What, if anything, happens to the speed of a fighter plane when it fires a cannon at an enemy plane in ...?

Question

front of it?

Answer 1

The law of conservation of momentum applies in all cases.
For a given system, the initial momentum must equal the final momentum.

In your example, we can define the "System" to be the combination fighter plane + bullet. If we call this initial momentum P, then at any time after the initial conditions, the sum of the momentums of all parts of our system must add up to P.
When the bullet is fired foward, the fighter plan slows down to compensate and conserve momentum.

Momentum (P) = mass (m) * velocity (v)
P = mv

The bullet's mass is very tiny compared to that of the fighter plane, so even a large increase in the bullet's velocity would only mean a small decrease in the resulting velocity of the plane.

If you call the bullet's mass, m_1 (with final velocity v_2) and the plane's mass m_2 (with final velocity V_2),
we know that,
P = m_1*v_1 + m_2*v_2
if P_1 goes up, then P_2 must go down by just that amount.
(delta P_1) = - (delta P_2)
m_1 * (delta v_1) = -m_1 * (delta v_2)
since we assume the mass stays constant,
(delta v_2) = - (m_1 * delta v_1) / m_2

Answer 2

The plane and "cannon" must follow the conservation of momentum. Since the combination of the two are progressing forward and the cannon fires its projectile forward with respect to the plane, there must be some sort of recoil in the opposite direction. If the projectile has a momentum of p, the plane's momentum must decrease by p, lowering its speed. However, this is dependent of the mass of the plane, the projectile, the speed of the projectile, and the fact that the plane may or may not burn more fuel to compensate for the difference.

Answer 3

The speed of the plane will be reduced according to the conservation of momentum equation. An interesting real world application of this is the A-10 "tank killer" Warthog. If the cannon is fired for too long, given the appropriate circumstances, it can actually stall the plane due to the reward momentum imparted by the relatively heavy, fast moving bullets.

Answer 4

Counter intutive...the round and the powder charge are all ready travelling at the speed of the fighter...as the round is fired its chemical reaction as it it propels the mass of the round down the barrell and out into the air PROBABLY gives a net LOSS to the aircraft..in V...but miniscule.. I'm guessing

Answer 5

I would imagine it slows down a bit, relative to the weight of the round fired and the speed it was fired at. I'm sure there's an equation that will demonstrate how much it will slow.

Answer 6

Some really erudite answers, but the closest was still a bit off.
Cannon fire slows the plane, with the slug serving as the propellant and Newton's "Equal and opposite reaction" rule applying.
In WWI, machine-gun firing aircraft could actually bring themselves to a standstill, stall and crash, simply through the thrust of their weapon (referred to as "recoil").
Cheers

Answer 7

a lot of thing will happen, but it will slow the plane down for sure, it will heat the air and metal around the cannon, possible the air around the plane, and other minor unoticeable effects

Answer 8

I'm sure it slows then goes back to its original speed, for every action there is an equal ( opposite)reaction.