Question 114 of 130

Two particles of masses $M$ and $2M are$moving with speeds of $10m{s}^{-1}$ and $5m{s}^{-1}$, as shown in the figure. They collide at the origin and after that they move along the indicated directions with speeds ${v}_{1}$ and ${v}_{2}$, respectively. The values of ${v}_{1}$ and ${v}_{2}$ are, nearly

A$6.5m{s}^{-1}$ and $3.2m{s}^{-1}$B$3.2m{s}^{-1}$ and $12.6m{s}^{-1}$C$13.02m{s}^{-1}$ and $19.7m{s}^{-1}$D$3.2m{s}^{-1}$ and $6.3m{s}^{-1}$

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Answer: (C) $13.02m{s}^{-1}$ and $19.7m{s}^{-1}$

From conservation of linear momentum,
Along $x$-axis:
${\text{10Mcos30}}^{0}+5\left(\text{2M}cos{45}^{0}\right)=\text{M}{v}_{1}cos{30}^{0}+2\text{M}{v}_{2}cos{45}^{0}$
$10 M \dfrac{\sqrt{3}}{2}+ 10M\dfrac{1}{\sqrt{2}}=M\dfrac{\sqrt{3}}{2} {v}_{1}+2M{v}_{2}\dfrac{1}{\sqrt{2}}$
$\dfrac{\sqrt{3}}{2} {v}_{1}+\sqrt{2}{v}_{2}=5(\sqrt{3}+\sqrt{2})$ … (1)
Along $y$-axis:
$-10M\dfrac{1}{2}+10M\dfrac{1}{\sqrt{2}}=M{v}_{1}\dfrac{1}{2}-2M{v}_{2}\dfrac{1}{\sqrt{2}}$
$\dfrac{ {v}_{1}}{2}-\sqrt{2}{v}_{2}=5(\sqrt{2}-1)$ … (2)
From equations (1) and (2)
${v}_{1}\left(\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\right)=5\left(\sqrt{3}+\sqrt{2}\right)+5(\sqrt{2}-1)$
${v}_{1}=13.02 m/s$
${v}_{2}=\dfrac{5\sqrt{3}+5\sqrt{2}-13.02\times \dfrac{\sqrt{3}}{2}}{\sqrt{2}}$
$=19.72 m{s}^{-1}$