Potential Energy & Conservative Forces #20

  • #1
Two blocks, each with a mass m = 0.348 kg, can slide without friction on a horizontal surface. Initially, block 1 is in motion with a speed v = 1.56 m/s; block 2 is at rest. When block 1 collides with block 2, a spring bumper on block 1 is compressed. Maximum compression of the spring occurs when the two blocks move with the same speed, v/2 = 0.780 m/s. If the maximum compression of the spring is 1.88 cm, what is its force constant?

I was figuring it out using hte conservatin of energy principle, but that isnt working:

1/2mu^2 + 1/2mu_2^2 + 1/2kx^2 = 1/2mv^2 + 1/2m_2v_2^2 + 1/2kx_2^2
 

Answers and Replies

  • #2
Doc Al
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Why isn't it working? What values are you using for the initial/final speeds of the masses and the compression of the spring?
 
  • #3
.5(0.348)(1.56^2) + .5(.348)(0^2) + .5k(0^2) = .5(.348)(1.56^2) + .5(.348)(v/2)^2 + .5k(1.88 x 10^-2)^2

those are all my numbers....is that right?
 
  • #4
Fermat
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Let both masses have the same velocity after impact.
 
Last edited:
  • #6
Fermat
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Did you use v/2 = 0.78 m/s instead of 1.56 on the rhs of the eqn ?
 
  • #7
yeah, i did, then you have to square it. I wrote my numbers up above that I was using in the equation.

YOu end up with -0.105862 = 706.88(k)
 
  • #8
Fermat
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.5(0.348)(1.56^2) + .5(.348)(0^2) + .5k(0^2) = .5(0.348)(1.56^2) = 0.4234464

What do you get?
 
  • #9
.5k(0^2) = 0

so how do I get anything when I can't even solve for k?
 
  • #10
Fermat
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The lhs of your eqn = 0.4234464

The rhs of your eqn = 0.217232 + 1.7672E-04 * k

Solve for k.

Check your arithmetic.
 
  • #11
i get

0.423 = 0.529308 + 1.7672 x 10^-4

Which gives a negative answer.


Using what you wrote above I get k = 1166.9 and that answer was not correct.
 
  • #12
Fermat
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My mistake: 0.217232 + 1.7672E-04 * k should be 0.2117232 + 1.7672E-04 * k

Which gives k = 1.2 KN/m

You had 0.529308 on the rhs. You will get that if you used v = 1.56 instead of v = 0.78.
Both masses have the same speed, 0.78 m/s.
 
  • #13
Doc Al
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UCrazyBeautifulU said:
.5(0.348)(1.56^2) + .5(.348)(0^2) + .5k(0^2) = .5(.348)(1.56^2) + .5(.348)(v/2)^2 + .5k(1.88 x 10^-2)^2

those are all my numbers....is that right?
No. As Fermat has already pointed out, both masses move with the same speed (0.780 m/s) when the spring is fully compressed.
 
  • #14
Thank you!
 

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