Cannon Question: Loose Cannon Velocity

[JessGold]

Homework Statement

A cannon of mass 5.80 x 10^3 kg is rigidly bolted to the Earth so it can recoil only by a negligible amount. The cannon fires an 85.0-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 by this loose cannon? (Hint: In both cases assume that the burning gunpowder imparts the same kinetic energy to the system.)

Homework Equations

The Attempt at a Solution

 

[CompuChip]

Two conservation laws apply here. One of them is already implicitly mentioned in the hint. The other one has to do with the masses and velocities.

 

[nlightner]

The velocity of a shell fired by the loose cannon can be calculated using the conservation of momentum principle. Since the burning gunpowder imparts the same kinetic energy in both cases, the initial momentum of the cannon-shell system when bolted to the Earth will be equal to the final momentum when the cannon is loose.

Using the equation for conservation of momentum, we can write:

m1v1 = m2v2

Where m1 and v1 are the mass and velocity of the cannon-shell system when bolted to the Earth, and m2 and v2 are the mass and velocity of the cannon-shell system when the cannon is loose.

Substituting the given values, we get:

(5.80 x 10^3 kg)(551 m/s) = (5.80 x 10^3 kg + 85.0 kg)(v2)

Solving for v2, we get:

v2 = 552.9 m/s

Therefore, the velocity of the shell fired by the loose cannon would be 552.9 m/s. This is slightly higher than the initial velocity when the cannon was bolted to the Earth due to the negligible recoil of the cannon.