Spring loaded collisionneed some guidance

  • Thread starter Juntao
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  • #1
Juntao
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A cart of mass m = 6 kg carrying a spring of spring constant k = 48 N/m and moving at speed v = 1.5 m/s hits a stationary cart of mass M = 12 kg. Assume all motion is along a line.

a) What is the speed of the center of mass of this system?
b) When the spring is at its maximum compression, with what speed are the carts moving in the lab frame?
c) How far will the spring be compressed?
d) After the collision is over, what is the velocity of m in the center-of-mass frame? (right is positive)
e) After the collision is over, what is the velocity of m in the lab frame?
f) After the collision is over, what is the velocity of M in the lab frame?
======================================================
Ok, I figured out a, and b, but now I'm stuck on c.

This is what I've done so far.

a) Initial momentum=final momentum
(m1v1+m2v2)=(m1+m2)Vcom

where m1=6kg
v1=1.5m/s
m2=12kg
v2=0
Vcom=velocity of center of mass

I figured out the answer of Vcom=.5m/s

b) I know that the answer is .5 m/s, but I'm sure why it is though. Can someone elaborate on this please?

c) This is where I get stuck so far. I used the idea of conservation of momentum and energy. In the end, I get this equation
KE=SPE or
.5(m1+m2)(Vcom)^2=.5kx^2
So I substituted my numbers, and the x I get isn't right. Is this the right way of apporaching it?
 

Answers and Replies

  • #2
Juntao
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Here is a pic of the thing.
 

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  • #3
NateTG
Science Advisor
Homework Helper
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Part B:

When the spring is fully compressed, then the carts are (momentarily) at rest with respect to each other right? What does that mean about their motion with respect to the center of gravity?

Part C:
You can calculate the energy before the colision, and when the spring is fully compressed. That shoud give you the answer.

Part D:
One way to do this would be to solve the colision as perfectly elastic, and then do the rest of the work. Perhaps you can come up with a more elegant solution.

Part E:
See part D

Part F:
See part D
 
  • #4
Doc Al
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Originally posted by Juntao
b) I know that the answer is .5 m/s, but I'm sure why it is though. Can someone elaborate on this please?
As long as the spring is still compressing, there is relative speed between the two carts. When maximum compression is reached, the two carts---momentarily---move as one: relative speed is zero. At that point, Vcart 1 = Vcart 2 = Vcm.
c) This is where I get stuck so far. I used the idea of conservation of momentum and energy. In the end, I get this equation
KE=SPE or
.5(m1+m2)(Vcom)^2=.5kx^2
So I substituted my numbers, and the x I get isn't right. Is this the right way of apporaching it?
Consider this: initial KE = KE(at max compression) + PE(of spring at max compression)
 

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