# Moomentum question

1. Dec 11, 2008

### devanlevin

a cannon with a mass of M is let free from the top of a slope with an incline of $$\vartheta$$ the friction between the cannon and the slope is given as $$\mu$$N.
after the cannon has passed a dstance of L it shoots a shell, with a mass of m horizontally, what is the minimal velocity of the shell so that the cannon will stop,?

im not sure if this means that M is the mass of the cannon and the shell, or of the cannon alone, but form the answer i think that they mean that it is just the mass of the cannon, and somehow they dont take the mass of the shell into consideration until its shot.

Umin=(M+m)/m(cos$$\vartheta$$-$$\mu$$sin$$\vartheta$$)*$$\sqrt{2Lg(sin$$\vartheta$$-$$\mu$$cos$$\vartheta$$)}$$

what i did was,
I)drew a diagram,-- http://picasaweb.google.com/devanlevin/DropBox?authkey=sbH95pBl_D8#5278474258602595442 [Broken]
II)found the cannons acceleration
III)found its velocity after it had passed L
IV)found the minimal Velocity of th shell

II)to find the acceleration

using$$\sum$$F=ma=mgsin$$\vartheta$$-$$\mu$$N

acceleration downhill a=g(sin$$\vartheta$$-$$\mu$$cos$$\vartheta$$)
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III)to find the cannons velocity

V^2=2a$$\Delta$$x

velocity downhill V=$$\sqrt{2Lg(sin$$\vartheta$$-$$\mu$$cos$$\vartheta$$)}$$

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IV) to find the minimum shooting velocity

(M+m)V=mu*cos$$\vartheta$$+MU (when MU=0)
u=V(M+m)/mcos$$\vartheta$$

which would be my final answer except that the answer according to the book is

u=V(M+m)/m(cos$$\vartheta$$-$$\mu$$sin$$\vartheta$$)

where does this extra bit come from? where have i gone wrong?

Last edited by a moderator: May 3, 2017
2. Dec 11, 2008

### Staff: Mentor

Realize that during the explosion of the cannon, an impulse is exerted perpendicular to the incline as well as parallel to it. That increases the normal force, and thus friction, resulting in a change in momentum parallel to the incline. (Momentum is not conserved.)

3. Dec 11, 2008

### devanlevin

so how do i approach that? if momentum is not conserved, what do i know? up till step (IV) have i done it all right?
what should my next steps be

4. Dec 11, 2008

### Staff: Mentor

Everything is fine up to there. Consider the impulse on the cannon perpendicular to the incline. (What's the cannonball's momentum in that direction?) Relate that to the change in momentum parallel to the incline due to the increased friction during the explosion.

5. Dec 11, 2008

### devanlevin

Consider the impulse on the cannon perpendicular to the incline
would that would be mu*sin$$\vartheta$$??

now what do i say? is the total momentum ==> (M+m)V=mu - mu*sin$$\vartheta$$??
is that what you are saying the loss of momentum is the cannonball's momentum in that direction?

how do i relate this to the friction?

6. Dec 11, 2008

### Staff: Mentor

Right.

Not quite, you have to figure out resultant change in momentum parallel to the incline.

And that's where friction comes in.

Consider: FΔt = mu sinθ

But here, F is the normal force. What additional friction force does that correspond to? Find the impulse parallel to the incline due to the increased friction force.

7. Dec 12, 2008

### devanlevin

how do i figure out resultant change in momentum parallel to the incline. what do i know to work this out?
i know that the momentum before was Mv parallel to the incline, and 0 perpendicular to the incline, and after that, was the total momentum [mu*sin$$\vartheta$$ + mu*cos$$\vartheta$$]?? just a guess:(

then what else do i know, - the F on y axis is N, which i need to find out, and the force on x axis is f and mgsin$$\vartheta$$, now i see that
Py=$$\int$$(N)dt
Px=$$\int$$(Mg-f)dt=$$\int$$(Mg)dt-$$\mu$$$$\int$$(N)dt

Px=Mg(dt)-$$\mu$$Py
then can i ignore the mg force assuming it does not affect the momentum that much
Px=-$$\mu$$Py

is this correct, still not sure what to do with it, problem being that momentum isnt conserved,
feel like im just writing things and running with them not really knowing what to do next

8. Dec 12, 2008

### Staff: Mentor

Find the impulse due to the increased friction. Reread my last post, near the end. How does friction relate to normal force?

9. Dec 12, 2008

### devanlevin

im not managing to connect the impulses to an actual force, i understand everything you are saying, but cant put it into equations,

i know that FΔt = mu sinθ, and that the force on the y axis is the normal force,
i know that the force on the x axis is the friction which is controlled by the normal force, therefore, the impulse on the x axis is $$\mu$$*the impusle on the y axis. but i dont know how to get anywhere from this information.
could you please show me your workings and your steps? has this has turned out to be a lot more complicated than i though, or am i just missing something?

10. Dec 12, 2008

### Staff: Mentor

That's really all there is to it, you just have to recognize what you have.

The impulse along the x axis is then -μmu sinθ (negative, since it acts in the direction of friction). And impulse is the change in momentum, so (in the x direction):
Initial momentum + impulse = final momentum
initial momentum + impulse = final momentum of cannon (0) + final momentum cannonball

Give it another shot.

11. Dec 12, 2008

### Staff: Mentor

Moomentum sounds like a property of a mooving cow.

12. Dec 13, 2008

### devanlevin

Give it another shot

getting there...

can i say ..
(M+m)V=mu + the impulse

and that the impulse is
x -μmu sinθ
y -mu sinθ

is this right??
from here i can find the magnitude of the impulse, and from the equation find u, although looking at this doesnt look like ill come to the right answer

13. Dec 13, 2008

### devanlevin

tried what you said

(M+m)V-μmu*sinθ=mu
(M+m)V=mu+μmu*sinθ
(M+m)V=mu(1+μ*sinθ)
from this i get

u=[(M+m)/m]*V/(1+μ*sinθ)

but correct the answer is
u=[(M+m)/m]*V/(cosθ-μ*sinθ)

14. Dec 13, 2008

### devanlevin

dont worry about it, i got it,, im such an idiot,,, see where ive gone wrong,, thanks for the help.