A 1050-kg sports car is moving westbound at 15.0 m/s on a level road when it collides with a 6320-kg truck driving east on the same road at 10.0 m/s. The two vehicles remain locked together after the collision. (a) What is the velocity (magnitude and direction) of the two vehicles just after the collision? (b) At what speed should the truck have been moving so that both it and the car are stopped in the collision? (c) Find the change in kinetic energy of the system of two vehicles for the situations of part (a) and part (b). For which situation is the change in kinetic energy greater in magnitude?

Solution 36E To solve this question, we shall have to apply the conservation of momentum principle before and after collision. Given, mass of the sports car m = 1050 sg Its speed v = s 15.0 m/s (negative sign taken for westbound motion) Mass of the truck m = 6320 kg t Its speed v = 10.0 m/s t They remain locked after the collision, hence they will have a common speed. Let this speed be V. (a) herefore, momentum before collision = Momentum after collision m vs+ s v = (m + m )V s t t t 1050 kg × ( 15.0 m/s) + 6320 kg × 10.0 m/s = (1050 + 6320) × V 47450 kg.m/s V = 7370 kg V = 6.44 m/s Since the speed value is positive, hence it will be eastbound. (b) Given that, both the car and the truck are stopped after the collision, therefore V = 0 Therefore, the initial momentum of the system will be zero. So, m v s msv = 0 t t 1050 kg × ( 15.0 m/s) + 6320 kg × v = 0 t v =t2.50 m/s Therefore, the speed of the truck should be 2.50 m/s for...