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Conservation of energy of a dart gun

  1. Oct 16, 2004 #1
    I have a two-part problem to solve. I think I solved it but in the class the professor told us that there's something wrong with the exercise.

    The spring constant of a toy dart gun is 1350 N/m. To cock the gun the spring is compressed 1 cm (0.01 m). The 5g (0.005 kg) dart, fired straight upward, reaches a maximum height of 24 m. g is 9.81 m/s^2. Determine the energy dissipated by air friction during the dart's ascent.

    My solution:

    E = -1/2 *(k) *(x)^2 + m*g*h
    =-0.5*1350* (0.01)^2 + 0.005*9.8*24
    = -0.0675 + 1.176 = 1.1085 J
    Part 2.

    What speed should the projectile have when it returns to its starting point?

    Solution: m*g*h = 0.5*m*v^2
    v = sqrt(2*g*h)
    =sqrt(470.4) = 21.69 m/s

    These make sense to me, but i don't know why the professor told me there's something wrong in the problems.
  2. jcsd
  3. Oct 16, 2004 #2


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    2) is inconsistent with 1) since you suddenly ignore air resistance..
    might this be what your prof. disliked?
    I would agree with him!
  4. Oct 16, 2004 #3
    For part A, the spring stores a certain amount of potential energy. When the gun is fired, the spring transfer all its potential energy to the bullet in the form of kinetic energy. As the bullet travels upwards this kinetic energy converts into gravitational potential energy. Do you see where the problem is?
  5. Oct 16, 2004 #4


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    Eeh, it seems that air friction has ADDED energy to the system, when I look a bit closer.
    this exercise is a complete mess.
  6. Oct 16, 2004 #5
    Although i think your reasoning for part B would be the correct one. I think you are supposed to ignore air friction on the way down because it says what speed *should* the bullet have.
  7. Oct 16, 2004 #6
    :smile: In both cases the air resistance is neglected. He said that the exercise doesn't make sense because the numbers come negative. This doesn't happen when I solve the problems. Maybe there's sth wrong in my reasoning? (The professor had different numbers, so maybe there was a typo in his own problem)
    I just want to if the way I have solved it makes sense.
  8. Oct 16, 2004 #7
    The air resistance would be negative because:

    V(spring) = V(gravitational) + friction

    Hence, friction would be V(spring) - V(gravitational). Since V(gravitational) > V(spring), friction is negative (ie, friction causes a net gain in energy). Which is the same thing arildno pointed out and I was implying.
  9. Oct 16, 2004 #8
    Hey Parth Dave!
    In fact I need to find the energy dissipated by air friction. So what I think is:

    Energy = Change in PE + Change in KE
    = (m*g*h final - m*g*h initial) + (0.5*m*Vfinal ^2 - 0.5*m*Vinitial^2)
    initial mgh = 0 (h = 0) and final 0.5*m*Vfinal = 0 (v=0)
    So we now have only: Energy = m*g*h final - 0.5*m*Vinitial^2
    0.5*m*Vinitial^2 = energy of spring = 0.5*k*x^2
    Finally: Energy = m*g*h final - 0.5*k*x^2

    Then for part 2, it seems suitable applying: mgh = 0.5mv^2
  10. Oct 17, 2004 #9
    :smile: The problems are not as difficult as you think. There's no air friction , but we need to find the energy lost by air friction during ascending. Also no wind or any other factor influencing the trajectory of the dart.
  11. Oct 17, 2004 #10
    Well by the conservation of energy:

    Ti + Vi = Tf + Vf + Friction (T = kinetic energy, V = potential) hence,
    Vi = 0
    Tf = 0
    Ti = Vf + friction
    friction = Ti - Vf

    Or from What you did:
    Ti + Vi = Tf + Vf + Friction
    0 = Change in T + Change in V + friction
    -friction = Change in T + Change in V

    Both methods are the same, but you just forgot the negative sign.
  12. Oct 17, 2004 #11
    Thanks Parth Dave,
    In fact the answer was positive.
    But the second Part is weird. As I told, I applied: mgh = 0.5mv^2 and hence I found v. BUt the answer was wrong. What might be the problem?
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