Car 1 has mass $1250 \mathrm{~kg}$ is initially traveling east with speed $15.0 \mathrm{~m} / \mathrm{s}$ when it collides inelastically with car 2 that has mass $1860 \mathrm{~kg}$ traveling north with speed $11.0 \mathrm{~m} / \mathrm{s}$, as shown. What angle $\theta$ does the car wreck's velocity travel with north of east after the collision? $58.1^{\circ}$ $47.5^{\circ}$ $42.5^{\circ}$ $74.6^{\circ}$

Step 1 ：Given that car 1 has a mass of \(1250 \mathrm{~kg}\) and is initially traveling east with a speed of \(15.0 \mathrm{~m} / \mathrm{s}\), and car 2 has a mass of \(1860 \mathrm{~kg}\) and is traveling north with a speed of \(11.0 \mathrm{~m} / \mathrm{s}\).

Step 2 ：The problem involves an inelastic collision between the two cars. The final velocity of the wrecked cars will be the vector sum of the initial velocities of the two cars.

Step 3 ：We can calculate the components of the final velocity in the east and north directions separately. The final velocity in the east direction is \(6.028938906752412 \mathrm{~m/s}\) and in the north direction is \(6.578778135048232 \mathrm{~m/s}\).

Step 4 ：The angle \(\theta\) with the east direction can then be calculated using the arctangent of the ratio of the north component to the east component of the final velocity. The calculated angle is \(47.49716242849709\) degrees.

Step 5 ：Rounding to the nearest tenth, the angle \(\theta\) that the car wreck's velocity travels with north of east after the collision is \(\boxed{47.5^\circ}\).