Block and bullet collision problem

[reb659]

Homework Statement

A Block of mass M is attached to a cable of length L. It is initially held at the horizontal position at rest. It is then released in the motion of a pendulum and collides with a bullet of mass m, traveling at a speed of Vi, at the lowest point of its trajectory. The bullet embeds in the block and stops its motion completely. What is the speed of the bullet just before it hits the block?

Homework Equations

conservation of linear momentum
cons. of energy

The Attempt at a Solution

I know cons. of linear momentum applies regardless of the collision type, so taking the initial to be the moment before the collision and the final to be after the collision I get:

mV1i = -MV2i

It seems like an inelastic collision, so cons. of energy does not apply during the collision. I'm a bit stuck here.

 

[Doc Al]

I know cons. of linear momentum applies regardless of the collision type, so taking the initial to be the moment before the collision and the final to be after the collision I get:

mV1i = -MV2i

Good. What's V2?

It seems like an inelastic collision, so cons. of energy does not apply during the collision.

That's true: Conservation of energy does not apply during the collision. But what about during the pendulum motion of the block?

 

[reb659]

I let V2i to be the speed of the block right before the collision.


Conservation of energy would still apply during the pendulum motion. So I was thinking I could take the initial moment to be when the block is horizontal and the final to be just before the collision, right before it gets hit by the bullet. so in addition to my first equation I'd get:

1/2m(V1i)^2+1/2M(V2i)^2-Mg(L)=0.

 

[Doc Al]

Conservation of energy would still apply during the pendulum motion.

Right. The mechanical energy of the block is conserved.

So I was thinking I could take the initial moment to be when the block is horizontal and the final to be just before the collision, right before it gets hit by the bullet. so in addition to my first equation I'd get:

1/2m(V1i)^2+1/2M(V2i)^2-Mg(L)=0.

Treat the swinging block by itself. (Don't involve the bullet.) What is its energy when the cable is horizontal? When vertical? (You're almost there.)

 

[reb659]

I don't really see why I would ignore the bullet. Isn't the energy of the bullet conserved too?

 

[Doc Al]

I don't really see why I would ignore the bullet. Isn't the energy of the bullet conserved too?

You can include the bullet's energy if you want to, but do it right. What's the change in the bullet's energy during the time that the block swings down? (I would advise against this approach.)

Better is just to realize that the block's energy is conserved. That's all you need to know to solve for V2i. Who cares about the bullet? You'll worry about the bullet during the collision.

 

[reb659]

Ok I see what you are saying now. Just for practice, would the change in energy of the bullet be zero?

For the block, the initial kinetic energy would be zero because its released from rest.
The final kinetic energy, before it hits the vertical point would be V2i since that's the moment during the motion I want. The change in gravitational potential energy would be -MgL because the initial is greater than the final. So I would get 1/2m(V2i)^2=MgL and from there I can solve for V2i.

 

[Doc Al]

Just for practice, would the change in energy of the bullet be zero?

That's what I would say, but I strongly suggest you don't "practice" such an approach.

You have no information about the bullet's change in energy during the time the block was falling. For all we know, the bullet's was fired one microsecond before the block reached it's lowest point. But the bullet is irrelevant to solving for the block's speed, which is what we are trying to do.

The right way is to apply conservation of energy to the block--we know all we need to know about its energy.

For the block, the initial kinetic energy would be zero because its released from rest.
The final kinetic energy, before it hits the vertical point would be V2i since that's the moment during the motion I want. The change in gravitational potential energy would be -MgL because the initial is greater than the final. So I would get 1/2m(V2i)^2=MgL and from there I can solve for V2i.

Good! That's the way.