Momentum of spring between two objects

  • Thread starter moonbase
  • Start date
  • #1
moonbase
21
0

Homework Statement


A massless spring of spring constant 20 N/m is placed between two carts on a frictionless surface. Cart 1 has a mass M1 = 5 kg and Cart 2 has a mass M2 = 2.5 kg. The carts are pushed toward one another until the spring is compressed a distance 1.2 m. The carts are then released and the spring pushes them apart. After the carts are free of the spring, what are their speeds?

Homework Equations


PEspring=0.5kx2
KE=0.5mv2=p2/2m
p=mv

The Attempt at a Solution


I plugged in the given values to find that the potential energy of the spring is 14.4 J. Since there is no friction, the energy is conserved and the resulting kinetic energy should be the same value, and plugging it into the equation in terms of momentum results in p=14.69 since momentum is also conserved. So when I plug that into p=mv, I get v1=2.94 and v2=5.88 but apparently these are incorrect. Am I skipping a step?
 

Answers and Replies

  • #2
issacnewton
983
22
Your calculations are not correct. check it
 
  • #3
moonbase
21
0
Well I'm trying to figure out where exactly I went wrong. What I did so far:

0.5*20*1.22=14.4 J

p2/2(7.5)=14.4 -> p2=216 -> p=14.69

v=p/m -> 14.69/5=2.94 m/s and 14.69/2.5=5.88 m/s

Any ideas?
 
  • #4
issacnewton
983
22
moon,

[tex]K=\frac{p^2}{2m_1}+\frac{p^2}{2m_2}[/tex]
 
  • #5
moonbase
21
0
Oh okay, I was combining them but I see now. Thank you!
 

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