Question

A railroad car of mass 3.00 ? 10^{4} kg moving at 3.00
m/s collides and couples with two coupled railroad cars, each of
the same mass as the single car and moving in the same direction at
1.20 m/s.

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

_______ m/s

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

_______ J

Answer #1

Solution:

Mass of the cars = m1 =m2 = m3 = m =3 x10^4 m/s

Velocity of car 1 = v1 = 3m/s

Mass of the coupled cars = m2+m3= 2m = 6 x 10^ kg

velocity of the coupled cars = v 2= 1.2 m/s

Since it is an inlastic collision, velocity of the coupled cars (all 3) is derived using the momentum conservation .

m v1 + 2mv2 = (m+2m) V

V = (mv1 + 2mv2) / (m+m+m) = m (v1+v2) / 3m

=> **Velocity of the three coupled cars = V** =
(3+1.2) / 3 = **1.40 m/s**

b) Initial kinetic energy = 1/2m1v1^2 + 1/2 (2m)v2^2 = 1/2 * (3x10^4)(3)^2 + 1/2 ( 2*3x10^4)(1.2)^2 =

**=** 13.5 x10^4 + 4.32 x 10^4 **= 1.78 x
10 ^{5} joules**

Final kinetic energy = 1/2 (3m)V^2 = 1/2 x 3 x 3 x10^4 ( 1.4)^2
= 8.82 x10^4 joules = **0.882 x 10^5 J**

**Loss in kinetic energy = Final - initial kinetic energy
= (0.882 -1.78) 10^5 J = -9 x10^5 J**

**= 9.0 0x 10 ^{5} joules (negative sign indicates
loss).**

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