1. A railroad car of mass 2.40E+4 kg moving with a speed of 3.10 m/s collides an

Question

1. A railroad car of mass 2.40E+4 kg moving with a speed of 3.10 m/s collides and couples with two other coupled railroad cars each of the same mass that are already moving in the same direction at a speed of 1.00 m/s.

a. What is the speed (in m/s) of the three coupled cars after the collision?

b.How much kinetic energy (in J) is lost in the collision?

2.A ring (2 kg, r = 2 m) rotates in a CW direction with initial angular velocity 20 s-1. A disk (8 kg, r = 2 m) rotates in a CCW direction with initial angular velocity 50 s-1. The ring and disk "collide" and eventually rotate together. Assume that positive angular momentum and angular velocity values correspond to rotation in the CCW direction.

a. What is the initial angular momentum Li of the ring+disk system?

b.What is the final angular velocity ?f of the ring+disk system?

Explanation / Answer

First question -

(1) (a) mass of the railroad car, m = 2.40 x 10^4 kg = 24000 kg

velocity, v1 = 3.10 m/s

velocity of the two railroad car, v2 = 1.00 m/s

Intial momentum, Pi = m*v1 + 2m*v2

suppose the final velocity of the railroad car is V.

So, final momentum, Pf = 3m*V

Apply conservation of momentum -

Pi = Pf

=> m*v1 + 2m*v2 = 3m*V

=> V = (v1 + 2v2) / 3 = (3.10 + 2*1.00) / 3 = 1.70 m/s.

(b) Initial kinetic energy, Ki = (1/2)*m*v1^2 + (1/2)*2m*v2^2

= 0.5*m*(3.1^2 + 2*1.0^2)

= 0.5*24000*11.61 = 139320 J

Final kinetic energy = (1/2)*3m*V^2 = 0.5*3*24000*1.7^2 = 104040 J

So. loss in kinetic energy in the collision = Ki - Kf = 139320 - 104040 = 35280 J

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