How Accurate Is the Physics Behind a Bullet Striking a Metal Plate?

[jisbon]

Homework Statement

As shown below, a metal plate is pivoted at point P. A 0.04kg bullet is shot into the plate (velocity = 220m/s) and is stuck in the plate. Determine maximum height the COM of the system rises after the collision. The breadth of plate = 40cm, height = 1m

Relevant Equations

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Here are my workings, and I was wondering if I'm correct so far.
Let $m$ be mass of bullet and $M$ be mass of plate.
COM:
$mu_{bullet} = (m+M)v$
$\frac{1}{2}(m+M)v^2=(m+M)gh +\frac{1}{2}I\omega ^2$
where I is the inertia, so using parallel axis theorem,
$I = \frac{1}{12}bh^3 + md^2 = \frac{1}{12}(0.4)1^3 + (5)(0.5-0.2)^2$
Is this correct so far?

 

[haruspex]

Homework Statement: As shown below, a metal plate is pivoted at point P. A 0.04kg bullet is shot into the plate (velocity = 220m/s) and is stuck in the plate. Determine maximum height the COM of the system rises after the collision. The breadth of plate = 40cm, height = 1m
Homework Equations: -

View attachment 250533
Here are my workings, and I was wondering if I'm correct so far.
Let $m$ be mass of bullet and $M$ be mass of plate.
COM:
$mu_{bullet} = (m+M)v$
$\frac{1}{2}(m+M)v^2=(m+M)gh +\frac{1}{2}I\omega ^2$
where I is the inertia, so using parallel axis theorem,
$I = \frac{1}{12}bh^3 + md^2 = \frac{1}{12}(0.4)1^3 + (5)(0.5-0.2)^2$
Is this correct so far?

Your first equation is wrong because it ignores the (unknown) reaction impulse from the hinge.
Your second is wrong because work will not be conserved. Never assume it is without good reason.
So that makes the laws of conservation of work and linear momentum of no help here. What does that leave?

 

[jisbon]

Your first equation is wrong because it ignores the (unknown) reaction impulse from the hinge.
Your second is wrong because work will not be conserved. Never assume it is without good reason.
So that makes the laws of conservation of work and linear momentum of no help here. What does that leave?

I'm not exactly sure the first equation (about the reaction impulse)
Regarding the second equation though, should I be using COAM (conservation of angular momentum instead?)

 

[haruspex]

should I be using COAM (conservation of angular momentum instead?)

Yes. Whenever you do that, you need to consider the best choice of axis. Which would you choose?

 

[jisbon]

Yes. Whenever you do that, you need to consider the best choice of axis. Which would you choose?

Axis? Not sure what you meant here.
But I crafted an equation, as was wondering if you could check it for me:
COAM:
$L_{initial bullet}+L_{initialplate}=L_{finalbullet}+L_{finalplate}$
$mvr + 0 = (I_{bullet}+I_{plate})\omega$
$(0.04)(220)(0.5) = (m_{plate}(0.5)^2 + \frac{1}{12}bh^3 + md^2)\omega$

 

[haruspex]

Axis? Not sure what you meant here.

Except in special cases, moment of inertia, angular momentum and torque are always in relation to a chosen axis. When dealing with them you should always state your choice of axis.
From your post above, you seem to have chosen the centre of the plate. The problem with that is that the unknown reaction impulse from the hinge will also have a moment about that axis, and you have left that of your equation.

There are two ways to proceed.
With the axis you have chosen, include the contribution from the hinge reaction (call the impulse from it J, say). Then write the equation for linear momentum, which will also feature J. Then you can combine the equations to eliminate J.
The neater method is to choose your angular momentum axis so that J makes no contribution. Then you do not need a second equation. About what axis would a force from the hinge have no moment?

 

[jisbon]

Except in special cases, moment of inertia, angular momentum and torque are always in relation to a chosen axis. When dealing with them you should always state your choice of axis.
From your post above, you seem to have chosen the centre of the plate. The problem with that is that the unknown reaction impulse from the hinge will also have a moment about that axis, and you have left that of your equation.

There are two ways to proceed.
With the axis you have chosen, include the contribution from the hinge reaction (call the impulse from it J, say). Then write the equation for linear momentum, which will also feature J. Then you can combine the equations to eliminate J.
The neater method is to choose your angular momentum axis so that J makes no contribution. Then you do not need a second equation. About what axis would a force from the hinge have no moment?

The axis I should have chosen is probably the pivot I will guess?

 

[haruspex]

The axis I should have chosen is probably the pivot I will guess?

Yes.

 

[jisbon]

Hi :)
So continuing to solve this question, here's what I have:

$$\begin{aligned}I_{B}W=\left( I_{B}+I_{p}\right) W_{f}\\ mv_{r}=\left( mv^{2}+\dfrac {1}{12}\left( a^{2}+b^{2}\right) \right) W_{f}\end{aligned} $$
Is this correct? If it is, how do I use the final angular velocity to find out the change in height of COM?
$$\left( 0.04\right) \left( 220\right) \left( 0.5\right) =\left( 0.04\left( 0.5\right) ^{2}+\dfrac {1}{12}\left( 1^{2}+0.4^{2}\right) \left( 5\right) \right) W_{f} $$

 

[haruspex]

$$\begin{aligned}I_{B}W=\left( I_{B}+I_{p}\right) W_{f}\\ mv_{r}=\left( mv^{2}+\dfrac {1}{12}\left( a^{2}+b^{2}\right) \right) W_{f}\end{aligned} $$

I assume $mv_r$ is supposed to be $mvr$, where r is the vertical distance from the hinge to the path of the bullet. Anyway, that matches what you had in post #5.
But the right hand side has me baffled. Where does mv² come from, what happened to the mass of the plate, what exactly do a and b represent and what axis are you using to get I_p?