# Conservation of momentum when a shell is fired from a cannon

When a shell is fired from a cannon free to move on a horizontal surface and inclined at an angle say 45 degree with horizontal. Does relative velocity of shell with respect to cannon depends on whether cannon is moving or not.That is, will the shell will have same velocity relative to cannon irrespective of the initial state of motion of cannon or not

• yes, the speed of the shell relative to the cannon will remain the same
– user83548
Oct 30, 2015 at 12:59

Let $\vec{p}_i$ and $\vec{p}_f$ denote the initial and final momenta of the cannon-shell system, respectively. Suppose we shoot the cannon horizontally (shooting at an angle will not change the argument) so that we can consider scalar quantities (or, equivalently, the components of the vectors in the horizontal direction).

Since no external forces are present, we may use conservation of momentum, which says that $\vec{p}_i = \vec{p}_f$, i.e., $p_i = p_f$ for our 1-dimensional problem.

Initially we imagine the system to be at rest, so $p_i = 0 = p_f$. On the other hand, we know that $p_f = m_\text{shell} v_\text{shell} - m_\text{cannon} v_\text{cannon}$. Equating the two expressions for the final momentum $p_f$ and solving for $v_\text{cannon}$, we conclude that

$v_\text{cannon} = \frac{m_\text{shell}}{m_\text{cannon}} v_\text{shell}$.

Therefore, the relative speed between the two objects is

$v_\text{relative} = v_\text{shell} - v_\text{cannon} = \left( 1 - \frac{m_\text{shell}}{m_\text{cannon}} \right) v_\text{shell}$.

Case 1:

If the cannon is "held fixed," we are imagining a situation in which the cannon is affixed firmly to the earth, so that the mass of the cannon can be taken to be very large. In this limit, $m_\text{shell}/m_\text{cannon}$ becomes very small, can $v_\text{relative} \sim v_\text{shell}$.

Case 2:

If the cannon is "free to move" or "recoil" we immediately see that $v_\text{relative} < v_\text{shell}$.

Case 3:

If we imagine a situation in which the cannon's speed is held at some fixed quantity (be it 0 or some value like, say, 100 m/s), we are really in Case 1 again, and the previous anaswer (of hiddenharmonizer) is valid.

You are correct. If the cannon was not moving and a shell was fired moving away from the cannon at 300m/s, it would still move away from the cannon at 300 m/s if the cannon itself was initially moving at 100 m/s. In the perspective of the first cannon, the shell from the second cannon would be moving away at 400 m/s.

This would be the same as throwing a baseball out the window of your car in the direction you are driving. You already have an initial velocity relative to the earth and you are adding to the baseball additional energy when you throw it.