Question

Two cars are approaching a perpendicular intersection without a stop sign. Car 1 has a mass mi =700kg and is heading north at v1 =15m/s. Car 2 has mass m2 =600kg and is heading west at an unknown speed V2. The two cars collide at the intersection and stick together as a result of the collision. The police report stated that after the collision, the two cars moved together in a direction 40° west of north and stopped after sliding d =35m from the collision point. 1. Find the speed of car 2 before the collision V2= 2. What is the speed of the cars immediately after the collision? V = 3. What is the coefficient of kinetic friction between the cars and the pavement? Mk =

Answer 1

40 (mit M₂) V M2 V2 mi V Use conservation of momentum in x. direction, -M₂.V₂ to --(mitm) v Sin 40°----(i) in y-direction, miv,to (M,+ m₂) v cos 40°. (ii) V = 700 x 15 (700 +600) x cos 40° 10.54367 mis in Ci), Putting this ean cm, tma) rsin 40° Vz 14.68424 m/s m2 (1) V2 = 14.68424 m/s (2) V = 10.54367 m 13 e in (3) Work done by friction change Kinetic energy Ź tron kumo) v? xd (M112) g => elk elk = (10.54367)2 V? agd 0.162 2* 9.8 x 35