How Does a Bullet's Momentum Determine the Force Stopping It in a Tree Trunk?

AI Thread Summary
The discussion focuses on calculating the force that stops a bullet penetrating a tree trunk using momentum and energy principles. A 4.78 g bullet traveling at 510.8 m/s penetrates 4.93 cm into the trunk, leading to a calculated stopping force of approximately 32,870.75 N. The second part of the discussion involves determining the time it takes for the bullet to stop, with a participant questioning the use of the equation vf^2 = vi^2 + 2at for this calculation. There is a correction regarding the numbers used in the first part of the problem. The conversation emphasizes the application of physics principles to solve real-world problems.
Ronaldo21
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Basic Momentum and Energy!

1. (part 1 of 2) A 4.78 g bullet moving at 510.8 m/s penetrates a tree trunk to a depth of 4.93 cm. a) Use work and energy considerations to find the magnitude of the force that stops the bullet.
Answer in units of N.

for this i did 1/2mv^2=w=fd
so 1/2(.00536)(681.8)^2=w=fd
1245.801323=fd
1245.801323=f(.0379)
and i solved for f and got 32870.74731

2. (part 2 of 2) b) Assuming that the frictional force is constant, determine how much time elapses between the moment the bullet enters the tree and the moment the bullets stops moving.
Answer in units of s.

for this one i just did vf^2=vi^2+2at. is that correct??

THANK YOU!
 
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The numbers in your solution to part one are not those in the question!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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