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Homework Help: Need Help Not the Answer Please

  1. Oct 28, 2004 #1
    Need Help Not the Answer Please!!!!

    A toy cannon uses a spring to project a 5.30-g rubber ball. The spring is originally compressed by 5.00 cm and has a force constant of 8.00 N/m. When the cannon is fired, the ball moves 15.0 cm through the horizontal barrel, and there is a constant force of 0.0320 N between the barrel and the ball.

    (a) With what speed does the projectile leave the barrel of the cannon?

    (b) At what point does the ball have maximum speed?

    (c) What is the maximum speed?

    Right now my class is dealing with potential energy and this is the first class for Physics for Engineers. I already solved letter (a) but I need help with (b) and (c). I was going to use the derivative to solve for (b) since the keyword is maximum, but I don't know if that's correct. Any advice on approaching this problem would be great. Don't worry it's not an exam question just a problem assigned from the book:)
  2. jcsd
  3. Oct 29, 2004 #2
    max speed, energy all kinetic. That should be just after the spring does all the work it possible can to accelerate it. So just when it leaves the barrel. After that point, it goes up and trades speed for potential energy in height.
  4. Oct 29, 2004 #3
    I don't need the height of the projectile after it leaves the cannon, just at what point inside the barrel of the cannon is the velocity the maximum. I know it's some point after the initial release from the spring because the longer the ball travels through the barrel the slower it is because the frictional force is constant.
  5. Oct 29, 2004 #4
    yeah, its when the spring stops supplying a force. That happens when the spring has relaxed its full amount. I think in your problem, that is when it just reaches the end of the barrel.
  6. Oct 29, 2004 #5
    In part (a) the speed of the projectile as it leaves the barrel is 1.40 m/s, and I found this out from taking the final energy=1/2 mv^2(final) subtracted from the initial energy=1/2kx^2(initial) and equating the change of energy to the negative of the frictional force*distance the projectile travelled.

    The answer is correct in the back of the book, yet I don't know how they got the 4.60 cm in part (b). Is there any way I can manipulate the energy equation to find the point at whic h the velocity of projectile inside the cannon is at its maximum?
  7. Oct 29, 2004 #6


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    You need to account for the amount work required to push the ball out of the barrel (you are given the force the barrel exerts on the ball) . Write an expression, which includes the energy of the spring and the energy lost to the barrel, solve for the velocity then maximize (I assume you have had some calculus!)
  8. Oct 30, 2004 #7
    [tex] 1/2 kX^2 = (.0320)x + 1/2 mv^2 + 1/2k(x^2) [/tex]

    X is the 5cm distance.
    x is the distance at any point from rest to the length of the barrel.
    k is the spring constant.
    m is the mass of the ball in kg.

    You have two variables.
    x, distance, and v velocity. Solve in terms of v.

    [tex] 1/2 kX^2 - .0320x - 1/2kx^2 = 1/2 mv^2 [/tex]
    [tex] 2/m(1/2 kX^2 - .0320x - 1/2kx^2)= v^2 [/tex]
    [tex] (kX^2 - .064x - kx^2)/m= v^2 [/tex]
    [tex] \sqrt([(kX^2 - .064x - kx^2)/m])= v [/tex]
    thus V as a function of x.

    take the derivative of this to find the critical points,

    [tex] ((kX^2 - .064x - kx^2)/m)^-^1^/^2 *[(.064+2kx)/-m] = dv [/tex]

    set dv equal to zero and solve for x. Id use a calculator for doing that though.

    that is the value for x to plug back into the equation. You know the distance, you can plug it in to back out the speed.
    Last edited: Oct 30, 2004
  9. Oct 30, 2004 #8

    In this case am I taking the derivative of the velocity with respect to the compression of the spring (x)?
  10. Oct 30, 2004 #9


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    x should be your distance from some reference point in the barrel. That is the independent variable you need to use in the max/min process.
  11. Oct 30, 2004 #10
    what anwser did the book give for part b? is it somewhere near 4cm?
    Last edited: Oct 30, 2004
  12. Oct 31, 2004 #11

    The book got 4.60 cm after initial release is that what you have?
  13. Oct 31, 2004 #12
    hey moszzad i think so let me double check right quick.
  14. Oct 31, 2004 #13

    Hey, when I solve it using your expression I come up with -0.400 cm. Is that what you get as well?
    If you take dV=0 then the final expression I come up with is 2kx=-.064, and when I solve for x, I get -.004m
    Last edited: Oct 31, 2004
  15. Oct 31, 2004 #14
    theres a mistake in it im trying to work out sorry for that moszzad i got your anwser too.
  16. Oct 31, 2004 #15
    That's ok I appreciate any help I can get. There were just two problems in this section that are giving me heart-ache and this happens to be one of them.
  17. Oct 31, 2004 #16
    i just cant find the problem write now, i changed the equation to read .064(x+.05) instead of just x. and evaluated it from x=-.05 to x=0 but its not working out as your book got. Im too tired to find the error right now sorry. maybe someone can check over my work to find a mistake.
  18. Nov 1, 2004 #17
    Here's my take on the problem...

    "The spring is originally compressed by 5.00 cm and has a force constant of 8.00 N/m."

    Therefore, the PE stored in the spring is 0.5*k*d^2 = 0.5*8*0.05^2 = 0.01 J

    "When the cannon is fired, the ball moves 15.0 cm through the horizontal barrel, and there is a constant force of 0.0320 N between the barrel and the ball."

    The total energy lost to friction is d*F = 0.15*0.032 = 0.0048 J

    The KE of the ball when it leaves the barrel is the initial PE that was in the spring, minus the total frictional loss, which is 0.01 - 0.0048 = 0.0052 J
    KE = 0.5*m*v^2, 0.0052 = 0.5*0.0053*v^2, v = 1.40 m/s
    The ball has the highest velocity just as the spring completes it 5 cm travel. At that point, the loss due to friction is 0.05*0.032 = 0.0016 J, so the ball's KE is 0.01 - 0.0016 = 0.0084 J, and its velocity is 1.78 m/s.
  19. Nov 2, 2004 #18

    except the book says that the point of maximum speed is at 4.6cm not 5cm

    but it should be @ 5cm right? i even worked it out solving
    .5kx^2 -.5mv^2 = -fk*d for v and then differentiating, and the only v' = 0 is at x=0 which is 5 cm from the beginning(i.e. relaxed state).

    so why is fastest point @ 4.6cm?

    p.s. mozzsad ur HW is due @ 11:50 tonight, right? lol, i cant get this one either.
    Last edited: Nov 2, 2004
  20. Nov 2, 2004 #19
    Post the solution to the problem your instructor gave please moszzad.
  21. Nov 2, 2004 #20
    heres the anwser. Draw a free body diagram. We define it such that at x=0, F=0. And at x=.05 F=value. Then we can easily see that the spring will not provide any acceleration after the force due to the spring is equal to the force due to friction. So setting the two equal we get,

    F=ma=Kx=friction, or kx=Ff or 8x=.032

    Solving for x, you will get .004.

    So the displacemenet of the spring is .05-.004 = .046, or 4.6cm.

    OK I got the anwser twice. You can also use the equations i provided. BUt make one correction, it should read .064(x+.05) instead of .064x. That fixes the problem. You will find that this has a range of values from -.05, fully compressed, to a value of 0, fully relaxed. The calculator should tell you again, x=-.04, which means that it moved a total of 4.6cm, or .046m, in agreement with the way I solved it the at the top of this post using a different method.
    Last edited: Nov 2, 2004
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