what is the total force exerted on the bullet while travelling down the 0.85m long barrel of the rifle? also find kinetic energy of the bullet as it leaves the muzzle.
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The impulse = F*t = the change in momentum. We know the momentum of the bullet when it leaves the muzzle = 0.025kg * 150 = 0.375kgm/s
How long was the bullet in the barrel?
Vf^2 = Vi^2+2ad => 150^2=0+2a*0.85 => a = 150^2/2*0.85= 13,235.3m/s^2
Vf = at => t = 150/13.253.3 = 11.33ms
F = 0.375/11.33ms = 33.09N
OR we could calculate the work done on the bullet while in the barrel and equate that to the kinetic energy at the muzzle = 0.5*0.025*150^2=28.125J
28.125J = F*0.85 => F = 33.09N
Same answer
How long was the bullet in the barrel?
Vf^2 = Vi^2+2ad => 150^2=0+2a*0.85 => a = 150^2/2*0.85= 13,235.3m/s^2
Vf = at => t = 150/13.253.3 = 11.33ms
F = 0.375/11.33ms = 33.09N
OR we could calculate the work done on the bullet while in the barrel and equate that to the kinetic energy at the muzzle = 0.5*0.025*150^2=28.125J
28.125J = F*0.85 => F = 33.09N
Same answer