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Ideal Spring Problem

  1. Dec 7, 2006 #1
    I worked this problem out, but my numbers make me think that my solution is wrong. Can you check this for me?

    A 5.00 g bullet moving with an initial speed of v0 = 420 m/s is fired into and passes through a 1.00 kg block. The block, initially at rest on a frictionless horizontal surface, is connected to a spring with a spring constant of 950 N/m.

    (a) If the block moves 5.00 cm to the right after impact, find the speed at which the bullet emerges from the block.
    (b) If the block moves 5.00 cm to the right after impact, find the mechanical energy lost in the collision.

    [​IMG]

    My attempt at a solution

    a) P(o) = P(f)

    m(1)v(1) + m(2)v(2) = (m(1) + m(2))*v(f)

    ^This is to find the velocity of the bullet and the block while the bullet is in the block - I'm thinking that it neglects the fact that the block has to accelerate, and that the spring is pushing back on the block. Anyway...

    v(f) = 2.09 m/s

    Then, to find the speed of the bullet when it exits the block:

    m(1)v(1) + m(2)v(2) = m(1)v(1) + m(2)v(2)

    Letting m(1) be the bullet, and m(2) be the block, v(2) = 2 m/s? Seems too slow.

    b) E(o) = E(f)

    m(1)v(1)^2 + m(2)v(2)^2 + kx(o)^2 = m(1)v(1)^2 + m(2)v(2)^2 + kx(f)^2

    ^The reason you don't see any (1/2)'s is because they all had (1/2)'s, so I multiplied the whole thing by 2.

    882 = 6.76

    Difference in mechanical energy = 882 - 6.76 = 875.2 J lost?

    Where'd all the energy go, anyway? Was it absorbed by the spring?
     
  2. jcsd
  3. Dec 8, 2006 #2

    SGT

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    You should use energy, instead of momentum, to take in consideration the spring.
    The bullet has an initial kinetic energy, that is transferred to the block and to the spring. I suppose that when the block attains its maximum displacement of 5 cm its velocity is 0. So all the energy is to be shared between the bullet and the spring.
     
  4. Dec 8, 2006 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Since there is an external force involved (the spring), momentum is NOT conserved. The work done in compressing a spring, with constant k, a distance x, is (1/2)kx2. Find the work done in compressing the spring (which is the answer to (b)) and subtract that from the kinetic energy of the bullet.
     
  5. Dec 8, 2006 #4
    When I did it this way, I got:

    a) 419 m/s
    b) 1.19 J

    And my friend (who got the problem right with different numbers) told me those values weren't in the ballpark. I should have mentioned that the problem came with this hint:

    "This problem will stretch your insight... and hooke your interest. If you don't force it and conserve your momentum... you should find the solution."

    My friend told me that he got part (a) using this equation:

    (1/2)mv(f)^2 = (1/2)kx^2

    ^Where I suppose the final potential energy of the bullet somehow equals the potential elastic energy of the spring?

    and v(f) = 21.8 m/s. How is that right?

    Then for part (b), he said to use the momentum equation, but I had to go and he didn't have time to set it up for me.

    How would momentum be conserved?
     
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