# Problem of conservation of linear mometum

1. Aug 25, 2014

### dk_ch

1. The problem statement, all variables and given/known data
N cannonballs each of mass m are fired at velocity V one after another from a cannon of mass M(including balls) lying on a movable carriage initially at rest . what will be the final velocity of cannon after the last ball is fired?

2. Relevant equations
Initial total momentum = 0
after the first ball is fired let the recoil velocity of the the cannon be v1 and its mass was then M-m
then by the law of conservation of linear momentum (M-m)v1 +mv = 0.
Now i am confused. My question is in what respect the velocity of the second ball is becoming v.

3. The attempt at a solution

Last edited: Aug 25, 2014
2. Aug 25, 2014

### Vibhor

Assuming the equation is in vector form ,it is fine .

Since nothing is mentioned ,it should be with respect to the ground .

3. Aug 25, 2014

### dk_ch

velocity (v) of the first ball was w.r.t ground because the cannon was in rest then. how does the second ball have the same velocity w.r.t ground ? Please give some light on further proceeding .

4. Aug 25, 2014

### jbriggs444

I think that you are expected to assume that the velocity of each cannonball is the same with respect to the cannon. [This would be measured against the cannon's velocity just prior to the shot]

You are correct that the second shot will be going slower (wrt the ground) than the first shot because it was fired by a moving cannon.

However, this assumption makes the problem significantly more difficult to solve than Vibhor's version.

5. Aug 25, 2014

### squareroot

Curiosity

Hey, i ve stumbled upon this thread, studied this problem and got some questions about it.

My first question is related to time.When studying this system can you consider that all the projectiles are shot at once? In that case if we take the momentum of the projectiles Pp and the momentum of the canon Pc we know that Pp + Pc=0,
so Nmv - Mv'=0 from where v'=NMv/M

Does this work?
Because if you take the time between shots in consideration, then we must ask how much does the canon decelerate between shots(due to friction) or is it a frictionless motion?

Anyway, I'm really curious about what the validity of my statement from above.

Thanks guys!(&gals *heh*)

6. Aug 25, 2014

### jbriggs444

If the total projectile mass is much less than the total cannon mass then it would be a reasonable approximation, though not 100% accurate.

7. Aug 25, 2014

### BvU

dear dk_ch,

Is it Ok if I first reply to sqrt (who actually shouldn't be rewarded like that for hijacking a thread!):
This exercise wasn't thought out on the battlefield, but from behind a desk. You are not supposed to assume too much. No friction is mentioned, so no friction is present. Cannonballs are fired one after another. It specifically says so.

Briggs is right: muzzle velocity is a property of a piece of ordnance. Which is usually attached to something heavy so it doesn't cause too much damage to own side. Not in this exercise.
Velocity is somewhat objectively measurable wrt the muzzle itself, eh....

We can assume shots are fired horizontally (else introduction of a $\cos\theta$ takes care of that).

dk_ch (swiss or danish ?) is spot on with (M-m)v1 +mv = 0 after the first shot. Let's travel with the gun carriage and hope nothing much is in our path...
write v1 = mV/(M-m). 1 is for the number of balls shot off. Light the fuse and experience another bump which gives us a momentum kick mV again, so we gain speed. This time we write v2 = v1 + mV/(M-2m).

Our joy ride continues faster and faster until vN = vN-1 + mV/(M-Nm) is the final speed.

Note we have used the Galileo transformation to add the speeds. Justified, since all the firing power in the world can't bring us anywhere near relativistic speeds.

Do the math. vN is proportional to V, so work out vN/V as a function of A and N (where A = M/m, known to be bigger than N). Still ends with an ugly expression, I suspect, but who knows.

8. Aug 25, 2014

### jbriggs444

I agree that the result will likely be ugly.

If the number of cannonballs is large there is a useful approximation that applies. I don't want to say much more.

9. Aug 26, 2014

### dk_ch

In this case it is assumed no frictional force is acting either on ball or on the carriage-cannon combination and the balls are horizontally projected with same nuzzle velocity v. The total mass of N balls is lees than mass of the cannon and the combination has got total mass M.

Considering conservation momentum, We can write after the 1st ball is shot if the cannon and carriage combination gets velocity v1 then
(M-m)v1=mv .........(1)

Then after the second ball is fired with nuzzle velocity v and the combination gets velocity v2 then by conservation law we have
(M-m)v1=(M-2m)v2-mv ........(2)
combining (1) and (2) we have 2mv = (M-2m)v2
similarly after N th ball is fired if the combination acquires velocity vN then
Nmv = (M-Nm)vN
hence vN =Nmv/{(M-Nm)vN}

Last edited: Aug 26, 2014
10. Aug 26, 2014

### ehild

The nuzzle velocity is v with respect to the cannon. Conservation of momentum is valid in the ground frame of reference. What is the velocity of the cannonball with respect to the ground?

ehild

Last edited: Aug 26, 2014
11. Aug 27, 2014

### dk_ch

dk_ch (swiss or danish ?) is spot on with (M-m)v1 +mv = 0 after the first shot. Let's travel with the gun carriage and hope nothing much is in our path...
write v1 = mV/(M-m). 1 is for the number of balls shot off. Light the fuse and experience another bump which gives us a momentum kick mV again, so we gain speed. This time we write v2 = v1 + mV/(M-2m).

Our joy ride continues faster and faster until vN = vN-1 + mV/(M-Nm) is the final speed.

The portion of statement given in BvU-s post #7 is quoted above in which it is not understood why
v2 = v1 + mV/(M-2m)

12. Aug 28, 2014

### ehild

The velocity of the ball is v with respect to the cannon. Its velocity with respect to the ground is Vn+v where Vn is the velocity of the cannon after the n-th shot. So the momentum kicked away by the next shot is m(Vn+v), and conservation of momentum dictates

$(M-n)V_n=(M-(n+1))V_{n+1}+m(V_n+v)$.

Doing it step by step: At the first shot, the velocity of the cannon is $V_1=-\frac{mv}{M-m}$

For the second ball, it has velocity $v'=V_1+v$. Applying conservation of momentum:

$(M-m)V_1=(M-2m)V_2+(V_1+v)m$.

Rearranging, collecting the terms with V1 you get

$V_2=V_1-\frac{mv}{M-2m}$

BvU denoted the speeds by Vn and v, but it might confuse one when applying conservation of momentum.
If you consider the ball's velocity negative v=-|v| and the velocity of the cannon positive, the equation becomes

$V_2=V_1+\frac{m|v|}{M-2m}$

ehild

13. Aug 29, 2014

### dk_ch

$(M-n)V_n=(M-(n+1))V_{n+1}+m(V_n+v)$.

Is this line correctly written ?

14. Aug 29, 2014

### ehild

Not quite. M is mass n is a number... m is missing.

ehild

15. Aug 29, 2014

### jbriggs444

ehild's analysis uses the ground frame. That works, but is not the way that I would have tackled the exercise. Adopt the frame in which the cannon is at rest after the first n-1 shots.

The starting momentum is zero. The ending momentum is divided between one cannonball and the cannon and its remaining ammo. Let $V_c$ denote the leftward velocity of the cannon and $V$ denote the rightward muzzle velocity of the ball (eliminating minus signs by burying them in the sign convention).

$(M-nm)V_c=mV$

or

$V_c=\frac{mV}{M-nm}$

Converting back to ground coordinates...

$V_n=V_{n-1} + \frac{mV}{M-nm}$

That's the sum of a harmonic series. The sum of a harmonic series is a worked problem.

Last edited: Aug 29, 2014