MacDougal87
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Hey guys, I've been a member of this site for a while now, looking at other people's homework problems and attempting them as extra practice. I find that reading about what other people have trouble with when attempting physics problems helps me avoid them myself, and it has worked pretty well so far.
Most of the time when I make a mistake in physics problems, and have the correct answer on hand, I am eventually able to figure out where I went wrong and apply the correct steps to get there.
However, this situation truly baffles me:
A .005-kg bullet traveling horizontally with speed 1,000-m/s strikes an 18-kg door, embedding itself 10 cm from the side opposite the hinges. The 1-m wide door is free to swing on its frictionless hinges.
Before it hits the door, what is the bullet's angular momentum relative to the door's axis of rotation?
(to distinguish between the two, m will be the mass of the bullet and M will be the mass of the door, the distance from the axis of rotation of the door is d)
The way I approached this problem I figured we'd need:
L=Iω
To get I; I modeled the bullet as a particle, and used I=md²
Using v=dω; ω=v/d
Substituting, I could use L=(md²)(v/d)
Which simplifies to L=mdv
Plugging in the given values, I calculated (.005kg)(.1m)(1,000m/s) = .5 kgm²/s
The answer in the back of the book however, gives the solution as 4.5 kgm²/s. The only way I can figure they might have gotten this solution was to factor in the mass of the door somewhere, maybe using the door's moment of inertia, but that doesn't make any sense to me since the problem specifically asks for the bullet's angular momentum before it hits the door. I can't seem to think of any other way to take this problem since the bullet is not interacting with anything else in the system at this moment.
Any suggestions would be great, I have a big exam coming up on Wednesday, and need all the practice I can get =)
Most of the time when I make a mistake in physics problems, and have the correct answer on hand, I am eventually able to figure out where I went wrong and apply the correct steps to get there.
However, this situation truly baffles me:
Homework Statement
A .005-kg bullet traveling horizontally with speed 1,000-m/s strikes an 18-kg door, embedding itself 10 cm from the side opposite the hinges. The 1-m wide door is free to swing on its frictionless hinges.
Before it hits the door, what is the bullet's angular momentum relative to the door's axis of rotation?
(to distinguish between the two, m will be the mass of the bullet and M will be the mass of the door, the distance from the axis of rotation of the door is d)
Homework Equations
The way I approached this problem I figured we'd need:
L=Iω
To get I; I modeled the bullet as a particle, and used I=md²
Using v=dω; ω=v/d
Substituting, I could use L=(md²)(v/d)
Which simplifies to L=mdv
The Attempt at a Solution
Plugging in the given values, I calculated (.005kg)(.1m)(1,000m/s) = .5 kgm²/s
The answer in the back of the book however, gives the solution as 4.5 kgm²/s. The only way I can figure they might have gotten this solution was to factor in the mass of the door somewhere, maybe using the door's moment of inertia, but that doesn't make any sense to me since the problem specifically asks for the bullet's angular momentum before it hits the door. I can't seem to think of any other way to take this problem since the bullet is not interacting with anything else in the system at this moment.
Any suggestions would be great, I have a big exam coming up on Wednesday, and need all the practice I can get =)
