Problem of conservation of linear mometum

[dk_ch]

Homework Statement

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?

Homework 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.

The Attempt at a Solution

So i can not proceed after that. Please help and remove my confusion.

 

[Vibhor]

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.

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

Now i am confused. My question is in what respect the velocity of the second ball is becoming v.

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

 

[dk_ch]

Check the signs.

what sign to be checked ?

With respect to the ground .

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 .

 

[jbriggs444]

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 ?

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.

 

[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 P_p and the momentum of the canon P_c we know that P_p + P_c=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*)

 

[jbriggs444]

My first question is related to time.When studying this system can you consider that all the projectiles are shot at once?

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

 

[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 v_N = v_N-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. v_N is proportional to V, so work out v_N/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.

 

[jbriggs444]

Still ends with an ugly expression, I suspect, but who knows.

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.

 

[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 v_N then
Nmv = (M-Nm)v_N
hence v_N =Nmv/{(M-Nm)v_N}

Please comment if not logical.

 

[ehild]

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)

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?
Read BvU-s post #7.

ehild

 

[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)
will someone explain it,please?

 

[ehild]

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

 

[dk_ch]

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

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

Is this line correctly written ?

 

[ehild]

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

Is this line correctly written ?

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

ehild

 

[jbriggs444]

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)
will someone explain it,please?

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.