A bullet collides perfectly elastically with one end of a rod

[mattlfang]

Homework Statement

A bullet with mass m, velocity v perfectly elastically collide with one end of a rod on a slippery plane and the bullet stop moving after the collision. Find the mass of the stick M

Relevant Equations

angular momentum, momentum, conservation of energy

A bullet with mass m, velocity v perfectly elastically, vertically collide with one end of a rod on a slippery plane and the bullet stops moving after the collision. Find the mass of the stick M

the bullet stops moving after an elastic collision, so all energy is transformed to the rod. There is a conservation of energy and conservation of momentum, but I feel like some conditions is still missing.

moment of inertia around its center I = M L²/12,

We have conservation of momentum where mv = I ω
we have conservation of energy where 0.5* mv² = 0.5 * I ω²

but we have three unknown M, L, ω but only two equations here?

 

[PeroK]

but we have three unknown M, L, ω but only two equations here?

$\omega$ will depend on the other variables.

Where does the bullet impact the stick? On the end?

 

[mattlfang]

$\omega$ will depend on the other variables.

Where does the bullet impact the stick? On the end?

yes, one end of the stick., I just made it more clear

 

[PeroK]

yes, one end of the stick., I just made it more clear

Why don't you try to solve the equations and see what happens?

 

[mattlfang]

Why don't you try to solve the equations and see what happens?

I got $v = \omega$ and $m = I$, but nothing about $L$ can help me solve for $M$?

 

[PeroK]

I got $v = \omega$ and $m = I$,

Those equations are both dimensionally invalid.

 

[mattlfang]

Those equations are both dimensionally invalid.

That's true, but what exactly are wrong with my equations in the OP?

 

[PeroK]

we have conservation of energy where 0.5* mv² = 0.5 * I ω²

This is not right. The KE of a rigid body is the sum of the KE of its CoM and its rotational KE about the CoM. In this case we have:$$\frac 1 2 mv^2 = \frac 1 2 MV^2 + \frac 1 2 I \omega^2$$

 

[mattlfang]

This is not right. The KE of a rigid body is the sum of the KE of its CoM and its rotational KE about the CoM. In this case we have:$$\frac 1 2 mv^2 = \frac 1 2 MV^2 + \frac 1 2 I \omega^2$$

But now we have 2 equations but four unknowns...

 

[PeroK]

But now we have 2 equations but four unknowns...

Who cares? Everything is unknown! Get solving!

 

[PeroK]

PS there are three equations: conservation of 1) energy; 2) linear momentum; 3) angular momentum (about a point of your choosing).

 

[mattlfang]

PS there are three equations: conservation of 1) energy; 2) linear momentum; 3) angular momentum (about a point of your choosing).

oh indeed...

so I should have

$mv = MV$
$mvL = I \omega$
$\frac{1}{2} m v^2 = \frac{1}{2} MV^2 + \frac{1}{2}I \omega^2$
$\omega \frac{L}{2} = V$

here the angular momentum is w.r.t to the other end of the rod...

 

[PeroK]

oh indeed...

so I should have

$mv = MV$
$mvL = I \omega$
$\frac{1}{2} m v^2 = \frac{1}{2} MV^2 + \frac{1}{2}I \omega^2$
$\omega \frac{L}{2} = V$

here the angular momentum is w.r.t to the other end of the rod...

First, this idea that you always need more equations than variables is a myth. Sometimes you do, but not always. In this case, we've got to hope that ultimately $M$ depends only on $m$ and $v$ and not on $L$. Note that $\omega$ and $V$ should be functions of $m, v, M, L$, but that should come out in the calculations. You don't need to see this in advance. You just get stuck in and see what happens.

Let's take your first equation. $$mv = MV \ \Rightarrow \ V = \frac m M v$$ and that immediately takes care of $V$.

Your second equation is not right. You chose badly! If we take AM about the centre of the stick, then$$mv \frac L 2 = I \omega$$Note that the AM about the other end of the stick has a component due to the linear momentum of the CoM of the stick.

Try getting a bit further now.

 

[mattlfang]

First, this idea that you always need more equations than variables is a myth. Sometimes you do, but not always. In this case, we've got to hope that ultimately $M$ depends only on $m$ and $v$ and not on $L$. Note that $\omega$ and $V$ should be functions of $m, v, M, L$, but that should come out in the calculations. You don't need to see this in advance. You just get stuck in and see what happens.

Let's take your first equation. $$mv = MV \ \Rightarrow \ V = \frac m M v$$ and that immediately takes care of $V$.

Your second equation is not right. You chose badly! If we take AM about the centre of the stick, then$$mv \frac L 2 = I \omega$$Note that the AM about the other end of the stick has a component due to the linear momentum of the CoM of the stick.

Try getting a bit further now.

indeed, my second one was wrong. yes ineed, I don't have to have 4 equations for four unknowns. Now I have figured out my problem. Thanks for the help.

$mv = MV$
$\frac{mvL}{2} = I \omega = \frac{M}{12} L^2 \omega = \frac{M L^2 \omega}{12}$
$\frac{mv^2}{2} = \frac{MV^2}{2} + \frac{I \omega^2}{2}$

from $\frac{mvL}{2} = \frac{M L^2 \omega}{12}$ and $mv = MV$
$6V = \omega L$

$\frac{mv^2}{2} = \frac{MV^2}{2} + \frac{I \omega^2}{2}$
we got $mv^2 = MV^2 + \frac{ML^2 \omega^2}{12}$

$L^2 \omega^2 = 36 V^2$

so $mv^2 = 4MV^2$

and $V = \frac{mv}{M}$ so $mv^2 = \frac{4m^2v^2}{M}$

$M = 4m$

Just one last question. Can we always argue that angular momentum and linear momentum conserve individually? Can some part of linear momentum convert into angular momentum?

 

[PeroK]

so $mv^2 = 7MV^2$ $M = 7m$

It's always best to double-check your answers. Does $M = 7m$ satisfy the conservation equations?

 

[PeroK]

PS all your work looks right, but you must have made a slip somewhere in the final calculation.

 

[mattlfang]

PS all your work looks right, but you must have made a slip somewhere in the final calculation.

yes just fixed... sorry

 

[PeroK]

Just one last question. Can we always argue that angular momentum and linear momentum conserve individually? Can some part of linear momentum convert into angular momentum?

They are actually both consequences of the other. I tend to think that conservation of linear momentum is a special case of angular momentum. In this case, if you took angular momentum about two different points (perhaps one end of the rod), then you would get the equivalent of taking AM and LM.

Note that taking AM about the centre of the rod "lost" the LM of the CoM of the rod. Taking it about another point would include it.

 

[jbriggs444]

Can some part of linear momentum convert into angular momentum?

To answer that question directly: No.
They are conserved separately. You cannot convert one to the other.

 

[PeroK]

To answer that question directly: No.
They are conserved separately. You cannot convert one to the other.

Nevertheless, we can solve this problem using AM about two axes and not directly invoking conservation of linear momentum.

Conservation of AM about the centre of the rod, as above, gives: $$\frac 1 2 Lm v = \frac 1 {12}ML^2 \omega$$And conservation of AM about the far end of the rod gives $$Lmv = \frac{MVL}{2} + \frac 1 {12}ML^2 \omega$$And we can obtain the linear momentum equation by subtracting the first equation from the first. Alternatively, we can combine those equations to get, as above $$L\omega = 6V$$ And then combine that with the conservation of energy to get the result.

 

[haruspex]

here the angular momentum is w.r.t to the other end of the rod...

Depends whether you mean about the initial position of the end of the rod or about the end of the rod however it moves before and after. In kinetics, be wary of taking moments about a point that is neither fixed in space nor the mass centre of the system under consideration.

 

[mattlfang]

Nevertheless, we can solve this problem using AM about two axes and not directly invoking conservation of linear momentum.

Conservation of AM about the centre of the rod, as above, gives: $$\frac 1 2 Lm v = \frac 1 {12}ML^2 \omega$$And conservation of AM about the far end of the rod gives $$Lmv = \frac{MVL}{2} + \frac 1 {12}ML^2 \omega$$And we can obtain the linear momentum equation by subtracting the first equation from the first. Alternatively, we can combine those equations to get, as above $$L\omega = 6V$$ And then combine that with the conservation of energy to get the result.

btw, I just realized, if we choose the conservation of AM wrt the end of the rod where the bullet hits, it's even simply since the angular momentum wrt to that end is 0.