A cannon of mass 6.50 x 10 3 kg is rigidly bolted to the earth so it can recoil

Question

A cannon of mass 6.50 x 103 kg is rigidly bolted to the earth so it can recoil only by a negligible amount. The cannon fires a 88.1-kg shell horizontally with an initial velocity of 551 m/s. Suppose the cannon is then unbolted from the earth, and no external force hinders its recoil. What would be the velocity of a shell fired from this loose cannon? (Hint: In both cases assume that the burning gunpowder imparts the same kinetic energy to the system.)

A cannon of mass 6.50 x 10^3 kg is rigidly bolted to the earth so it can recoil only by a negligible amount. The cannon fires a 88.1-kg shell horizontally with an initial velocity of 551 m/s. Suppose the cannon is then unbolted from the earth, and no external force hinders its recoil. What would be the velocity of a shell fired from this loose cannon? (Hint: In both cases assume that the burning gunpowder imparts the same kinetic energy to the system.)

Explanation / Answer

Since the cannon is bolted to the ground the recoil is absorbed by the cannon plus the earth so there is effectively no recoil and no loss of energy. All energy goes into kinetic energy of the shell and this is also the energy produced by the gunpowder.

KE = (1/2)m*v0^2

v0 = velocity of the shell with no recoil = 551 m/s
M = mass of cannon
m = mass of shell
V = velocity of recoil
v = velocity of shell

Now let us do the second case. Since the cannon is free to move we must have conservation of momentum.

M*V = m*v
Solve this for V in terms of v: V = (m/M)v

The sum of the kinetic energies must equal what we did above so:
(1/2)M*V^2 + (1/2)m*v^2 = (1/2)m*v0^2
M*V^2 + m*v^2 = m*v0^2
(m^2/M)*v^2 + m*v^2 = m*v0^2 .... using the equation for V
m*v^2(m/M + 1) = m*v0^2
v = v0*SQRT[1/(m/M) + 1)] = v0*SQRT[M/(m + M)]

Check by using this v to calculate V and the kinetic energies of the cannon from recoil and that of the shell. The sum of these should equal the initial kinetic energy for the case of no recoil.
KE1 = (1/2)m*v^2 = (v0^2/2)m*M[1/(m + M)] = (v0^2/2)*M/(1 + M/m)

V = (m/M)v = v0*SQRT[(m^2/M)/(m + M)]
KE2 = (1/2)M*V^2 = (v0^2/2)m^2/(m + M) = (v0^2/2)m/(1 + M/m)

KE1 + KE2 = (v0^2/2)*M/(1 + M/m) + (v0^2/2)m/(1 + M/m)
KE1 + KE2 = (v0^2/2)*(M + m)/(1 + M/m)
KE1 + KE2 = (1/2)m*v0^2

And this is the kinetic energy in the case of no recoil so this all looks good.

M/(m + M) = 0.9866

So the velocity of the shell allowing for cannon recoil is only a slight bit lower than in the case when the cannon is constrained to not recoil. The velocity will 0.9866 of the recoilless velocity.

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