Harmonic Motion: Collision Calculation

[wilson_chem90]

A new car, car 1, has a spring-loaded rear bumpe with a force constant of 8.4 x 10^5 N/m. While car 1 is parked, a second vehicle, car 2 (with a mass of 1.5 x 10^3 kg), travels at a constant speed of 18 km/h (5 m/s), hitting car 1 in the rear bumper.

a) Calculate the kinetic energy of car 2

b) Calculate the distance that car 1's bumper will compress if car 2 comes to a complete stop after striking it.

I found the answer for a, its 18,750 J but for b) I keep trying different equations and my units don't equal to x, which is metres obviously.

I've tried these equations thus far:

Et = 1/2mv^2 + 1/2 kx^2
1/2mvi^2 + 1/2kxi^2 = 1/2mvf^2 + 1/2kxf^2
and W = 1/2kx^2

 

[belliott4488]

Okay, when car 1's bumper is completely compressed, what is car 2's kinetic energy?

If it's changed (yes, it has), where has the energy gone?

 

[wilson_chem90]

shouldn't it be the same? because it was transferred to car 2?

 

[belliott4488]

Nope - Car 2 has come to a stop, right? So how much kinetic energy does an object have if it's not moving?

By the way, as I understand your problem statement we are to assume that Car 1 never moves, either it's infinitely heavy or it just has really good brakes and tires. If it moved, you couldn't answer the question without knowing how fast it was moving.

 

[wilson_chem90]

i mean the energy from the compressed bumper would transfer into car 2

 

[wilson_chem90]

well it would have 0 kinetic energy cause its not moving.

 

[belliott4488]

Yes - at the moment when the bumper spring is compressed the most, car 2 has come to a stop (maybe it will bounce back afterwards, but we're not asked about that). So you're right - its kinetic energy is now zero.

But total energy is conserved, right? So that energy is now stored somewhere. One of your equations describes that. (Think of what is doing the work of stopping the car.)

 

[wilson_chem90]

soo its the W = 1/2kx^2 equation. I rearranged it to be x = {2/k} I'm not sure if that's right

 

[belliott4488]

soo its the W = 1/2kx^2 equation. I rearranged it to be x = {2/k} I'm not sure if that's right

I don't think so ...

x^2 is x squared, right? So you need a couple of things - first, you haven't told me what W is (although I think you know), but once you have that, can you get x by itself to solve for it? (This is algebra at this point.)

 

[belliott4488]

Maybe it would help to write this:

W = 1/2kx^2 = k/2 x^2

I didn't do anything, but it's a little easier to read.

 

[wilson_chem90]

well W = 0 right? seeing how there is no work being done cause its at rest.

but if you have W in the equation it would be x = {k/2} ( {} = square root)

 

[belliott4488]

well W = 0 right? seeing how there is no work being done cause its at rest.

but if you have W in the equation it would be x = {k/2} ( {} = square root)

"( {} = square root)"! OH! Now I get it ...

But you're still not quite there. W is the amount of work that was done in moving the spring from its equilibrium position, i.e. where it started, to its current position. That means that it represents the amount of energy that has been stored in the spring by compressing it. Think about it: if you squeeze a stiff spring a lot, then you know that it's got potential energy stored in it - if you release it suddenly you can launch a ball across the room with it.

Back to the problem - the total energy can't change, so the kinetic energy that car 2 had at the start must still be somewhere - and you know that the spring has potential energy stored in it once it's compressed ... see where to go now?

I'll bet you do, so I'm calling it a night. Good luck!

 

[wilson_chem90]

OHHH i get it now! so the amount of work done is equal to the kinetic energy of the second car, so then the equation would be W = k/2x^2 and rearranged to x = {2(W/k)}
... please be right lol

 

[ideasrule]

OHHH i get it now! so the amount of work done is equal to the kinetic energy of the second car, so then the equation would be W = k/2x^2 and rearranged to x = {2(W/k)}
... please be right lol

That's right. Now calculate W, the kinetic energy of the second car.

For future questions about conservation of energy, remember this handy formula:

Initial potential energy + initial kinetic energy = final potential energy + final kinetic energy

 

[wilson_chem90]

sickk thank you, its my last question. so happy its done