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A particle of mass $10 \mathrm{~kg}$ moving eastwards with a speed $5 \mathrm{~ms}^{-1}$ collides with another particle of the same mass moving north-wards with the same speed $5 \mathrm{~ms}^{-1}$. The two particles coalesce on collision. The new particle of mass $20 \mathrm{~kg}$ will move in the north-east direction with velocity
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The correct answer is:
$\quad(5 / \sqrt{2}) \mathrm{ms}^{-1}$
Here $\hat{i} m v+\hat{j} m v=2 m \vec{V}$
That is $\overrightarrow{\mathrm{V}}=\frac{\mathrm{v}}{2}(\hat{\mathrm{i}}+\hat{\mathrm{j}})$
Hence $V=\frac{v}{2} \times \sqrt{2}=\frac{v}{\sqrt{2}} \cdot\left[\right.$ Here $\left.v=5 \mathrm{~ms}^{-1}\right]$
$$
\text { So, } V=\frac{5}{\sqrt{2}} \mathrm{~ms}^{-1}
$$
That is $\overrightarrow{\mathrm{V}}=\frac{\mathrm{v}}{2}(\hat{\mathrm{i}}+\hat{\mathrm{j}})$
Hence $V=\frac{v}{2} \times \sqrt{2}=\frac{v}{\sqrt{2}} \cdot\left[\right.$ Here $\left.v=5 \mathrm{~ms}^{-1}\right]$
$$
\text { So, } V=\frac{5}{\sqrt{2}} \mathrm{~ms}^{-1}
$$
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