Spring Loaded Collision


by moonbase
Tags: collision, loaded, spring
moonbase
moonbase is offline
#1
Mar15-12, 11:22 PM
P: 21
1. The problem statement, all variables and given/known data
A cart of mass m = 3 kg carrying a spring of spring constant k = 46 N/m and moving at speed v = 2.8 m/s hits a stationary cart of mass M = 8 kg. Assume all motion is along a line. What is the maximum amount the spring will be compressed?

2. Relevant equations
PEspring=0.5kx2
KEinitial=0.5mv2

3. The attempt at a solution
A previous part of this question asked me to calculate the mechanical energy in the center of mass frame, which I correctly found to be 8.55 J. I then had to find the total potential energy of the system when the spring is fully compressed, which I correctly calculated to also be 8.55 J. But when I plug this into the spring energy equation, the value I get for x is apparently incorrect.

0.5(46)x2=8.55 -> x=0.61

I must be missing a step here. Anyone see what I did wrong?
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PeterO
PeterO is offline
#2
Mar15-12, 11:37 PM
HW Helper
P: 2,316
Quote Quote by moonbase View Post
1. The problem statement, all variables and given/known data
A cart of mass m = 3 kg carrying a spring of spring constant k = 46 N/m and moving at speed v = 2.8 m/s hits a stationary cart of mass M = 8 kg. Assume all motion is along a line. What is the maximum amount the spring will be compressed?

2. Relevant equations
PEspring=0.5kx2
KEinitial=0.5mv2

3. The attempt at a solution
A previous part of this question asked me to calculate the mechanical energy in the center of mass frame, which I correctly found to be 8.55 J. I then had to find the total potential energy of the system when the spring is fully compressed, which I correctly calculated to also be 8.55 J. But when I plug this into the spring energy equation, the value I get for x is apparently incorrect.

0.5(46)x2=8.55 -> x=0.61

I must be missing a step here. Anyone see what I did wrong?
How incorrect was your answer?
By my calculation, there was only about 7.15 J of energy stored in the spring? I may have been wrong - I did not have access to a calculator and couldn't be bothered using excel.
moonbase
moonbase is offline
#3
Mar15-12, 11:46 PM
P: 21
The program doesn't tell me how far off I am, but it told me that 8.55 in indeed correct for the spring's energy. Though when I inquire about it, it says "Look at this in the center of mass frame and consider how there is 8.55 J of mechanical energy. What does this say about the velocity of the carts in the center of mass frame when the spring is fully compressed?"

I can't seem to find the problem, I tested it and know it's not a sig fig issue.

PeterO
PeterO is offline
#4
Mar15-12, 11:54 PM
HW Helper
P: 2,316

Spring Loaded Collision


Quote Quote by moonbase View Post
The program doesn't tell me how far off I am, but it told me that 8.55 in indeed correct for the spring's energy. Though when I inquire about it, it says "Look at this in the center of mass frame and consider how there is 8.55 J of mechanical energy. What does this say about the velocity of the carts in the center of mass frame when the spring is fully compressed?"

I can't seem to find the problem, I tested it and know it's not a sig fig issue.
If you ignore the reference to the centre of mass frame, and proceed in the usual way, you can calculate the energy before - only the 3 is moving - then use concervation of momentum to find the common velocity that occurs when the spring is at maximum compression. From that find the Kinetic energy at the time. You then know how much energy is stored in the spring.
I didn't get 8.55 [as I said without a calculator so I may have slipped].
If the answer is different, and leads to the correct compression there is an interesting puzzle to ponder.
moonbase
moonbase is offline
#5
Mar16-12, 08:28 AM
P: 21
Yea I tried that too, the kinetic energy being 11.76 J and the compression being 0.715 m, but that didn't work either. It also didn't accept it when I tried it with what you calculated (0.558 m) so I'm really not sure what's wrong here. If it helps, here's how I got 8.55:

vcm=[3(2.8)+8(0)]/(3+8)=0.7636 m/s
v1,cm=2.8-0.7636=2.036 m/s
v2,cm=0-0.7636=-0.7636 m/s
KEtotal=0.5(3)(2.0362)+0.5(8)(-0.76362)=8.55
moonbase
moonbase is offline
#6
Mar16-12, 01:05 PM
P: 21
Nevermind, I realized that the answer is zero. The compression in the center of mass frame is like momentum, the total is zero because the values of both sides cancel out. Thanks for your help though!


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