How Far Will the Spring Be Compressed?

[pleasehelpme6]

A cart of mass m = 2 kg carrying a spring of spring constant k = 48 N/m and moving at speed v = 3.2 m/s hits a stationary cart of mass M = 10 kg. Assume all motion is along a line.

a) What is the speed of the center of mass of this system? A: .5333 m/s

b) When the spring is at its maximum compression, with what speed are the carts moving in the lab frame? A: .5333 m/s

c) How far will the spring be compressed?
This is where I have trouble. I know potential energy of spring = 1/2kx^2, but how do i find initial kinetic energy?

The equation i tried was the following:

(1/2)*m*(Vi+Vcm) = (1/2)*k*x^2

i also tried (1/2)*m*(Vi) = (1/2)*k*x^2, but neither of these worked.

any suggestions?

 

[Doc Al]

c) How far will the spring be compressed?
This is where I have trouble. I know potential energy of spring = 1/2kx^2, but how do i find initial kinetic energy?

The equation i tried was the following:

(1/2)*m*(Vi+Vcm) = (1/2)*k*x^2

i also tried (1/2)*m*(Vi) = (1/2)*k*x^2, but neither of these worked.

any suggestions?

What's the KE of the system before the collision? Thus, what's the total mechanical energy before the collision?

What's the KE of the system at the point of maximum compression? (Use the result of question (b).)

 

[pleasehelpme6]

What's the KE of the system before the collision? Thus, what's the total mechanical energy before the collision?

What's the KE of the system at the point of maximum compression? (Use the result of question (b).)

Before collision, KE = 1/2*(2)*(3.2)^2 = 10.24
At max compression, KE = 12*(2)*(3.2+0.5333)2?

Here, what i don't understand is which mass and which velocity i use...

or would it be, at max compression...
KE = 1/2*2*(.5333)^2?

 

[Doc Al]

Before collision, KE = 1/2*(2)*(3.2)^2 = 10.24

Right. The only thing moving is the 2kg mass.

At max compression, KE = 12*(2)*(3.2+0.5333)2?

Why are you adding the speeds?

Here, what i don't understand is which mass and which velocity i use...

or would it be, at max compression...
KE = 1/2*2*(.5333)^2?

You're getting warmer. But you only found the KE of the 2kg mass. What about the 10kg mass? (Hint: At the point of maximum compression, everything is moving at the same speed.)

 

[pleasehelpme6]

Right, I tried using the following already:

KE = PE
(1/2)mv^2 = (1/2)kx^2
substituting in my numbers...

(1/2)(12)(.5333)^2 = (1/2)48x^2, and i got x = 2.6667. but that's not the right answer.

is there something I am still missing?

 

[Doc Al]

Right, I tried using the following already:

KE = PE
(1/2)mv^2 = (1/2)kx^2
substituting in my numbers...

(1/2)(12)(.5333)^2 = (1/2)48x^2, and i got x = 2.6667. but that's not the right answer.

is there something I am still missing?

For some reason you are setting the KE equal to the PE. Instead, use energy conservation:
KE_i + PE_i = KE_f + PE_f

Initially, there is only KE. (The spring hasn't been compressed yet.)

At the point of maximum compression, there is both KE and spring PE. You found the KE, so set up an equation and solve for the spring PE. (That will tell you the amount of compression.)

 

[pleasehelpme6]

Hmm...

so before collision, Total Energy = 1/2*(2)*(3.2)^2 = 10.24
After, Total Energy = (1/2)(12)(.5333)^2 + (1/2)48x^2

So set those equal and solve?

 

[Doc Al]

Yep!

 

[pleasehelpme6]

Yes! That worked. The answer I got for max compression distance was 0.596 m.
Thanks so much, I actually understand it now.

 

[Doc Al]

Sweet.