Question

A locomotive of mass 2.1 x 104 kg travelling at 2.0 m/s runs into an identical stationary locomotive.
a) If the collision is perfectly inelastic, what is the speed of the two trains after the collision?
b) If the collision is perfectly inelastic, what fraction of the energy is lost after the collision?
c) If the collision is perfectly elastic, find the motions of the locomotives after the collision.
d) If the collision is perfectly elastic and lasts for 1.5 s, find the average force exerted on the stationary locomotive during the collision.

Answer 1

if they stick together:
original momentum = 2.1*10^4 * 2

final momentum = 4.2*10^4*v
so
v = 1 m/s

initial ke = (1/2)m v^2
final ke = (1/2)(2m)(v/2)^2
final/initial = [2v^2/4]/v^2
=1/2 so you lose half

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LOL if perfectly elastic, first one stops, second continues at original speed:)
F = change of momentum/time = mv/1.5

Answer 2

Given:
M1 = 2.1*10^4kg, V1 = 2 m/s.
M2 = 2.1*10^4kg, V2 = 0.

a. M1*V1 + M2*V2 = M1*V + M2*V.
2.1*10^4*2 + M2*0 = 4.2*10^4*V,
4.2*10^4 = 4.2*10^4*V,
V = 1 m/s each.

b. Before the collision:
KE = 0.5M1*V1^2 + 0.5M2*V2^2.
KE = 0.5*2.1*10^4*2^2 + 0.5*2.1*10^4*0 = 4.2*10^4 Joules.
After the collision:
KE = 0.5M1*V^2 + 0.5*M2*V^2.
KE = 0.5*2.1*10^4*1^2 + 0.5*2.1*10^4*1^2 = 2.1*10^4 Joules.

KE Lost = 4.2*10^4 - 2.1*10^4 = 2.1*10^4 Joules.

Fraction Lost = 2.1*10^4/4.2*10^4 = 1/2.

Answer 3

c. M1*V1 + M2*V2 = M1*V3 + M2*V4.
2.1*10^4*2 + M2*0 = 2.1*10^4*V3+2.1*10^4*V4,
4.2*10^4 = 2.1*10^4*V3 + 2.1*10^4*V4,
Divide both sides by 2.1*10^4:
Eq1: V3 + V4 = 2.

V3 = (V1(M1-M2) + 2M2*V2)/(M1+M2).
V3 = (2(0) + 2M2*0)/(4.2*10^4) = 0 = Velocity of M1 after the collision.

In Eq1, replace V3 with 0 and solve for V4:
0 + V4 = 2.
V4 = 2 m/s = Velocity of M2 after the collision. INTERESTING!