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Inelastic Collisions and Dissipation of Energy

  1. Dec 10, 2007 #1
    1. The problem statement, all variables and given/known data

    Block A of mass, mA = 2.0 kg, is moving on a frictionless surface with a velocity of 5.0 m/s to the right and another block B of mass, mB = 8.0 kg, is moving with a velocity of 3.0 m/s to the left, as shown in the diagram below. The two block eventually collide.

    d. If the initial velocity of Block A is 4.0 m/s to the right and -30 J is dissipated in the collision, what was the initial velocity of Block B?

    3. The attempt at a solution

    mavai + mbvbi = (ma+mb) vf
    (0.5)mavai² + (0.5)mbvbi² = Ei
    (0.5)(ma+mb)vaf² = Ef
    Ef – Ei = -30 J

    (4.0) + (4.0)vbi = (5.0) vf
    (16) + (4.0)vbi² = Ei
    (5.0)vf² = Ef
    Ef – Ei = -30 J

    I don't know how to go from here. Any help?
  2. jcsd
  3. Dec 10, 2007 #2
    I don't know where to go from here, either. The question as you have given it gives two velocities for blocks A and B -- and you've made your query harder to understand by missing out the formulae you are using and poor formatting of your attempt.
  4. Dec 10, 2007 #3
    About the poor formating, yes, I noticed it's pretty bad. I pasted directly from Word because I'm still not familiar with the formating thing we have for this forum.
  5. Dec 10, 2007 #4

    Momentum is conserved, as you have shown. This can be used to show how the before and after kinetic energies are related-- see


    I think this may help a bit. If you take a look at the formulas you have written already:

    [tex] E_f - E_i = -30 [/tex]

    [tex] m_av_{ai} + m_bv_{bi} = (m_a + m_b)v_f [/tex]

    [tex] \frac{1}{2}m_a{v_{ai}}^2 + \frac{1}{2}m_b{v_{bi}}^2 = E_i [/tex]

    [tex] \frac{1}{2}(m_a + m_b){v_f}^2 = E_f [/tex]

    You have four equations and four unknowns ([itex] E_i , E_f, v_{bi}, v_f[/itex]) so you should be able to solve this. I would substitute my expressions for [itex] E_i and E_f [/itex] into the first equation, and then use the second equation to substitute in for either [itex] v_f or v_{bi} [/itex].

    Finally, check your interpretation of the first equation. You stated that "-30J are dissipated"-- does this mean that 30J are actually gained in the collision? Or were they meaning that 30J were dissipated (lost)? If the former, your equation has a sign error. If the latter (which is most likely), you should be ok.

    Solve it with the letters, and put the numbers in last.
  6. Dec 10, 2007 #5

    No amount of conservation of momentum and energy equations is going to solve the riddle of which velocity to use! Has something significant been lost along with parts a, b and c of the question?
  7. Dec 10, 2007 #6
    Alright, I solved it.
    I ended up isolating just the Vf² and the Vf into one equation so I just solved it with the quadratic formula. So I got two final velocities, one positive and one negative. Since the initial velocity of block B has to be negative, Vf had to be negative too.

    By dissipated I mean the energy is lost, that's why I put negative.
    So the initial velocity for Block B is -1.4 m/s.
    I guess it makes sense, could you guys confirm?

    Thank you.
  8. Dec 10, 2007 #7

    That's not what I got, but I just did the problem real quick on a napkin, and may have made a mistake. What value did you get for Vf? You can at least check if your answer "makes sense".

    More importantly, why does the initial velocity of block B have to be negative? And even if it is, this does not imply that Vf is negative, necessarily-- what if Vb is very very slightly negative? (What if its zero?)

    Lastly, there is a simpler way to do this problem. Did you check out that link?

    EDIT: Actually, re-looking at that link, it looks like its only valid for the case of Block B being motionless. So that's no good for here, necessarily.
    Last edited: Dec 10, 2007
  9. Dec 11, 2007 #8
    Yeah, I checked that link and saw that one of the masses was motionless, so the situation is a bit different from this one.
    I redid the problem and got two answers: 0.9 m/s or -13 m/s for the initial velocity of block B and two final velocities for both objects as 1.5 m/s or -9.5 It makes sense I guess, because block A is traveling about twice as faster and hits the bigger block B making Block B travel a bit faster while block A has energy dissipated due to the collision.
    Is that what you got?
    Last edited: Dec 11, 2007
  10. Dec 11, 2007 #9
    Not quite what I got. From a physical standpoint, you're right, the answers kind of make sense. In the first case, the big block is moving slowly to the right, and the small block comes up and hits it, and the two now move slightly faster than Block B, as expected. In the second case, the massive block flies in fast, to the left, strikes block A, reverses it, and they continue left more slowly. That's all well and good.

    You can check, however, to see if your answers are correct or not. They have to satisfy your initial equations-- conservation of momentum, and your energy equation. Let's have a look:

    First, vb = 0.9 m/s and vf = 1.5 m/s:

    [tex] m_av_{ai} + m_bv_{bi} = (m_a + m_b)v_f [/tex]

    [tex] \ \Rightarrow (2){(4)} + (8){(0.9)} = (2 + 8){(1.5)} [/tex]

    [tex] \ \Rightarrow 8 + 7.2 = 15 [/tex]

    [tex] \ \Rightarrow 15.2 = 15 [/tex]

    So that's good, the slight difference is due to rounding. Now for the energy equation:

    [tex] \frac{1}{2}m_a{v_{ai}}^2 + \frac{1}{2}m_b{v_{bi}}^2 = E_i [/tex]

    [tex] \frac{1}{2}(m_a + m_b){v_f}^2 = E_f [/tex]

    [tex] E_i - E_f = 30 [/tex]

    [tex] \ \Rightarrow \frac{1}{2}m_a{v_{ai}}^2 + \frac{1}{2}m_b{v_{bi}}^2 - [ \frac{1}{2}(m_a + m_b){v_f}^2] = 30 [/tex]

    [tex] \ \Rightarrow \frac{1}{2}(2){(4)}^2 + \frac{1}{2}(8){(0.9)}^2 - \frac{1}{2}(2 + 8){(1.5)}^2 = 30[/tex]

    [tex] \ \Rightarrow 16 + 3.24 - 11.25 = 30[/tex]

    [tex] \ \Rightarrow 19.24 - 11.25 = 30[/tex]

    [tex] \ \Rightarrow 7.99 \not= 30[/tex] So there's a problem here.

    You can do the same kind of thing for your other set of answers, to see what fits and what doesn't.
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