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  • Two spheres A and B of masses m1 and m2 respectively collide. A is at rest initially and B is moving with velocity v along x-axis. After collision B has a velocity v/2 in a di

Two spheres A and B of masses m1 and m2 respectively collide. A is at rest initially and B is moving with velocity v along x-axis. After collision B has a velocity v/2 in a direction perpendicular to the original direction. The mass A moves after collision in the direction:

Option: 1

same as that of B


Option: 2

Opposite to that of B


Option: 3

\theta=tan-1(1/2) to the x-axis


Option: 4

\theta= \tan^{-1}(\frac{-1}{2}) to the x-axis


Answers (1)

best_answer

\\ \text{According to law of conservation of linear momentum along x-axis, we get} \\ \\ m_{1} \times 0+m_{2} \times v=m_{1} v^{\prime} \cos \theta$ \\ \\ $m_{2} v=m_{1} v^{\prime} \cos \theta$ \\ \\ or $\cos \theta=\frac{m_{2} v}{m_{1} v^{\prime}} \ldots .( i )$

According to law of conservation momentum along y-axis, we get

\begin{array}{c} m_{1} \times 0+m_{2} \times 0=m_{1} v^{\prime} \sin \theta+m_{2} \frac{v}{2}-m_{2} \frac{v}{2} \\ =m_{1} v^{\prime} \sin \theta \\ \sin \theta=-\frac{m_{2} v}{2 m_{1} v^{\prime}} \dots(2) \end{array}

\\ \text{ Divide (ii) by (i), we get} \\ \tan \theta=-\frac{1}{2}$ $\theta=\tan ^{-1}\left(-\frac{1}{2}\right)$ to the $x$ -axis

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manish painkra

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