A 1500-kg car moving east at \(17 \mathrm{m} / \mathrm{s}\) collides with a 1800-\(\mathrm{kg}\) car moving south at \(15 \mathrm{m} / \mathrm{s}\) and the two cars stick together. (a) What is the velocity of the cars right after the collision? (b) How much kinetic energy was converted to another form during the collision?

First, we can find the initial momentum of both cars, before the collision. We can use the formula for momentum: \(p=mv\), where \(p\) is the momentum, \(m\) is the mass, and \(v\) is the velocity. Car 1 (1500 kg): \(p_1 = m_1v_1 = 1500\,\text{kg} \times 17\,\text{m/s} = 25500\,\text{kg}\,\text{m/s}\) Car 2 (1800 kg): \(p_2 = m_2v_2 = 1800\,\text{kg} \times 15\,\text{m/s} = 27000\,\text{kg}\,\text{m/s}\)

According to the conservation of linear momentum, the total momentum before the collision should be equal to the total momentum after the collision. Since both of the cars are moving in different directions, we need to consider the momentum conservation in both the x and y directions separately. Total initial momentum in x-direction (east): \(p_{1x} = 25500\,\text{kg}\,\text{m/s}\) Total initial momentum in y-direction (south): \(p_{2y} = 27000\,\text{kg}\,\text{m/s}\) Total mass after the collision: \(m_{tot} = m_1 + m_2 = 1500\,\text{kg} + 1800\,\text{kg} = 3300\,\text{kg}\) Since the total initial momentum should be equal to the total final momentum, we can find the final velocity components by dividing the initial momentum components by the total mass: \(v_{fx} = \frac{p_{1x}}{m_{tot}} = \frac{25500\,\text{kg}\,\text{m/s}}{3300\,\text{kg}} = 7.73\,\text{m/s}\) \(v_{fy} = \frac{p_{2y}}{m_{tot}} = \frac{27000\,\text{kg}\,\text{m/s}}{3300\,\text{kg}} = 8.18\,\text{m/s}\)

To find the final velocity after the collision, we need to combine the final velocity components found in the previous step. We can use the Pythagorean theorem to find the magnitude of the final velocity, and calculate the angle with respect to the x-axis (east) using trigonometry. \(v_{f} = \sqrt{v_{fx}^2 + v_{fy}^2} = \sqrt{(7.73\,\text{m/s})^2 + (8.18\,\text{m/s})^2} = 11.26\,\text{m/s}\) \(\theta_f = \text{tan}^{-1}\frac{v_{fy}}{v_{fx}} = \text{tan}^{-1}\frac{8.18\,\text{m/s}}{7.73\,\text{m/s}} = 46.9^{\circ}\) So the final velocity right after the collision is \(11.26\,\text{m/s}\) at an angle of \(46.9^{\circ}\) south of east.

To find how much kinetic energy was converted to another form, we need to calculate the initial and final kinetic energies. Initial kinetic energies: \(KE_1 = \frac{1}{2}m_1v_1^2 = \frac{1}{2}(1500\,\text{kg})(17\,\text{m/s})^2 = 217125\,\text{J}\) \(KE_2 = \frac{1}{2}m_2v_2^2 = \frac{1}{2}(1800\,\text{kg})(15\,\text{m/s})^2 = 202500\,\text{J}\) Final kinetic energy: \(KE_f = \frac{1}{2}m_{tot}v_f^2 = \frac{1}{2}(3300\,\text{kg})(11.26\,\text{m/s})^2 = 210336.6\,\text{J}\)

Finally, to find the converted kinetic energy, we can subtract the final kinetic energy from the initial kinetic energies. Converted kinetic energy = \((KE_1 + KE_2) - KE_f = (217125\,\text{J} + 202500\,\text{J}) - 210336.6\,\text{J} = 209288.4\,\text{J}\) Hence, \(209288.4\,\text{J}\) of kinetic energy was converted to another form during the collision.