Inelastic collisions in two dimensions: Two automobiles traveling at right angles to each other collide and stick together. Car A has a mass of 1200 kg and had a speed of 25 m/s before the collision. Car B has a mass of 1600 kg. The skid marks show that, immediately after the collision, the wreckage was moving in a direction making an angle of 40° with the original direction of car A. What was the speed of car B before the collision, assuming that any other unbalanced forces are negligible?

Question

Quizplus · Accepted Answer

Explanation:Momentum is conserved in both the x and y directions. In the x direction, the total momentum before the collision is 1200 kg * 25 m/s = 30000 kg m/s. After the collision, the total momentum is (1200 kg + 1600 kg) * v_x = 2800 kg * v_x. Equating these two expressions, we get v_x = 10.71 m/s. In the y direction, the total momentum before the collision is 0 kg m/s. After the collision, the total momentum is (1200 kg + 1600 kg) * v_y = 2800 kg * v_y. Equating these two expressions, we get v_y = 14.29 m/s. The speed of car B before the collision is then sqrt(v_x^2 + v_y^2) = 17.89 m/s, which is closest to choice A, 16 m/s.

Inelastic Collisions in Two Dimensions: Two Automobiles Traveling at Right