Question 366: Centre of Mass and Rotational Motion

Question

A particle of mass $10 \mathrm{~kg}$ moving eastwards with a speed $10 \mathrm{~m} / \mathrm{s}$ collides with another particle of the same mass moving northward with the same speed $10 \mathrm{~m} / \mathrm{s}$. The two particles coalesce on collision. The new particle of mass $20 \mathrm{~kg}$ will move in the north-east direction with velocity.

Accepted Answer

Here, initial momentum of particle $=$ Final momentum of particles,
$m v \hat{\mathbf{i}}+m v \hat{\mathbf{j}}=2 m v$
Where, $\mathbf{v}=\frac{v}{2}(\hat{\mathbf{i}}+\hat{\mathbf{j}})$
Hence, $v=\frac{v}{2} 	imes \sqrt{2}=v / \sqrt{2} \quad\left(\because v=10 \mathrm{~ms}^{-1}ight)$
So, $v=\frac{10}{\sqrt{2}} \mathrm{~ms}^{-1}$

Answer

$10 \mathrm{~m} / \mathrm{s}$

Answer

$5 \mathrm{~m} / \mathrm{s}$

Answer

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

Answer

None

Centre of Mass and Rotational Motion

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