In a parking lot, an SUV of mass 2000 kg driving at 5 m/s collides head on with

Question

In a parking lot, an SUV of mass 2000 kg driving at 5 m/s collides head on with a sports car of mass 1000 kg driving in the opposite direction. The bumpers lock and the cars skid with locked brakes on the wet pavement (kinetic friction coefficient 0.1) for 2 m before they come to rest. We are interested in the velocity of the sports car before the collision.

(A) The problem naturally breaks into two parts.What are the parts?
(B)Which conservation laws apply?
(C) Determine the velocity of the sports car before the collision. Please use g = 10 m/s2and make sure your units work out.

Explanation / Answer

c) by conservation of momentum m1v1 - m2v2 = (m1+m2)v here m1 is mass of SUV v1 is its velocity m2 is mass of sports car v2 is its velocity v is their combined velocity after collision 2000(5) - 1000v1 = 3000v now distance they both traveled = 2m force on them due to friction = -(0.1)(mg) so acc = f/m = -0.1g so by kinematics equation v2^2 - v^2 = 2as 0 - v^2 = -2*.1*10*2 v = 2 so v1= (10000 - 6000)/1000 = 4 so velocity of sport car = 4m/sec a) two parts are in first conservation of momentum in second newton second law and kinematics equation b) conservation of momentum is applied here

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