Question

An 18000 kg freight car rests against a spring bumper at the end of a railroad track. The spring has constant k=3.2×10^5 N/m. The car is hit by a second car of 9400 kg mass moving at 8.0 m/s , and the two couple together.A) Find the maximum compression of the spring.B) Find the speed of the two cars when they rebound together from the spring.

Answer 1

A) Maximum compression of spring = 0.803 meters

B) Speed of both cars as they rebound = 2.745 m/s

Step-by-step explanation:

Let's first find the initial speed at which both cars (coupled together) move against the spring.

We can calculate this using the conservation of momentum principle:

Momentum before the crash = Momentum after the crash

(M1 * V1) + (M2 * V2) = (M1 + M2) * V3

Here, M1 = mass of first freight car = 18000 kg

V1 = speed of first freight car = 0 m/s

M2 = mass of second car = 9400 kg

V2 = speed of second car = 8 m/s

V3 = Speed of both cars combined (after collision)

Solving for V3, we get:

18000 * 0 + 9400 * 8 = 27400 * V3

V3 = 2.745 m/s (Speed of both cars after collision)

A) To find the maximum compression, we just need to set the kinetic energy of both cars equal to the potential energy of the fully compressed spring. This is:

Kinetic energy = 0.5 * (18000 + 9400) * 2.745^2

Kinetic energy = 103229 J

Potential energy of spring = 0.5 * k * x^2

where k = spring constant

and x = displacement of spring.

So we have:

Potential energy = kinetic energy

0.5 * 3.2 * 10^5 * x^2 = 103229

x^2 = 0.64518

x = 0.803 meters (spring is compressed by this distance)

B) The speed of the cars when they rebound the spring (assuming no losses in energy) will be equal to the speed they had when they had just started compressing the spring. This is (as calculated before) 2.745 m/s