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Homework Help: A spring gun

  1. Dec 1, 2007 #1


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    1. The problem statement, all variables and given/known data
    The spring constant of a experimental spring gun is 350N/m. The spring is compresed a distance of 80cm from it's equillibrium position and a 1.5kg musket ball is placed in the barrel against the compressed spring. When the gun is fired the spring extends the length of the barrel, propelling the ball out of the barrel. The barrel makes a angle with 30 deg witht he horizontal, and the coefficient of kinetic friction between the ball and the barrel is 0.57. Assume the spring is massless.

    a) Draw a free body diagram for the musket ball after the ball has been fired and the ball is 10cm below the spring's equillibrium position.

    b) What is the acceleration of the ball at this position?

    c) Use the work-energy theorem to determine the speed of the musket ball as it leaves the end of the barrel.

    d) What maximum height does the ball reach after it leaves the barrel?

    e) draw a free-body diargram for the musket ball at it's maximum height
    http://img209.imageshack.us/img209/605/balljd0.th.jpg [Broken]

    2. Relevant equations

    K+ Ug + Us = K + Ug + Us

    3. The attempt at a solution

    a) free body diagram of ball at position 10cm below the equillibrium point after it has been fired.
    I really don't know but I assume that 10 cm below the equillibrium of the spring that the only force on the ball is the mass and the force of gravity on the ball. (see picture that I drew)

    b)What is acceleration of ball at this position?

    Would it just be the acceleration due to gravity? (9.81m/s^2)

    c) Use the Work-energy to determine the speed of the musket ball as it leaves the end of the barrel?

    Kf+ Ug + Us = K + Ug + Us- fkd

    I know I would have to use this but I also have to include friction but I'm not sure how...(it would have to be angled)

    [tex]1/2 mv^2 + mgysin \theta = 1/2mv^2 + mgysin theta + 1/2kx^2- \mu_kmg [/tex]

    I need help on the equation set up.:confused:

    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Dec 1, 2007 #2
    use [tex]W_{net}=\Delta KE[/tex] make sure to include work done by everything.. including the spring
  4. Dec 1, 2007 #3


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    Oh..well I fixed that.

    [tex]1/2 mv^2 + mgysin \theta = 1/2mv^2 + mgysin theta + 1/2kx^2- \mu_kmg [/tex]

    but I still am not sure about the angles and where they are needed...I think I would need that for the distance the spring is compressed right? so the distance would be xcos theta for the distance that the spring stretches?

    Would I need the angle anywhere else?

    And one more question: how do I know which side do I put the friction sign?

    according to your equation my set up would be incorrect right?
  5. Dec 1, 2007 #4
    conservation of energy equation does not apply here because friction is not a conservative force. The problem states that you use work-energy which is my equation. work is a scalar quantity and the equation will provide you with the correct sign.

    remember that W = F * D * Cos(Theta) where theta is the angle between the force and distance vectors. Both F and D are scalar and should be positive. Work done by a spring is [tex]W_s=\frac{1}{2} k(x_i^2-x_f^2)[/tex]
    Last edited: Dec 1, 2007
  6. Dec 1, 2007 #5


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    Actually my book has a equation like this which would include the energy in there...

    [tex] (K + U_g + U_s)_f = (K + U_g + U_s)_i - f_kd + \sum W_{other forces} [/tex]

    should I use this for the equation???

    and where would I have the angles???

    can anyone help me?
    Last edited: Dec 2, 2007
  7. Dec 2, 2007 #6
    you draw the x-axis along the barrel so you only need angles to calculate the gravitational potential energy (to determine the height difference)... using work done instead of cons of energy for gravity makes it less confusing, but using that equation is fine too.
  8. Dec 2, 2007 #7


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    When the ball is fired the gun or barrel must recoil.
    When the gun is fired the spring extends the length of the barrel, propelling the ball out of the barrel. What is the meaning of this? To calculate the part c, I think, this information is needed.
  9. Dec 2, 2007 #8


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    so it isn't the same as what you were saying...

    however for the distance final and initial if I use the barrel direction as x

    would the height difference be using sin 30 deg - 0? (taking compression point as 0 height?)

    let me guess...it is conservation of momentum where there is momentum in x and y direction. But how would this be relevent to finding the final velocity of the ball when fired out of barrel??

    I don't see how that would fit into the equation..

    [tex] (K + U_g + U_s)_f = (K + U_g + U_s)_i - f_kd + \sum W_{other forces} [/tex]
  10. Dec 2, 2007 #9
    yes the height difference would be length*sin30
  11. Dec 3, 2007 #10


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    I was very busy but now I'm back.

    This is what I came up with from the work-energy equation.

    [tex] (K + U_g + U_s)_f = (K + U_g + U_s)_i - f_kd + \sum W_{other forces} [/tex]

    [tex] 0.5mv_f^2 + mg*d*sin30 + 0 = 0 + 0 + 0.5kx sin30 - \mu_k mgd sin30 + 0 [/tex]

    [tex] vf= 9.00m/s [/tex]

    Is this reasonable for the velocity??

    I think it's alright but...not sure if my equation is correct.

    d) max height reached

    vi= 9.0m/s

    vyi= 9.0m/s sin 30 = 4.5m/s

    vxi= 9.0m/s cos 30= 7.79m/s

    [tex]vy_{max}= 0 [/tex]

    t= ?

    [tex]vf_y= vi_y + a_yt [/tex]

    0= 4.5m/s +(-9.81m/s^2)t

    t= .4587s

    y= vy(t) + 0.5 ay t^2

    y= 4.5(.4587) + 0.5(9.81m/s^2)(.4587s)^2

    y= 3.096m

    e) free body of ball at max height
    (mg pointing down below ball at max height)

    does this look reasonable? Not sure about the velocity that I found though.

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