Answer:
a. Speed of cannon = 0.902 m/s b. speed of cannonball = 29.44 m/s
Explanation:
Here is the complete question
A revolutionary war cannon, with a mass of 2090 kg, fires a 16.7 kg ball horizontally. The cannonball has a speed of 113 m/s after it has left the barrel. The cannon carriage is on a flat platform and is free to roll horizontally. What is the speed of the cannon immediately after it was fired?
Answer in units of m/s
The same explosive charge is used, so the total energy of the cannon plus cannonball system remains the same. Disregarding friction, how much faster would the ball travel if the cannon were mounted rigidly and all other parameters remained the same?
Answer in units of m/s
Solution
From the law of conservation of momentum,
momentum of cannon = momentum of cannonball.
Let m₁,v₁ and m₂,v₂ represent the masses and velocities of the cannon and cannonball respectively.
So, m₁v₁ = m₂v₂.
The speed of the cannon is thus v₁ = m₂v₂/m₁
m₁ = 2090 kg, m₂ = 16.7 kg and v₂ = 113 m/s
v₁ = m₂v₂/m₁ = 16.7 × 113/2090 m/s = 1887.1/2090 m/s = 0.902 m/s
Since the same charge is used, and the cannon mounted rigidly, the total kinetic energy of cannon + cannon ball = kinetic energy of cannonball.
Since all other parameters remain the same,
1/2m₁v₁² + 1/2m₂v₂² = 1/2m₂v₃²
m₁v₁² + m₂v₂² = m₂v₃²
2090 × 0.902² + 16.7 × 113² = 16.7v₃²
1700.43 + 12769 = 16.7v₃²
14469.43 = 16.7v₃²
v₃² = 14469.43/16.7 = 866.43
v₃ = √866.43 = 29.44 m/s
So the speed of the cannon ball is now 29.44 m/s
Post answer:
Completed question
A revolutionary war cannon, with a mass of 2090 kg, fires a 16.7 kg ball horizontally. The cannonball has a speed of 113 m/s after it has left the barrel. The cannon carriage is on a flat platform and is free to roll horizontally. What is the speed of the cannon immedi-ately after it was fired?
Answer in units of m/s
The same explosive charge is used, so the total energy of the cannon plus cannonball system remains the same. Disregarding friction, how much faster would the ball travel if the cannon were mounted rigidly and all other parameters re-mained the same?
Answer in units of m/s
Explanation:
1. Mass of cannon, =2090kg
Mass of ball fires=16.7kg
The speed of the ball after being fired, = 113m/s.
Using the recoil of a gun.
Mass of cannon × recoil velocity = mass of ball fires × velocity of ball
2090×v=16.7×113
Then, v=16.7×113/2090
v=0.903m/s
2. The second problem deals with the conservation of energy and all the energy are Kinetic energy
If the explosive where mounted rigidly then, the recoil velocity will be zero, I.e the cannon won't move
K.E of cannon=1/2mv²
K.E=0.5×2090×0.903²
k.E=851.94J
K.E of ball=1/2mv²
K.E=0.5×16.7×113²
K.E=106,621.15J
Total energy =106,621.15+851.94
Total energy =107,473.09J
Now all this energy is transfer to the ball alone because the canon is mounted
Total energy =1/2mv²
107,473.09=0.5×16.7×v²
v²=107,473.09/(0.5×16.7)
v²=12,871.029
v=√12871.029
v=113.45m/s
The velocity if the ball increases because the canon was mounted
