# Recoiling Canon

1. Nov 4, 2004

### Mivz18

Recoiling Cannon

Problem:

A circus cannon, which has a mass M = 4500 kg, is tilted at q = 50°. When it shoots a projectile at v0 = 40 m/s with respect to the cannon, the cannon recoils along a horizontal track at vcannon = 1.5 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?

I am absolutely confused as to where to begin or how to go about this problem. Suggestions anyone???

Last edited: Nov 5, 2004
2. Nov 5, 2004

### Mivz18

So far I figure out that the Velocity of the projectile with respect to the ground is 38.5 m/s . After that I am not sure where to go. I think the equation :

(momentum of projectile)cosx - (momentum of cannon) = 0

I assume I use the velocity of the projectile with respect to the ground, but how do I find the angle to the horizontal with respect to the ground??

3. Nov 5, 2004

### Duarh

Well clearly the angle the projectile makes with the horizontal is not going to be constant throughout (the projectile's gonna fall down one day, after all). Think about it this way - the x component of the projectile's velocity is unchanging (assuming no air resistance), while the y component is changing at a rate of -g. You can use the initial angle to calculate the initial x and y components of the projectile's velocity, then you can write a time expression for the y component (you know its initial value and the acceleration -g, after all). Then the initial x component of the velocity over the y component at any point in time will give you the cotan of the angle the projectile makes with the horizontal.

You are correct in noting that the x component of the projectile's momentum will be equal to the x-direction momentum of the cannon after the projectile is shot out, so you can just plug the mass of the cannon and its recoil velocity in there and determine the mass of the projectile in that way.

If you can get through what I've outlined above, you should then be able to do part c on your own.

Edit: I believe that the note '40 m/s with respect to the cannon' was meant to indicate that the projectile was shot at an angle, _not_ that the initial velocity was supposed to be 38.5 m/s instead of 40 m/s. The initial velocity with respect to the ground (x component of velocity) should be calculated by trigonometry.

Last edited: Nov 5, 2004
4. Nov 5, 2004

### Mivz18

Can't you calculate the initial velocity in reference to the ground by adding the horizontal velocities as vectors?? Where the cannon projectile is shot at 40 m/s in reference to the cannon and the cannon moves back in reference to the ground at 1.5 m/s, then you just do 40 + (-1.5) = 38.5 ???

5. Nov 5, 2004

### Duarh

That would be assuming the velocity was given with respect to the cannon _after it had recoiled_ (accelerated to -1.5 m/s). There's no particular reason to think that's the case. I expect the reason they said 'with respect to the cannon' is that you'd understand 40 m/s was the speed in the "diagonal", not the horizontal, direction.

So the initial velocity with respect to the ground would be calculated by trig (using the angle of 50 degrees) from 40 m/s.

6. Nov 5, 2004

### Mivz18

Well, I've made it to part C) . I can't seem to find the correct answer according to the online program I'm using. I obtained 51.67 degrees for the angle to the horizontal with respect to the ground by taking the cotan of Vx/Vy. Then for part B) got 278.81 kg from the equation : 0 = MpVp - McVc .

When doing part C) I use the same equation. Since the angle in the x direction is 0, the cos(0) = 1 to where the speed relative to the ground is the same as relative to the cannon. so, I plug in my numbers: (278.81 kg)(40 m/s) = (4500 kg)(Vc)
I obtain 2.478 m/s for Vc, however, it is marked wrong. Am I doing this part correctly??

7. Nov 5, 2004

### Duarh

First of all, the angle with respect to the horizontal is not going to be constant throughout. Initially, of course, it is 50 degrees, since that's what the problem gave you - it is not particularly useful to take the cotan of vx/vy for the initial velocity components, because that's only going to give you those 50 degrees back, only with rounding errors (51.67 is not new information, it's only the same old 50 degrees distorted through manipulations). The actual expression for the angle should involve time, since the angle is changing.

In part B, though, you only need the initial angle provided by the problem to calculate the initial x component of the projectile's velocity and then plug it into the momentum equation (mv(xcomp)+MV=0, where v, V- opposite vectors in the x direction).

If you used 51.67 for the angle instead of 50, that might've caused the deviation.

8. Nov 6, 2004

### Mivz18

But how do I solve for part C? The online program OK'd the part A and B answers that it was looking for. I'm just having trouble with part C. So far, since the net momentum is zero, and the momentum of the projectile is equal and opposite of the momentum of the cannon, I get MpVp = -McVc . Then with the numbers I obtain, I get :

(278.81)(40-Vc) = -(4500)(Vc)

Then, when I solve for Vc, I get -2.642 m/s . However, the program says no, so what am I doing wrong??

9. Nov 6, 2004

### Duarh

Why are you multiplying your mass by (40-Vc) instead of just 40? Momentum is conserved with respect to the frame of reference in which both of the objects were unmoving previously, so the Vc term shouldn't be there (the projectile moves with 40 m/s in one direction, the cannon with 1.5 m/s in the other). Also, I'd be interested in seeing how exactly you got 278.81 kg in the first part of the problem; my calculations indicate the actual number should be more than 15 kg less. I believe the source of confusion is that you're using 40-Vc=38.5 m/s, instead of 40 m/s, for v0, and perhaps also 51.67 for the angle instead of 50. The program might have accepted 278.81 as a close-enough answer or perhaps you gave the inaccurate variables to it as starting parameters, I don't know since I don't know what program you're using.

>It fires the same projectile at the same speed relative to the cannon.
I believe this is again the source of confusion. It is _possible_ that, with this line, whoever was designing the problem meant that 40m/s was the speed the projectile had with respect to the _recoiling_ cannon; I highly doubt that, however, since that would make the problem ambiguous (in part a, for instance, the cannon is moving in the x plane, while the projectile is rising at 50 degrees). I believe 40 m/s is the velocity with respect to the cannon's position before it started recoiling (ie, with respect to the rest frame). Therefore, all your momentum equations should probably look like this: MpVp=McVc, instead of Mp(Vp-Vc)=McVc

Try recalculating the first part with this method (vo=40 m/s, angle=50 deg), then plug the answers into part c and see if it works out.

10. Nov 6, 2004

### Mivz18

Here is the link to the program I'm using. The first numbers I am confident are correct. I went over them with my professor earlier today. https://tycho-s.physics.wisc.edu/cgi/courses/shell/phys170/fall04/tma.pl?07/recoiling_cannon [Broken]

BTW, I calculated part B by MpVp = McVc .

Using trig in part A, I found that the initial velocity in the x direction is 24.21 m/s. Despite it is the horizontal direction, I didn't have to use an angle to calculate it. Therefore, 0 = Mp(24.21 m/s) + (4500 kg)(1.5 m/s) , then solve for Mp and you get 278.81 kg .

Last edited by a moderator: May 1, 2017
11. Nov 6, 2004

### Duarh

Hmm, I wonder how you got that value of 24.21, all I get is cos(50)x40=25.7 m/s, which is slightly different - but if your prof was ok with your numbers, whatever. I can't access that site, however (asks for a NetID). I'd just recommend seeing whether using cos(50)x40= 25.7 instead of 24.21 for the initial x direction velocity (to calculate the mass of the projectile) will ultimately give you the right answer for part C. Or at least explain why/how you get 24.21 m/s.

Edit: Hmm, it does appear I was mistaken about what the problem intends to ask, and 40 m/s is in fact the velocity with respect to the already-recoiling cannon. I'm still of the opinion the formulation is too vague, but sorry for confusing posts.

Last edited: Nov 6, 2004
12. Nov 6, 2004

### Staff: Mentor

sign problems

You are messing up your signs. If you are taking Vc as the magnitude of the cannon's speed (a postive number), then:
speed of projectile with respect to cannon is: 40 m/s
speed of cannon with respect to ground is: - Vc
speed of projectile with respect to ground is: 40 - Vc

momentum of cannon: 4500(-Vc)
momentum of projectile: m(40 - Vc)

Conservation of momentum => total momentum = zero =>
4500(-Vc) + m(40 - Vc) = 0

Now solve for Vc.

13. Nov 6, 2004

### Mivz18

I still get the same answer; whether my answer is positive, or negative, it is still wrong. Also, my signs weren't mixed I don't believe, because when I solve for Vc with the way you set it up, I get the same answer. I'm so lost in physic's space!!

14. Nov 6, 2004

### Mivz18

nevermind, I found my miscalculation....DUH!