Mastering Physics, Finding the Compresssion of a Spring using Momentum.

[theoblivin8r]

This is another problem from the same homework set that I was working on earlier. Again I believe conservation of momentum is the key formula.

Homework Statement

An 8000kg freight car rests against a spring bumper at the end of a railroad track. The spring has constant k=3.2*10^5. The car is hit by a second car of 7600kg mass moving at 6.8 m/s, and the two cars couple together.

1.) What is the maximum compression of the spring?

and

2.) What is the speed of the two cars together when they rebound from the spring?

Homework Equations

M1V1+M2V2=(M1+M2)Vfinal
F=-kx.

The Attempt at a Solution

I am having trouble bringing the compression of the spring into the equation. In order to find the compression I need the Force that the collision of the cars exert on it using F=m*a. I can kind the mass easily it the combination of both of the cars or 8000kg+7600kg=15600kg. How do I find the acceleration. I know that acceleration is change of velocity over time. I could find the velocity of the coupled cars using the conservation of momentum equation but I still don't have time. This is were I reach my dead end in my logic. A nudge in the right direction would help a ton. Thanks.

 

[ƒ(x)]

Do you remember the formula for momentum?

 

[ƒ(x)]

Being no expert, as I am only taking physics now myself, I would find how much kinetic energy the moving car has.

 

[ƒ(x)]

Ok:

Part 1: Think kinetic energy and potential elastic energy.

Part 2: Think conservation of energy

My answers are below in a spoiler tag for reference.

Spoiler

1)

KE = (1/2)mv²
KE = (1/2)(7600)(6.8)² = 175712 J

PE_Elastic = (1/2)kx²
And, since all of the kinetic energy from the moving car will be transferred to the spring:
PE = 175712 J
175712 = (1/2)(3.2 x 10⁵)x²
1.048 m = x

2)

I'm not sure about this method, but it is rather elegant so I'll post it.

Since all of the initial kinetic energy of the moving car is being transferred to the spring and then into the two cars

KE_{Moving car} = KE_{Both cars after the spring has decompressed}

(1/2)(7600)(6.8²) = (1/2)(7600+8000)(v_f)²

4.764 m/s = v_f

 

[ideasrule]

This is actually an energy question, not a momentum question. It's easy to prove that momentum ISN'T conserved: when the spring reaches its maximum compression, the cars must have a velocity of 0. The whole system now has 0 momentum, which isn't equal to its initial momentum of (humongous mass)*(terrific speed). For momentum to be conserved, you'd have to consider the Earth as part of the system.

So, energy's roughly conserved and the trains have 0 speed when the spring reaches min. length. What's the min. length?

 

[theoblivin8r]

F(x) your answers were incorrect according to Mastering Physics. I am having trouble understand how momentum does not come into play in this problem since it is in my momentum chapter. I blame my professor. Anyways, it is getting late and I don't have time to finish the problem. Thanks for your help though.

 

[ideasrule]

Momentum does come into play, just not in the final step. After the two cars collide and stick together, you have to determine their final speed before you can use E=1/2mv^2 to calculate kinetic energy. That final speed can only be arrived at using the conservation of energy. (Sorry if I meant to imply that momentum isn't used at all.)