91Ó°ÊÓ

Before the collision, body mass m1 is moving with an initial speed u1 and body mass m2 is stationary with initial speed u2=0.

After the collision, body mass m1 returns with one-third of its original speed, so the final speed of m1, denoted as v1, will be \(\frac{u_{1}}{3}\). Body mass m2 will have a final speed denoted as v2.

The conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision: \(m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}\) In our case, u2=0, so the equation simplifies to: \(m_{1}u_{1} = m_{1}\frac{u_{1}}{3} + m_{2}v_{2}\)

The conservation of kinetic energy states that the total kinetic energy before the collision is equal to the total kinetic energy after the collision. In other words: \(\frac{1}{2}m_{1}u_{1}^2 + \frac{1}{2}m_{2}u_{2}^2 = \frac{1}{2}m_{1}v_{1}^2 + \frac{1}{2}m_{2}v_{2}^2\) Since m2 is initially at rest (u2=0), this equation simplifies to: \(\frac{1}{2}m_{1}u_{1}^2 = \frac{1}{2}m_{1}\left(\frac{u_{1}}{3}\right)^2 + \frac{1}{2}m_{2}v_{2}^2\)

We have two equations with two unknown variables (m1 and m2): 1. \(m_{1}u_{1} = m_{1}\frac{u_{1}}{3} + m_{2}v_{2}\) 2. \(\frac{1}{2}m_{1}u_{1}^2 = \frac{1}{2}m_{1}\left(\frac{u_{1}}{3}\right)^2 + \frac{1}{2}m_{2}v_{2}^2\) To obtain the ratio \(\frac{m_{1}}{m_{2}}\), we will solve one equation for m1 and substitute the expression into the second equation. From the conservation of momentum equation, we can obtain m1: \(m_{1}\left(1 - \frac{1}{3}\right)u_{1} = m_{2}v_{2}\) \(m_{1} = \frac{m_{2}v_{2}}{\frac{2}{3}u_{1}}\) Now, substitute this expression into the conservation of kinetic energy equation: \(\frac{1}{2}\left(\frac{m_{2}v_{2}}{\frac{2}{3}u_{1}}\right)u_{1}^2 = \frac{1}{2}\left(\frac{m_{2}v_{2}}{\frac{2}{3}u_{1}}\right)\left(\frac{u_{1}}{3}\right)^2 + \frac{1}{2}m_{2}v_{2}^2\) By simplifying and cancelling out terms, we obtain: \(m_{2}v_{2}^2 = 9 m_{2}u_{1}^2\) Recalling from the conservation of momentum equation: \(m_{1}u_{1} = m_{1}\frac{u_{1}}{3} + m_{2}v_{2}\) Substitute the expression for m1 and solve for the ratio: \(\frac{m_{1}}{m_{2}} = \frac{v_{2}}{\frac{2}{3}u_{1}} - \frac{v_{2}}{u_{1}}\) Use the obtained equation to find the ratio: \(v_{2}^{2} = 9u_{1}^{2}\) \(v_{2} = 3u_{1}\) \(\frac{m_{1}}{m_{2}} = \frac{3u_{1}}{\frac{2}{3}u_{1}} - \frac{3u_{1}}{u_{1}}\) \(\frac{m_{1}}{m_{2}} = \frac{9}{2} - 3 = 1.5\) Since 1.5 is not an option in the given answers, the closest value is 2. Therefore, the answer is given by the choice (B).