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A particle of mass $10 \mathrm{~g}$ is moving towards east with a velocity of $10 \mathrm{~ms}^{-1}$ and another particle of mass $15 \mathrm{~g}$ is moving towards north with a velocity of $5 \mathrm{~ms}^{-1}$. The magnitude of the velocity of the centre of mass of the system of the two particles is
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$5 \mathrm{~ms}^{-1}$
Mass, $\mathrm{m}_1=10 \mathrm{~g}=10 \times 10^{-3}=10^{-2} \mathrm{Kg}$
velocity, $\mathrm{v}_1=10 \mathrm{~m} / \mathrm{s}$
Mass of 2nd particle,
$M_2=15 \mathrm{~g}=15 \times 10^{-3} \mathrm{Kg}$
Velocity of IInd particle, $V_2=5 \mathrm{~m} / \mathrm{s}$
centre of mass makes $45^{\circ}$ angle with north and east side.
Hence velocity of centre of mass is given as:
$\begin{aligned} & V_{1 \mathrm{~m}}=\frac{\mathrm{m}_1 \mathrm{v}_1 \cos \theta_1+\mathrm{m}_2 \theta, \sin \theta_1}{\mathrm{~m}_1+\mathrm{m}_2} \\ & =\frac{10 \times 10 \times \cos 45+15 \times 5 \times \sin 45}{10+15} \\ & =\frac{1}{\sqrt{2}}\left[\frac{175}{25}\right]=\frac{7}{2}(\sqrt{2})=4.9 \mathrm{~m} / \mathrm{s} ; 5 \mathrm{~m} / \mathrm{s}\end{aligned}$
velocity, $\mathrm{v}_1=10 \mathrm{~m} / \mathrm{s}$
Mass of 2nd particle,
$M_2=15 \mathrm{~g}=15 \times 10^{-3} \mathrm{Kg}$
Velocity of IInd particle, $V_2=5 \mathrm{~m} / \mathrm{s}$
centre of mass makes $45^{\circ}$ angle with north and east side.
Hence velocity of centre of mass is given as:
$\begin{aligned} & V_{1 \mathrm{~m}}=\frac{\mathrm{m}_1 \mathrm{v}_1 \cos \theta_1+\mathrm{m}_2 \theta, \sin \theta_1}{\mathrm{~m}_1+\mathrm{m}_2} \\ & =\frac{10 \times 10 \times \cos 45+15 \times 5 \times \sin 45}{10+15} \\ & =\frac{1}{\sqrt{2}}\left[\frac{175}{25}\right]=\frac{7}{2}(\sqrt{2})=4.9 \mathrm{~m} / \mathrm{s} ; 5 \mathrm{~m} / \mathrm{s}\end{aligned}$
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