speed $3 \mathrm{u}, 2 \mathrm{u}$ and $\mathrm{u}$ respectively as shown in figure. The three masses collide at the same point at $\mathrm{P}$ and stick together. The velocity of resulting mass will be:

Question

Aicharya · Accepted Answer

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 . $ Given $\mathrm{m}_1=\mathrm{m}, \mathrm{m}_2=2 \mathrm{~m}$ and $\mathrm{m}_3=3 \mathrm{~m}$ and $\mathrm{u}_1=3 \mathrm{u}, \mathrm{u}_2=2 \mathrm{u}$ and $\mathrm{u}_3=\mathrm{u}$ Let the velocity when they stick $=\overrightarrow{\mathrm{v}}$ $ \text { 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} $

Three masses $\mathrm{m}, 2 \mathrm{~m}$ and $3 \mathrm{~m}$ are moving in $\mathrm{x}-\mathrm{y}$ plane…

Solution