Question

A 2000-kg truck is sitting at rest (in neutral) when it is rear-ended by a 1000-kg car going 20m/s .After the collision, the two vehicles stick together.

1)What is the final speed of the car-truck combination?

2)What is the kinetic energy of the two-vehicle system before the collision?

3)What is the kinetic energy of the system after the collision?

4) Based on the results of the previous parts, what can you conclude about which type of collision this is? Elastic, Inelastic, or totally inelastic?

5) Calculate the coefficient of restitution for this collision.

Answer 1

Part A.

Using Momentum conservation:

Pi = Pf

m1u1 + m2u2 = M*V

m1 = mass of car = 1000 kg

m2 = mass of truck = 2000 kg

u1 = initial speed of car = 20 m/sec

u2 = initial speed of truck = 0 m/sec

M = Mass of car + truck = 1000 + 2000 = 3000 kg

V = final speed of coupled car + truck = ?

Using these values:

1000*20 + 2000*0 = 3000*V

V = 20000/3000

V = 6.67 m/sec = final speed of coupled car and truck

Part B.

Kinetic energy before the collision will be:

KEi = (1/2)*m1*u1^2 + (1/2)*m2*u2^2

KEi = (1/2)*1000*20^2 + (1/2)*2000*0^2

KEi = 200000 J

Part C.

Kinetic energy after the collision will be:

KEf = (1/2)*M*V^2

KEf = (1/2)*3000*6.67^2

KEf = 66733.35 J

Part D.

From above we can see that kinetic energy does not remain conserved and also car and truck are interlocked after collision, So this is totally inelastic collision

Part E.

Coefficient of restitution is given by:

e = (V2 - V1)/(U1 - U2)

Since in this case V2 = V1, since both objects are stick together, So

V2 - V1 = 0,

which means

e = coefficient of restitution = 0

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