A \(100-\mathrm{g}\) ball collides elastically with a \(300-\mathrm{g}\) ball that is at rest. If the \(100-\mathrm{g}\) ball was traveling in the positive \(x-\) direction at \(5.00 \mathrm{m} / \mathrm{s}\) before the collision, what are the velocities of the two balls after the collision?

We are given the following information: - The mass of ball 1 (\(m_1\)): \(m_1 = 100 \mathrm{g} = 0.1 \mathrm{kg}\) - The mass of ball 2 (\(m_2\)): \(m_2 = 300 \mathrm{g} = 0.3 \mathrm{kg}\) - The initial velocity of ball 1 (\(v_{1i}\)): \(v_{1i} = 5.00 \mathrm{m} / \mathrm{s}\) - The initial velocity of ball 2 (\(v_{2i}\)): \(v_{2i} = 0 \mathrm{m} / \mathrm{s}\) (at rest) We are asked to find the final velocities of both balls after the collision: - The final velocity of ball 1 (\(v_{1f}\)) - The final velocity of ball 2 (\(v_{2f}\))

Using the conservation of momentum and kinetic energy, we can set up two equations: 1. Total initial momentum equals total final momentum \(m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}\) 2. Total initial kinetic energy equals total final kinetic energy \(\displaystyle\frac{1}{2}m_1v_{1i}^2 + \displaystyle\frac{1}{2}m_2v_{2i}^2 = \displaystyle\frac{1}{2}m_1v_{1f}^2 + \displaystyle\frac{1}{2}m_2v_{2f}^2\) Calculate the initial total momentum and total kinetic energy: Initial total momentum: \(P_{initial} = m_1v_{1i} + m_2v_{2i} = 0.1 \mathrm{kg} \cdot 5.00 \mathrm{m} / \mathrm{s} + 0.3 \mathrm{kg} \cdot 0 \mathrm{m} / \mathrm{s} = 0.5 \mathrm{kg \cdot m} / \mathrm{s}\) Initial total kinetic energy: \(K_{initial} = \displaystyle\frac{1}{2}m_1v_{1i}^2 + \displaystyle\frac{1}{2}m_2v_{2i}^2 = \displaystyle\frac{1}{2}\cdot 0.1 \mathrm{kg} \cdot (5.00 \mathrm{m} / \mathrm{s})^2 + 0 = 1.25 \mathrm{J}\)

Use the conservation of momentum equation given in step 2: \(0.5 \mathrm{kg\cdot m}/\mathrm{s} = 0.1\mathrm{kg} \cdot v_{1f} + 0.3\mathrm{kg} \cdot v_{2f}\) ...(1) Now, use the conservation of kinetic energy equation given in step 2: \(1.25 \mathrm{J} = \displaystyle\frac{1}{2} \cdot 0.1 \mathrm{kg} \cdot v_{1f}^2 + \displaystyle\frac{1}{2} \cdot 0.3 \mathrm{kg} \cdot v_{2f}^2\) ...(2) Solve these two equations simultaneously. We can express one of the variables in terms of the other from Equation (1) and substitute it into Equation (2). From Equation (1): \(v_{2f} = \displaystyle\frac{0.5 - 0.1v_{1f}}{0.3}\) Substitute this expression for \(v_{2f}\) into Equation (2): \(1.25 = \displaystyle\frac{1}{2} \cdot 0.1 \cdot \left(v_{1f}\right)^2 + \displaystyle\frac{1}{2} \cdot 0.3 \cdot \left(\displaystyle\frac{0.5 - 0.1v_{1f}}{0.3}\right)^2\) Solve the above equation for \(v_{1f}\). Then substitute this value back into our expression for \(v_{2f}\) from Equation (1) to find \(v_{2f}\): \(v_{1f} = 2.5 \mathrm{m}/\mathrm{s}\) \(v_{2f} = 1.6667 \mathrm{m}/\mathrm{s}\)

The final velocity of ball 1 (100-g ball) after the collision is \(v_{1f} = 2.5 \mathrm{m}/\mathrm{s}\). The final velocity of ball 2 (300-g ball) after the collision is \(v_{2f} = 1.6667 \mathrm{m}/\mathrm{s}\).