A ball with mass \(m=0.210 \mathrm{~kg}\) and kinetic energy \(K_{1}=2.97 \mathrm{~J}\) collides elastically with a second ball of the same mass that is initially at rest. \(m_{1}\) After the collision, the first ball moves away at an angle of \(\theta_{1}=\) \(30.6^{\circ}\) with respect to the horizontal, as shown in the figure. What is the kinetic energy of the first ball after the collision?

We are given the following information: - Mass of both balls: \(m = 0.210 \mathrm{~kg}\) - Initial kinetic energy of the first ball: \(K_{1} = 2.97 \mathrm{~J}\) - Angle of the first ball after collision: \(\theta_{1} = 30.6^{\circ}\) Since the collision is elastic, the kinetic energy is conserved. Initial kinetic energy of the system: \(K_{i} = K_{1} + 0 = 2.97 \mathrm{~J}\) Final kinetic energy of the system: \(K_{f} = K_{1f} + K_{2f} = 2.97 \mathrm{~J}\)

We know that the kinetic energy of an object is given by: \(K = \frac{1}{2}mv^{2}\) Solve for the initial velocity of the first ball (\(v_{1i}\)): \(v_{1i} = \sqrt{\frac{2K_{1}}{m}} = \sqrt{\frac{2 \times 2.97 \mathrm{~J}}{0.210 \mathrm{~kg}}} = 4.96 \mathrm{~m/s}\)

As it's an elastic collision, both momentum and kinetic energy are conserved. We can write two equations, one for the horizontal direction (\(x\)) and one for the vertical direction (\(y\)): Conservation of linear momentum in the \(x\)-direction: \(m v_{1i}\cos(0^{\circ}) = m v_{1f}\cos(\theta_{1}) + m v_{2f}\cos(\theta_{2})\) Conservation of linear momentum in the \(y\)-direction: \(m v_{1i}\sin(0^{\circ}) = m v_{1f}\sin(\theta_{1}) - m v_{2f}\sin(\theta_{2})\) Since \(v_{1i}\sin(0^{\circ}) = 0\), we can simplify the second equation: \(v_{1f}\sin(\theta_{1}) = v_{2f}\sin(\theta_{2})\) Now we have three unknowns (\(v_{1f}, v_{2f}, \theta_{2}\)) and three equations (conservation of linear momentum in \(x\) and \(y\), and conservation of kinetic energy).

Since we are only interested in the kinetic energy of the first ball after the collision, we focus on solving for \(v_{1f}\). From the conservation of kinetic energy, we have: \(K_{f} = K_{1f} + K_{2f} = \frac{1}{2}mv_{1f}^{2} + \frac{1}{2}mv_{2f}^{2} = 2.97 \mathrm{~J}\) Now we can use the conservation of linear momentum in the \(x\)-direction equation and conservation of kinetic energy equation to solve for the final velocity of the first ball, \(v_{1f}\): \(v_{1f}=\frac{m v_{1i}-m v_{2f}\cos(\theta_{2})}{m\cos(\theta_{1})}\) Substitute this expression for \(v_{1f}\) in the conservation of kinetic energy equation and solve for \(v_{1f}\): \(v_{1f} = 3.11 \mathrm{~m/s}\)

We have the final velocity of the first ball, so we can now calculate its final kinetic energy: \(K_{1f} = \frac{1}{2}mv_{1f}^{2} = \frac{1}{2} \times 0.210 \mathrm{~kg} \times (3.11 \mathrm{~m/s})^2 = 1.01 \mathrm{~J}\) The kinetic energy of the first ball after the collision is \(1.01 \mathrm{~J}\).