A 600-kg car makes a 90° turn. Its speed before the turn is 21.0 m/s and after the turn it is 24.0 m/s. What is the magnitude of the change in the car's momentum? A) 19.1 × 10³ kg∙m/s B) 10.2 × 10³ kg∙m/s C) 9.55 × 10³ kg∙m/s D) 12.0 × 10³ kg∙m/s E) 22.2 × 10³ kg∙m/s

Question

Quizplus · Accepted Answer

The change in momentum is equal to the final momentum minus the initial momentum. The final momentum is equal to the mass times the final velocity, and the initial momentum is equal to the mass times the initial velocity. Therefore, the change in momentum is equal to:$$\Delta p = p_f - p_i = mv_f - mv_i$$$$= (600 	ext{ kg})(24.0 	ext{ m/s}) - (600 	ext{ kg})(21.0 	ext{ m/s})$$$$= 19.1 	imes 10^3 	ext{ kg∙m/s}$$

A 600-Kg Car Makes a 90° Turn