A railroad car of mass 253 104 kg is moving with a speed of

A railroad car of mass 2.53 104 kg is moving with a speed of 3.94 m/s. It collides and couples with three other coupled railroad cars, each of the same mass as the single car and moving in the same direction with an initial speed of 1.97 m/s.

(a) What is the speed of the four cars after the collision?

(b) How much mechanical energy is lost in the collision?

Solution

m1 = 2.53 x 10^4 kg
u1 = 3.94 m/s

Two cars, same mass as first: (2.53 x 10^4)*2 = 5.06 x 10^4 kg

m2 = 5.06 x 10^4 kg
u2 = 1.97 m/s

(a) velocity

Equation:

v * (m1 + m2) = m1u1 + m2u2

Solve for post-impact velocity

v * [ (2.53 x 10^4 kg) + (5.06 x 10^4 kg) ] = [ (2.53 x 10^4 kg) * (3.94 m/s) ] + [ (5.06 x 10^4 kg) * (1.97 m/s) ]
v * [ 75900 kg ] = [ 99682 kg-m/s ] + [ 99682 kg-m/s ]
v * [ 75900 kg ] = [ 199364 kg-m/s ]
v = [199364 kg-m/s ] / [ 75900 kg ]

Post-impact velocity = 2.63 m/s

(b) energy

KE = 0.5 * m * v^2

KE of 1st car
KE = 0.5 * (2.53 x 10^4 kg) * (3.94 m/s)^2
KE = 196373.5 J

KE of the 2 cars
KE = 0.5 * (5.06 x 10^4 kg) * (1.97 m/s)^2
KE = 98186.77 J

So KE of pre-impact system:

(196373.5 J) + (98186.77 J) = 294560.27 J

KE of 3 car combination post-impact
KE = 0.5 * (75900 kg) * (2.63 m/s)^2
KE = 262496.355 J

Energy loss:

(294560.27 J) - (264496.355 J) = 30063.915 J