How is the Barrel Length of a Human Cannon Calculated?

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SUMMARY

The calculation of the barrel length for a human cannonball involves determining the force acting on the cannonball and the work-energy principle. A human cannonball with a mass of 70 kg experiences an impulse of 4.0 x 10^3 N•s over 0.35 seconds, resulting in an average force of 11,428.5 N. Using the work-energy equation, the barrel length is calculated to be 10 meters, based on the kinetic energy derived from the impulse and force applied.

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  • Understanding of impulse and momentum (Δp = J)
  • Knowledge of the work-energy principle (Work = ΔKE)
  • Familiarity with basic physics formulas for force and energy
  • Ability to perform calculations involving mass, velocity, and displacement
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  • Study the relationship between impulse and momentum in physics
  • Learn about the work-energy theorem and its applications
  • Explore advanced topics in dynamics, such as projectile motion
  • Investigate real-world applications of physics in entertainment, such as circus acts
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Physics students, educators, and professionals interested in mechanics, as well as anyone involved in designing or analyzing circus performances involving human cannonballs.

Mary1910
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Human cannonballs have been a part of circuses for years. A human cannonball with a mass of 70kg experiences an impulse of 4.0 x 10^3 N•s for 0.35 s.

a) Calculate the force acting on the human cannonball.

F=(J)/(Δt)
=(4000 N•s)/(0.35s)
=11428.5 N
=1.14 x 10^4 N

b)How long was the barrel of the cannon?(Assume the force is applied only for the period of time that the cannonball is in the cannon.)

Im not to sure about part b), At first I thought that I should be using F•d=ΔEk, but I don't think so. Although I don't think I know of any other formulas for this type of problem that would incorporate displacement.

Any help would be appreciated. Thank you.
 
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Mary1910 said:
Human cannonballs have been a part of circuses for years. A human cannonball with a mass of 70kg experiences an impulse of 4.0 x 10^3 N•s for 0.35 s.

a) Calculate the force acting on the human cannonball.

F=(J)/(Δt)
=(4000 N•s)/(0.35s)
=11428.5 N
=1.14 x 10^4 N

b)How long was the barrel of the cannon?(Assume the force is applied only for the period of time that the cannonball is in the cannon.)

Im not to sure about part b), At first I thought that I should be using F•d=ΔEk, but I don't think so. Although I don't think I know of any other formulas for this type of problem that would incorporate displacement.

Any help would be appreciated. Thank you.
Actually that's the average force (and that's what the question should have asked you for).

Your idea was fine! Work = ΔKE seems as reasonable as any way to do this.
 
SammyS said:
Your idea was fine! Work = ΔKE seems as reasonable as any way to do this.

Thank you! Could you just let me know if this is correct?

b)

J=Δp
Δp=mΔv

4.0 x 10^3 N•s=70kgΔv
Δv=(4.0 x 10^3 N•s)/(70kg)
Δv=57m/s

Ek=½mvf^2-½mvi^2
=½(70kg)(57m/s)^2-½(70kg)(0m/s)^2
=113715-0
=1.14 x 10^5 J

F•d=ΔEk
d=(ΔEk)/(F)
=(1.14 x 10^5)/(1.14 x 10^4)
=10m

Therefore the barrel of the cannon is 10m :smile:
 
Yes, with significant digits, it's 10m or arguably it's 10.0m
 
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