Final answer:
The 3000-kg cannon will have a recoil velocity of 2.25 m/s when it fires a 15.0-kg shell at 480 m/s. The kinetic energy of the cannon is 7575 J. The vertical component of momentum does not affect the horizontal recoil due to the cannon mount.
Step-by-step explanation:
To calculate the recoil velocity of a 3000-kg cannon firing a 15.0-kg shell at 480 m/s at a 20.0° angle, we apply the law of conservation of momentum. Initially, the system's momentum is zero since the cannon is at rest, and after the shell is fired, the total momentum must remain zero. Focusing on the horizontal component, the momentum of the shell can be found using its mass, the cosine of the firing angle, and its velocity.
Horizontal momentum of the shell = mass of shell × (velocity of shell × cos(20.0°))
Pshell = 15.0 kg × (480 m/s × cos(20.0°))
Pshell = 15.0 kg × 450.4 m/s
Pshell = 6756 kg·m/s
The cannon's recoil velocity (vcannon) is in the opposite direction with equal momentum to conserve momentum:
3000 kg × vcannon = 6756 kg·m/s
vcannon = 6756 kg·m/s / 3000 kg
vcannon = 2.25 m/s
The direction of the recoil is exactly opposite to the horizontal component of the shell's velocity. The kinetic energy of the cannon can then be calculated:
KE = 0.5 × mass of cannon × (recoil velocity)2
KE = 0.5 × 3000 kg × (2.25 m/s)2
KE = 0.5 × 3000 kg × 5.06 m2/s2
KE = 7575 J
The vertical component of momentum that is imparted to the cannon when it is fired results in an upward and downward force due to gravity, and as the cannon is mounted to only recoil horizontally, this vertical component does not affect the recoil velocity.