91Ó°ÊÓ

To analyze the problem, we can draw a diagram of the initial state before the collision, and the final state after the collision. The first mass has an initial velocity \(v\), while the second mass is at rest. Before collision: 1. Mass 1: mass = \(m\), velocity = \(v\). 2. Mass 2: mass = \(m\), velocity = \(0\). After collision: 1. Mass 1: mass = \(m\), velocity = \(\frac{v}{\sqrt{3}}\), and direction = perpendicular to the initial direction. 2. Mass 2: mass = \(m\), velocity = \(?\).

Conservation of momentum states that the total initial momentum of the system should be equal to the total final momentum. The momentum of each mass can be written as \(p = m \times v\). Since mass 1 moves in a different direction after the collision, we should split the equations into two components (x-component and y-component). Let's consider the initial direction of mass 1 as the positive x-direction. Initial momentum (x-component): 1. Mass 1: \(m \times v\). 2. Mass 2: \(0\). Final momentum (x-component): 1. Mass 1: \(0\) (since it moves perpendicular to the initial direction). 2. Mass 2: \(m \times v_x\), where \(v_x\) is the x-component of mass 2's velocity. Initial momentum (y-component): 1. Mass 1: \(0\) (since it is initially moving along the x-axis) 2. Mass 2: \(0\) Final momentum (y-component): 1. Mass 1: \(m \times \frac{v}{\sqrt{3}}\) 2. Mass 2: \(m \times v_y\), where \(v_y\) is the y-component of mass 2's velocity. We have two conservation of momentum equations: 1. In the x-direction: \(m \times v = m \times v_x\) 2. In the y-direction: \(m \times \frac{v}{\sqrt{3}} = m \times v_y\)

Now, we will solve the conservation of momentum equations for the velocity components of mass 2. X-component: Since mass 1 and mass 2 are identical, we can cancel out "\(m\)" in the equation, and we get: \(v = v_x\) Y-component: Similarly, we cancel out "\(m\)": \(\frac{v}{\sqrt{3}} = v_y\)

The speed of mass 2 can be found by taking the square root of the sum of the squares of the x and y components of its velocity (\(v_x\) and \(v_y\)): \(v_2 = \sqrt{v_x^2 + v_y^2}\) Substitute the values we got in step 3: \(v_2 = \sqrt{v^2 + \left(\frac{v}{\sqrt{3}}\right)^2}\) \(v_2 = \sqrt{v^2 + \frac{v^2}{3}}\) Combining the terms: \(v_2 = v\sqrt{\frac{4}{3}}\) Now, multiplying both numerator and denominator with \(\sqrt{3}\): \(v_2 = \frac{2}{\sqrt{3}}v\) The speed of the second mass after collision is \(\boxed{\text{(D) } \frac{2}{\sqrt{3}}v}\).