A 900-kg car traveling east at 15.0 m/s suddenly collides with a 750-kg car traveling north at 20.0 m/s.The cars stick together after the collision.In what direction does the wreckage move just after the collision?

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Quizplus · Accepted Answer

The correct answer is E.The total momentum of the system before the collision is:$$P_i = m_1v_1 + m_2v_2$$$$P_i = (900 	ext{ kg})(15.0 	ext{ m/s})\hat{i} + (750 	ext{ kg})(20.0 	ext{ m/s})\hat{j}$$$$P_i = (13500 	ext{ kg m/s})\hat{i} + (15000 	ext{ kg m/s})\hat{j}$$The total momentum of the system after the collision is:$$P_f = (m_1 + m_2)v_f$$$$P_f = (900 	ext{ kg} + 750 	ext{ kg})v_f$$$$P_f = (1650 	ext{ kg})v_f$$The law of conservation of momentum states that the total momentum of a system remains constant in the absence of external forces. Therefore, we have:$$P_i = P_f$$$$(13500 	ext{ kg m/s})\hat{i} + (15000 	ext{ kg m/s})\hat{j} = (1650 	ext{ kg})v_f$$Solving for $v_f$, we get:$$v_f = \frac{(13500 	ext{ kg m/s})\hat{i} + (15000 	ext{ kg m/s})\hat{j}}{1650 	ext{ kg}}$$$$v_f = (8.18 	ext{ m/s})\hat{i} + (9.09 	ext{ m/s})\hat{j}$$The direction of the wreckage just after the collision is:$$	heta = 	an^{-1}\left(\frac{9.09 	ext{ m/s}}{8.18 	ext{ m/s}}ight)$$$$	heta = 48.0^\circ 	ext{ N of E}$$

A 900-Kg Car Traveling East at 15