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Momentum Problem

  • Thread starter mikep
  • Start date
  • #1
43
0
can someone please help me with this problem?
A 10kg cart with low friction wheels has a 4000N/m
spring on one end. The cart is pressed against the
wall so that the spring is compressed 0.50m. The cart
is released and rolls on the surface only to hit
another identical cart. The two carts lock together
and run into a wall on the other side of the room
where a 4000N/m spring on the second cart compresses
until the carts stop. How far does the second spring
compress as the carts stop?
I think i need to break it up into three parts. For
the first one i need to find the force of the s[ring
then the collision of hte two carts and then the
compression of the spring. So for the first part i
did:
PEs = (1/2) k(x^2) = (1/2)(4000N/m)(0.50^2) = 1000J
is that right for the first part?
 
Last edited:

Answers and Replies

  • #2
cepheid
Staff Emeritus
Science Advisor
Gold Member
5,192
36
Check your math. I think you forgot to divide by 2. And the unit for energy is the joule. I think your strategy is essentially correct. Now that you know the initial potential energy of the cart, you know its final kinetic energy as it is released (from conservation of energy). Therefore, you know its velocity when it hit the second cart. From conservation of momentum, you can calculate the velocity of the two combined carts after the collision. That will give you their initial kinetic energy when they hit the spring. Since this kinetic energy is converted to elastic potential energy, you can calculate the displacement of the spring.
 
  • #3
43
0
oh you're right its 500J. ok so i use that and i did 500 = (1/2)m(v^2) v = 10m/s then i did the conservation of momentum. (10)(10) + (10)(0) = (10 + 10) V V = 5m/s (1/2)m(V^2) = (1/2)k(x^2) x = 0.25 as the displacement. does this look right?
 

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