A 1050 kg sports car is moving westbound at 14.0 m/s on a level road when it col

Question

A 1050 kg sports car is moving westbound at 14.0 m/s on a level road when it collides with a 6320 kg truck driving east on the same road at 14.0 m/s . The two vehicles remain locked together after the collision.

a)   What is the velocity (magnitude) of the two vehicles just after the collision?

b) At what speed should the truck have been moving so that it and car are both stopped in the collision?

c) Find the change in kinetic energy of the system of two vehicles for the situations of part A.

d) Find the change in kinetic energy of the system of two vehicles for the situations of part C.

Explanation / Answer

a)

conservation of momentum

momentum before collision = momentum after collision

6320*14 - 1050*14 = (6320 + 1050)*V

V = (6320*14 - 1050*14)/(6320 + 1050) = 10.01 m/s

b)

so 1050*14 = 6320*V

V = 1050*14/6320 = 2.32 m/s

c)

KE = (1/2)mv^2

the change in kinetic energy = (1/2)*1050*14^2 + (1/2)*6320*14^2 - (1/2)*(1050+6320)*10.01^2 = 353022.63 J

d)

delta KE = (1/2)*1050*14^2 + (1/2)*6320*2.32^2 - 0 = 119908.38 J

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