How is the Barrel Length of a Human Cannon Calculated?

AI Thread Summary
The discussion focuses on calculating the force and barrel length of a human cannonball. 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 1.14 x 10^4 N. To determine the barrel length, the impulse is used to find the change in velocity, which is calculated to be 57 m/s. The kinetic energy is then derived, leading to the conclusion that the barrel length is 10 meters. The calculations confirm the accuracy of the results, emphasizing the relationship between force, displacement, and energy.
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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