A car with inertia 1200kg is driving down a one-way street at ?5 m/s . An unobse

Question

A car with inertia 1200kg is driving down a one-way street at ?5 m/s . An unobservant driver of a 1800kg pickup truck is driving the wrong way down the street at a speed of +3 m/s , and runs head-on into the car.

(a) What kind of collision is this (totally inelastic, inelastic, or elastic), if the velocity of the pickup is ?1.5 m s after they collide? You must prove your answer.

(b) What is the change in internal energy of the system for this collision?

(c) What is the coefficient of restitution for this collision?

(d) What happens to the change in internal energy of a system if the coefficient of restitution gets smaller? For what kind of collision is the most energy converted?

Explanation / Answer

a)Momentum conservation

1200 * -5 + 1800 *3 =1800 * -1.5 + 1200 * V

V = 1.75 m/sec

So velocity of approach,Va = 5+3 =8 m/sec

Velocity of separation,Vs = 1.75 + 1.5 = 3.25 m/sec

So coefficient of restitution,e =Vs/Va= 3.25/8 = 0.40

So the collision is inelastic as e = 0.4 (ans)

b)Change in energy = (0.5*1200*25 + 0.5*1800*9) - (0.5 * 1800*2.25 + 0.5*1200*1.75*1.75) =19237.5 J (ans)

c) As found in 1st part,e=0.40 (ans)

d)As e gets lower, change in energy increases

And in perfectly elastic collision(e=1), all the energy gets converted (ans)

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