# Momentum of a projectile and ship

## Homework Statement

A battle ship (4.09*10^7 kg) fires a salvo of 3 rounds from its foward turret in the direction of the bow at an angle of 10 degrees above the horizontal. Each projectile weighs 1220 kg, each barrel is 20.9 m long and the muzzle velocity of the projectile is 770 m/sec. Acceleration of each projectile down the barrel is 1.418 x 10^4 m/s2 at 0.0543 seconds.

If the momentum must be conserved in the x direction, 1) what's the velocity of the ship in the x direction and 2) in what direction is it moving along the x axis after the projectiles have left the ship's guns?

## Homework Equations

p(linear momentum of an object)=mv
M1V1 + M2V2=0

## The Attempt at a Solution

1) I'm not sure whether I should get the velocity of the projectile in the x direction, but I just used it's velocity in the x direction in solving the problem. I did the trigonometric work on paper.

M1V1 + M2V2=0
V2= -M1/M2(V1)
V2= -1220/4.09*10^7(758.3)
V2= -.0226 m/sec

2) The ship recoils to the left of the x axis.

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"1.418 x 104 m/s2 at 0.0543 seconds" what do you mean by that? Is it an increase of 1.418 x 104 m/s every 0.0543 seconds?

"1.418 x 104 m/s2 at 0.0543 seconds" what do you mean by that? Is it an increase of 1.418 x 104 m/s every 0.0543 seconds?
sorry, its suppsoe to be 1.418*10^4; 1.418*10^4 is the acceleration of each projectile in the barrel and .0543 sec is the time it takes for the projectile to travel down the barrel.

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Ok but how is this useful if we are given the muzzle velocity?

Ok and the muzzle velocity is the velocity at which the projectile leaves the barrel? Or is it the initial velocity down the barrel?
i was told that it was supposidly the final velocity/velocity at which the projectile leaves the barrel.

okay; so does that mean the muzzle velocity is the the velocity I would use in the momentum equation for V1, but instead the velocity vectors in the x direction which is 758.3 m/sec?

Ok, it actually does affect the problem. What we are interested in is the moment when the projectile leaves the ship. This moment would be right when the projectile begins to travel down the barrel. So with the information given, we must calculate the initial velocity. We have

$${v_{i}}^{2} = {v_{f}}^{2} - 2ad$$

We know vf, a and d, so we can calculate vi. Once we get vi, we calculate its horizontal component and use it in the conservation of momentum formula.

Ok, it actually does affect the problem. What we are interested in is the moment when the projectile leaves the ship. This moment would be right when the projectile begins to travel down the barrel. So with the information given, we must calculate the initial velocity. We have

$${v_{i}}^{2} = {v_{f}}^{2} - 2ad$$

We know vf, a and d, so we can calculate vi. Once we get vi, we calculate its horizontal component and use it in the conservation of momentum formula.
$${v_{i}}^{2}$$= (770)^2 - 2(14180)(20.9m)
Vi=13.27 m/sec!!!!

thank you, sir.

I do have one more question, if you don't mind...
What is the total momentum of the 3 projectiles in the x direction parrell to the water?
Would I just multiply the momentum of one of the projectiles by three or is it zero since total momentum is always zero??

If the three projectiles are in the same direction, their momentum adds ups since the sign of momentum depends on the sign of the velocity. If you add the momentum of the ship into the equation, the sum would be 0. In any explosion or implosion (in this case it's an explosion - things get separated) the momentum gained in a direction by a fragment is equal to the momentum gained by the opposite fragment in the other direction. This is basically the law of conservation of momentum.

If the three projectiles are in the same direction, their momentum adds ups since the sign of momentum depends on the sign of the velocity. If you add the momentum of the ship into the equation, the sum would be 0. In any explosion or implosion (in this case it's an explosion - things get separated) the momentum gained in a direction by a fragment is equal to the momentum gained by the opposite fragment in the other direction. This is basically the law of conservation of momentum.
so is the following correct? (I now used the new initial velocity in the x-direction:
M1V1(3)=?
(1220)(13.07)(3)= 47830.34 J

?

Yes. The ship would have the opposite momentum.

Yes. The ship would have the opposite momentum.
If I had the power to bestow upon you a salmagundi of riches beyond belief, you would be the first person to share in this wealth.
THANK YOU!

If I had the power to bestow upon you a salmagundi of riches beyond belief, you would be the first person to share in this wealth.
THANK YOU!
Not to rain on ayones parade, from the earlier thread you posted, I wrote:

"This is right, but perhaps a more straightforward approach assumes vf=770, and Vi=0,

Given that Vf^2-Vi^2=2ax, then a=770^2/(2*20.9)
a=14184m/s^2 and t would just be the distance divided by ave velocity:
20.9/385=0.0543s Different ways to skin a cat. Best to have many knives.:"

In other words you just have introduced some roundoff error and vinit=0, this was the assumption after all, and acceleration was calculated on this basis.

Again, this operates under several assumptions that acceleration is linear from the detonation of the charge, to the time the shell leaves the barrel. It is entirely possible that this is not the case due to friction, leakage of gas, the fact that the volume occupied by the gas is becoming larger as the shell progresses down the barrel, etc.

The question about which velocity to consider is an interesting one. If we ignored the ship altogether, I think most would agree that say in the case of the rifle, its the muzzle velocity (Vf) that determines the momentum, even though the physical recoil may be experienced sooner.

So I would disagree, and suggest by extension to the case with the rifle, that the ships negative velocity*mass=3*cos(10)*770*mass(shell).

Not to rain on ayones parade, from the earlier thread you posted, I wrote:

"This is right, but perhaps a more straightforward approach assumes vf=770, and Vi=0,

Given that Vf^2-Vi^2=2ax, then a=770^2/(2*20.9)
a=14184m/s^2 and t would just be the distance divided by ave velocity:
20.9/385=0.0543s Different ways to skin a cat. Best to have many knives.:"

In other words you just have introduced some roundoff error and vinit=0, this was the assumption after all, and acceleration was calculated on this basis.

Again, this operates under several assumptions that acceleration is linear from the detonation of the charge, to the time the shell leaves the barrel. It is entirely possible that this is not the case due to friction, leakage of gas, the fact that the volume occupied by the gas is becoming larger as the shell progresses down the barrel, etc.

The question about which velocity to consider is an interesting one. If we ignored the ship altogether, I think most would agree that say in the case of the rifle, its the muzzle velocity (Vf) that determines the momentum, even though the physical recoil may be experienced sooner.

So I would disagree, and suggest by extension to the case with the rifle, that the ships negative velocity*mass=3*cos(10)*770*mass(shell).
you do indeed have a very strong point when noting this new initial velocity has turned the momentum problem into a bit of an error. However, i am a bit perplexed as to whether the equation you presented above clearly depicts the shell's (or projectile's) velocity if one multiplies along with it 3 and cos(10). is such thing possible to correctly calculate the momentum?

you do indeed have a very strong point when noting this new initial velocity has turned the momentum problem into a bit of an error. However, i am a bit perplexed as to whether the equation you presented above clearly depicts the shell's (or projectile's) velocity if one multiplies along with it 3 and cos(10). is such thing possible to correctly calculate the momentum?
actually, I would like to answer my own question...
Multiplying cos(10) and (770) gives one the velocity in the x direction, which is what I want. Multiplying three simply takes into account the 3 projectiles.

You are as well a life saver, denverdoc and you too are deemed greatful to partake of these chimerically hypothetical riches.

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any of u guys know how to do electric potential?

any of u guys know how to do electric potential?
negatory, sir.

well, i think your point is well taken, we are ignoring the vertical component of the momentum, as this would simply try to push the ship deeper in the water, which would be opposed by buoyancy, etc. It is essential tho that we only consider the x component of the shells velocity, as one fired vertically would have no tendency to cause the boat to recoil along the x axis, thus the cos 10 degree factor. As to algebraically adding the volley of shells, I can see some concern on your part that this isn't fair and proper. One reason might be that the ship has already come to rest between shots. But this is physics homework where all kinds of outlandish assumptions are made routinely in the service of introducing principles. In other words, for the purpose of this problem since there is no consideration of friction, we can superimpose the three shots whether minutes apart or a single shell of 3600Kg was shot with the same muzzle velocity, or whether three guns all aimed in parallel lines were fired simultaneously. (coure then you might actually get some torque and cause the ship to rotate--couple chapters down the line, not to worry). Hope this didn't confuse matter furthers, if you have some specific concerns re the legitimacy, holler.

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well, i think your point is well taken, we are ignoring the vertical component of the momentum, as this would simply try to push the ship deeper in the water, which would be opposed by buoyancy, etc. It is essential tho that we only consider the x component of the shells velocity, as one fired vertically would have no tendency to cause the boat to recoil along the x axis, thus the cos 10 degree factor. As to algebraically adding the volley of shells, I can see some concern on your part that this isn't fair and proper. One reason might be that the ship has already come to rest between shots. But this is physics homework where all kinds of outlandish assumptions are made routinely in the service of introducing principles. In other words, for the purpose of this problem since there is no consideration of friction, we can superimpose the three shots whether minutes apart or a single shell of 1600Kg was shot with the same muzzle velocity, or whether three guns all aimed in parallel lines were fired simultaneously. (coure then you might actually get some torque and cause the ship to rotate--couple chapters down the line, not to worry). Hope this didn't confuse matter furthers, if you have some specific concerns re the legitimacy, holler.
so would you say that the ship, for my homework's sake, is moving left of the x direction after all of this firing of shells has been made?

denverdoc, I don't think they gave information on acceleration and distance if we were not meant to consider the initial velocity. This said, I think the whole idea of a projectile accelerating down a barrel is flawed. The projectile gets sent by absorbing some momentum of an explosion that happens at the lower end of the barrel. I believe the projectile stops accelerating there and then, as in a instantaneous collision.

denverdoc, I don't think they gave information on acceleration and distance if we were not meant to consider the initial velocity. This said, I think the whole idea of a projectile accelerating down a barrel is flawed. The projectile gets sent by absorbing some momentum of an explosion that happens at the lower end of the barrel. I believe the projectile stops accelerating there and then, as in a instantaneous collision.
Ordinarily, I would agree, but in this case my sense was it more of a mega problem embracing all basic mechanics already covered... Hey if you read my post I'm with you about the linear acceleration assumption down the length of a gun barrel, but would argue that acceleration does take place after the initial explosion--if you look at some really high speed photos of a bullet exiting a gun, the gas is on the slugs heels.

Now its entirely possible that the bullet reaches a peak velocity somewhere short of the muzzle end, and the gas catches up. But the types of powders used iirc (not knowing much about guns) are supersonic. It may be a stretch, but it is like saying your average stroke in an internal combustion engine culminates with inertia of the piston.

I was watching the military channel late last night(hey education should come from many sources) and was intrigued by these anti-tank rifles whose sole mission was to serve up ultrahigh velocity armor piercing bullets--the barrel lengths were like a yard or more long, and not for reasons of accuracy. I'm therefore guessing that the expanding gas is still doing some useful work for the entire length or at least some fraction thereof.

In that context, I would also call attention to the fact that handguns rarely come close to the mach 2 and 3 muzzle velocities of a well tooled rifle.

so would you say that the ship, for my homework's sake, is moving left of the x direction after all of this firing of shells has been made?
well the problem says nothing about motion of the ship. If at rest, it would move infinitessimally in the opposite direction of the munitions direction.