Question

A truck with a mass of 1510 kg and moving with a speed of 14.5 m/s rear-ends a 671 kg car stopped at an intersection. The collision is approximately elastic since the car is in neutral, the brakes are off, the metal bumpers line up well and do not get damaged. Find the speed of both vehicles after the collision in meters per second. m/s vcar truck m/s

Answer 1

Mass of Truck (M) = 15 10 kg mass of car (m) = 671 Before collision Truck speed (U,)=14.5m/s Car speed (4)=0 After Collision Let, Truck speed =ų Car speed = Vz V 14.5 m/s & & Truck Fi и, о Truck Before Collision Atter Collision Given Collision is elastic, It means i momentum conserve ci RiE conserve 2- cis conservation of momentem Momentum before collision Momenfem After Collision Mu, muz = Myt me 15lox 14.5 + 671 X0 = |2|ox + 671 x 1, 21895 = 15104, + 671VZ Va = 21895 - 1510 vi 671 (ii) KEconservation- -CÚ 2 K.E before collision K.E ofler Collision Ź mugit į m (4_)* = M (V)?+ Ź m(« )? 1 1510X(14.5)*+ 0 = 11510(132) +671 (0) — (11

1) 317477.5 = 151000) + 671(0,5) Put value of a 317477.5 = 15101083) + 661 (21895 -15104.)? (67182 213027 402.5 = 1013210 v2+(21895-151049). 3293310142 - 6612290049 +266363622.520 solve this a-degree equation (By calculator) Vi = 550 m/s ANS speed of truck offer collision from equation (i), Vz= 21095- 1510x5.50 671 N 13469.2 671 y = 20.07 m/s ANS & -20.1 m/s y Speed of car ofter collision