Bullet hits block connected to wheel and bounces off

[Sunfun]

Homework Statement

There is a small block of steel with mass 2.4kg and is firmly attached to a low mass wheel a distance .3m from its center. The wheel is free to rotate about a vertical axis and is at rest. A .0023kg bullet traveling with an initial speed of 305m/s in the horizontal plane strikes the block and bounces off the block and has final speed of 210m/s in the horizontal plane.

Homework Equations

a) what is the angular momentum of the bullet about the center of the wheel just before the collision?
b) what is the rotational speed of the block of steel and wheel just after the collision?
c) how much mechanical energy is lost in this collision?

The Attempt at a Solution

L=r x p
so L=sin(67)(.3)(.0023)(305)?

 

[V0ODO0CH1LD]

The momentum transferred by the bullet onto the block of steel is $(v_icos23^{\circ} - v_fcos18^{\circ})m_b$ where $v_i$ and $v_f$ are the initial and final velocities of the bullet, respectively, and $m_b$ is the mass of the bullet. Can you work out the rest?

 

[Sunfun]

So if that is the momentum, how is that related tot he rotational speed?

 

[gneill]

So if that is the momentum, how is that related tot he rotational speed?

You might take a moment to reflect upon what quantities are conserved for a closed system.

 

[Sunfun]

Isn't the angular momentum conserved?

 

[V0ODO0CH1LD]

The wheel can only move by rotating around the vertical axis perpendicular to the plane of the wheel it self. So the only way to transfer momentum to that block of steel which is attached to the wheel is in the direction perpendicular to the radius of the wheel. Since that bullet lost some momentum to the block during the collision you can assume that the momentum was transferred in the horizontal direction, because that is the only direction the block can move anyway. So the equation basically looks like:

(v_icos23^{\circ} - v_fcos18^{\circ})m_{bullet} = v_{block}m_{block}

 

[gneill]

Isn't the angular momentum conserved?

Yes it is!

 

[Sunfun]

So for the mechanical energy, does the bullet have kinetic energy initial=the rotational energy of the block minus the kinetic energy of the bullet final?

 

[gneill]

So for the mechanical energy, does the bullet have kinetic energy initial=the rotational energy of the block minus the kinetic energy of the bullet final?

The collision may or may not be perfectly elastic. But regardless, angular momentum is always conserved. I'd suggest working with angular momentum.

 

[Sunfun]

So how would that give the mechanical energy?

 

[gneill]

So how would that give the mechanical energy?

Once you have the final motions of each component, you can determine the final mechanical energy.

 

[Sunfun]

So it is the difference between the initial kinetic energy and the final kinetic energy+rotational energy?

 

[gneill]

So it is the difference between the initial kinetic energy and the final kinetic energy+rotational energy?

Sure. Have you already worked out the rotation of the wheel?

 

[Sunfun]

Yes I have. Thank you so much!

 

[haruspex]

The momentum transferred by the bullet onto the block of steel is $(v_icos23^{\circ} - v_fcos18^{\circ})m_b$ where $v_i$ and $v_f$ are the initial and final velocities of the bullet, respectively, and $m_b$ is the mass of the bullet. Can you work out the rest?

I think you mean $v_i$ and $v_f$ are the initial and final speeds of the bullet, and it should be plus, not minus.
In general, it would be better to work with angular momentum for a problem like this, as gneill suggests. That's because there could be a reactive impulse from the wheel's axle. But because the block is considered a point mass and the wheel is considered massless, there is no reactive impulse and the equations produce the same result.