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Recoiling problem help

  1. Nov 14, 2003 #1
    recoiling problem help :(

    I've included a picture also.

    A circus cannon, which has a mass M = 4000 kg, is tilted at theta = 40°. When it shoots a projectile at v0 = 90 m/s with respect to the cannon, the cannon recoils along a horizontal track at vcannon = 1 m/s with respect to the ground.
    ----------------------------------------------------

    a) At what angle to the horizontal does the projectile move with respect to the ground?
    b) What is the mass of the projectile?
    c) The cannon is now lowered to shoot horizontally. It fires the same projectile at the same speed relative to the cannon. With what speed does the cannon now recoil with respect to the ground?

    ================
    a) At first I thought the angle with respect to the ground is just 40 degrees, but it isn't. I still don't know how to do this, but my buddy figured out that the answer is 40*sin (1)+40=, but even he doesn't know why it works. Can somebody show me how to do this correctly?

    b) Well, using the conservation of momentum, momentum of cannon=momentum of the projectile
    so
    4000kg(1m/s)=m(90m/s*cos(40.7)
    m= 58.6 kg

    c) I'm stuck. I figured that since the cannon is lowered to shoot horizontally, the angle would be 0, so I tried to equate the momentum of the cannon with the projectile to get:
    4000kg*V=(58.6kg)(90 m/s*cos (0))
    to get V=1.32m/s, but its not working out for some reason. Whats up with that?
     

    Attached Files:

  2. jcsd
  3. Nov 14, 2003 #2

    jamesrc

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    Gold Member

    a. For the first part, it should just be a relative velocity problem. I get a slightly different answer than you do, so I'm a bit hesitant to write this out... You can do one of two things, both of which will give you the same answer:
    1. You can find the angle with respect to the ground by finding the vertical and horizontal components (the vertical component will be unaffected by the change in reference frame). The horizontal component with respect to the ground should be
    vpg,x = vcg,x + vpc,x
    where the boldfaced quantities are vectors and the subscripts p,g,and c stand for projectile, ground, and cannon, respectively. The order of subscript 'ab' implies 'a' with respect to 'b', and these are all in the x-direction.
    2. You can forget about resolving into components and use the law of cosines, then the law of sines, to find the angle. (use the same formula without the x subscripts to see the relationship between the vectors).

    Either way, I get 40.41 degrees. If that's not the answer, there's clearly an error above, though I'm not sure how egregious it is.

    b. For the second part, I get 58.87 kg - same way you got your answer (conservation of momentum --> MV = mv, where V and v are the x-direction velocities of the cannon and cannon ball, respectively , with respect to the ground) with some potential error propagation.

    c. Here (again, if I'm understanding this problem in the first place) you have to remember that it says relative again. If the cannon ends up with speed V with respect to the ground, then the cannon ball's speed will be (90-V) with respect to the ground. Conservation of momentum:

    MV = m(90-V) --> V = 90m/(M+m)

    Using my value for m, that comes out to V = 1.305 m/s

    Hope this helps...
     
    Last edited: Nov 14, 2003
  4. Nov 15, 2003 #3
    Very cool. Everything is working out finally.
     
  5. Nov 17, 2003 #4
    Hi, its me again. I have the general idea of why part a works, but I need help writing it out mathematically.

    For the law of sines way:
    I first figured out the height of the triangle, where base is 1m/s, hyp is 90m/s, so the height is 89.24m/s

    so with law of sines:

    sin 40 sin y
    ------ = -----
    89.24 1


    and I get y=.41 degrees...so then I added 40 to.41 to get 40.41 degrees, but is that the right way of using the law?

    Then the other way of doing it would be breaking it into components like Jamesrc said.
    vpg,x = vcg,x + vpc,x

    so would vpg,x = 1+90=91 m/s?
    then vpg,y= 91 sin 40=58.49?

    then do tan^-1(91/58.49) but that doesn't give me the right answer. Can someone show me the correct way of setting up part a. Yea, I know its a bit of work, but I'd highly appreciate it.
     
  6. Nov 17, 2003 #5

    jamesrc

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    Gold Member

    What you're calling y is the angle at the top of the triangle formed by the three vectors. The third angle of that triangle (the one I went for first, but that doesn't matter) is then (180 - 40 - y). The angle of interest (the answer to this question) is the supplement to that angle, 180- (180-40-y) = 40 + y. I don't know if that's how you reasoned it out, but that is the correct answer.

    You made a slight error when breaking it up into components:
    vpg,x = -1 + 90*cos(40) (remember these are vectors; that's where the negative sign comes from on the 1). That should fix you up...
     
  7. Nov 17, 2003 #6

    Doc Al

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    Staff: Mentor

    Let's use components. The components of the ball's velocity (in m/s) with respect to the cannon:

    90 cos40 (x)
    90 sin40 (y)

    The velocity of the cannon is:

    -1 (x)

    Resultantx = 90cos40 - 1 = 67.94
    Resultanty = 90sin40 = 57.85

    tanθ = y/x = 0.851; thus θ = 40.41
    I get a magnitude of 89.2 m/s; at angle of 40.41 degrees to the horizontal.

    Your use of the law of sines is correct and gives the same answer.
     
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