A car with a mass of \(1700 \mathrm{kg}\) is traveling directly northeast $\left(45^{\circ} \text { between north and east) at a speed of } 14 \mathrm{m} / \mathrm{s}\right.\( \)(31 \mathrm{mph}),$ and collides with a smaller car with a mass of \(1300 \mathrm{kg}\) that is traveling directly south at a speed of \(18 \mathrm{m} / \mathrm{s}\) \((40 \mathrm{mph}) .\) The two cars stick together during the collision. With what speed and direction does the tangled mess of metal move right after the collision?

Solution: 1. Understand the momentum conservation principle: The total momentum of the system is conserved in a collision if there are no external forces acting on the system. 2. Resolve velocities into components: Break down the velocities of both cars into their x and y components. 3. Calculate initial and final momenta: Determine the initial momenta along the x and y directions for each car and conserve them after the collision. 4. Calculate the magnitude and direction of the final velocity: Use the Pythagorean theorem to find the magnitude, and the arctangent function to find the direction. Calculations: 1. For larger car: - \(v_{1x} = 14 \cos(45^{\circ}) = 9.90 \frac{\mathrm{m}}{\mathrm{s}}\) - \(v_{1y} = 14 \sin(45^{\circ}) = 9.90 \frac{\mathrm{m}}{\mathrm{s}}\) 2. For smaller car: - \(v_{2x} = 0 \frac{\mathrm{m}}{\mathrm{s}}\) - \(v_{2y} = -18 \frac{\mathrm{m}}{\mathrm{s}}\) 3. Initial momenta: - \(p_{1ix} = 1800 \times 9.90 = 17820 \frac{\mathrm{kg \cdot m}}{\mathrm{s}}\) - \(p_{1iy} = 1800 \times 9.90 = 17820 \frac{\mathrm{kg \cdot m}}{\mathrm{s}}\) - \(p_{2ix} = 0 \frac{\mathrm{kg \cdot m}}{\mathrm{s}}\) - \(p_{2iy} = 1200 \times (-18) = -21600 \frac{\mathrm{kg \cdot m}}{\mathrm{s}}\) 4. Final momenta and velocity components: - \(p_{fx} = 17820 + 0 = 17820 \frac{\mathrm{kg \cdot m}}{\mathrm{s}}\) - \(p_{fy} = 17820 - 21600 = -3780 \frac{\mathrm{kg \cdot m}}{\mathrm{s}}\) - \(v_{fx} = \frac{17820}{3000} = 5.94 \frac{\mathrm{m}}{\mathrm{s}}\) - \(v_{fy} = \frac{-3780}{3000} = -1.26 \frac{\mathrm{m}}{\mathrm{s}}\) 5. Magnitude and direction: - \(v_{f} = \sqrt{5.94^2 + (-1.26)^2} = 6.07 \frac{\mathrm{m}}{\mathrm{s}}\) - \(\theta = \arctan \frac{-1.26}{5.94} = -12.05^{\circ}\) (measured clockwise from east) The final velocity of the combined wreckage after the collision is 6.07 m/s, 12.05° south of east.