Maximum change in the kinetic energy of a particle

[TheSodesa]

Homework Statement

Show that the largest possible change in the kinetic energy , $\Delta E_{kin}$, of a particle of mass $m$ running into another particle of mass $M$ at rest in the lab coordinate system is
$$\Delta E_{kin} = \frac{-4AE_{kin}}{(1+A)^{2}}$$, where $A = \frac{M}{m}$.

Calculate the limit $$\lim_{M \to m} \Delta E_{kin}$$ and plot $\Delta E_{kin}(A)$.

Homework Equations

Conservation of energy.
Conservation of momentum.
Rutherford's experiment?

The Attempt at a Solution

I'm going to assume, that both energy and momentum are conserved.

Energy:
$$E_{mi} + \stackrel{0}{E_{Mi}} = E_{mf} + E_{Mf}\\ \iff \\ |\Delta E_{kin}| = |E_{mf} - E_{mi}| = |-E_{Mf}| = \frac{1}{2}Mv^{2}$$

Momentum:
\begin{align*}
x) &&mu_{m} &= mv_{m}\cos{\theta} + Mv_{M}\cos{\varphi}\\
y) &&0 &= mv_{m}\sin{\theta} + Mv_{M}\sin{\varphi}
\end{align*}

Adding these equations together and solving for $v_{M}$ yields
\begin{align*}
v_{M}
&= \frac{mu_{m} - mv_{m}(\sin{\theta} + \cos{\theta})}{M(\sin{\varphi} + \cos{\varphi})} &| \sin{x} + \cos{x} = \sqrt{2}\sin{(x + \frac{\pi}{4})}\\
&= \frac{m(u_{m} - \sqrt{2}v_{m}\sin{(\theta + \frac{\pi}{4})})}{\sqrt{2}M\sin{(\varphi + \frac{\pi}{4})}}
\end{align*}

Plugging this into the energy expression:
\begin{align*}
\Delta E_{kin}
&= \frac{1}{2}M \left( \frac{m(u_{m} - \sqrt{2}v_{m}\sin{(\theta + \frac{\pi}{4})})}{\sqrt{2}M\sin{(\varphi + \frac{\pi}{4})}} \right)^{2} \\
&= \frac{m^{2}(u_{m} - \sqrt{2}v_{m}\sin{(\theta + \frac{\pi}{4})})^{2}}{4M\sin^2 (\varphi + \frac{\pi}{4})} \\
\end{align*}

Now the numerator is at it's largest, when (this bit is irrelevant)
$$sin(\theta + \frac{\pi}{4}) = 0 \\ \iff\\ \theta + \frac{\pi}{4}rad = 0^{\circ} + n \times 180^{\circ}\\ \iff\\ \theta = 180^{\circ} - \frac{\pi}{4}rad = 180^{\circ} - 45^{\circ} = 135^{\circ}$$

Then, if we differentiate with respect to $\varphi$,
\begin{align*}
\frac{d \Delta E_{kin}}{d \varphi}
&= \frac{d}{d \varphi} \frac{m^{2}(u_{m})^{2}}{4M\sin^{2} (\varphi + \frac{\pi}{4})} &&| \text{ Let }\frac{m^{2}(u_{m})^{2}}{4M} = C \\
&= \frac{d}{d \varphi} C \sin^{-2}(\varphi + \frac{\pi}{4})\\
&= -2 C \cos(\varphi + \frac{\pi}{4})\sin^{-3}(\varphi + \frac{\pi}{4})\\
&= 0
\end{align*}

when

$$cos(\varphi + \frac{\pi}{4}) = 0\\ \iff\\ \varphi + \frac{\pi}{4} = \frac{\pi}{2}\\ \iff\\ \varphi = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} = 45^{\circ}$$

Alright, now we have the angles (which I'm not convinced by). Now what? Plugging them back into the energy expression isn't going to help me simplify it.

I feel like I've gone down a wrong path. Help, please?

 

[Simon Bridge]

How are you uncertain about your calculations so far?
You have gone to a lot of trouble to calculate some angles ... why? What do these angles mean?

Don't you need an expression for $\Delta E_{kin}$ in terms of $E_{kin}$?

 

[TheSodesa]

How are you uncertain about your calculations so far?
You have gone to a lot of trouble to calculate some angles ... why? What do these angles mean?

Don't you need an expression for $\Delta E_{kin}$ in terms of $E_{kin}$?

Well, I was hoping I could get rid of the angles somehow. The expression that I arrived at
$$\Delta E_{kin} = \frac{m^2 (u_{m}^2 - \sqrt{2} v_{m} \sin^2 (\theta + \frac{\pi}{4}))}{\sqrt{2} M \sin^2 (\varphi + \frac{\pi}{4})}$$

does have $m$ and $u_{m}$ in it, meaning if I could only get rid of the other variables (angles), I might be able to sort a $\frac{1}{2}mu_{m}^2$ in there somewhere.

Looking at the photo that I posted, the angles themselves signify which directions the two particles move in after the collision; $\theta$ is the direction of $m$ and $\varphi$ the direction of $M$. My uncertainty stems from the fact that I'm not sure I've made the right assumptions thus far. Because the question asked us to find the largest possible $\Delta E_{kin}$, we want the numerator to be as large as possible and the denominator to be as small as possible. We can sort out the numerator easily enough:
$$m^2 (u_{m}^2 - \sqrt{2} v_{m} \sin^2 (\theta + \frac{\pi}{4}))$$
is at it's largest when $\sin^2 (\theta + \frac{\pi}{4}) = 0$ (because it can't be negative). However, we can't just make the $\sin^2 (\varphi + \frac{\pi}{4})$ in the denominator equal zero for obvious reasons. We could differentiate the function to find the extrema, but that only allows us to find the angle $\varphi$, that I'm not sure what to do with.

There's my problem.

 

[TheSodesa]

Well, I was hoping I could get rid of the angles somehow. The expression that I arrived at
$$\Delta E_{kin} = \frac{m^2 (u_{m}^2 - \sqrt{2} v_{m} \sin^2 (\theta + \frac{\pi}{4}))}{\sqrt{2} M \sin^2 (\varphi + \frac{\pi}{4})}$$

does have $m$ and $u_{m}$ in it, meaning if I could only get rid of the other variables (angles), I might be able to sort a $\frac{1}{2}mu_{m}^2$ in there somewhere.

Looking at the photo that I posted, the angles themselves signify which directions the two particles move in after the collision; $\theta$ is the direction of $m$ and $\varphi$ the direction of $M$. My uncertainty stems from the fact that I'm not sure I've made the right assumptions thus far. Because the question asked us to find the largest possible $\Delta E_{kin}$, we want the numerator to be as large as possible and the denominator to be as small as possible. We can sort out the numerator easily enough:
$$m^2 (u_{m}^2 - \sqrt{2} v_{m} \sin^2 (\theta + \frac{\pi}{4}))$$
is at it's largest when $\sin^2 (\theta + \frac{\pi}{4}) = 0$ (because it can't be negative). However, we can't just make the $\sin^2 (\varphi + \frac{\pi}{4})$ in the denominator equal zero for obvious reasons. We could differentiate the function to find the extrema, but that only allows us to find the angle $\varphi$, that I'm not sure what to do with.

There's my problem.

Hang on...

If $\varphi = 45^{\circ}$, then $\sin^2 (\varphi + \frac{\pi}{4}) = 1$ and
\begin{align*}
\Delta E_{kin}
&= \frac{m^2 (u_{m}^2 - \sqrt{2} v_{m} \sin^2 (\theta + \frac{\pi}{4}))}{\sqrt{2} M \sin^2 (\varphi + \frac{\pi}{4})} && |\text{ We assume the sine in the numerator = 0}\\
&= \frac{m^2 u_{m}^2}{\sqrt{2} M}\\
&= \frac{\sqrt{2}m^2 u_{m}^2}{2M} && |\frac{m}{M} = A^{-1}\\
&= \frac{\sqrt{2}m u_{m}^2}{2A}\\
&= \frac{\sqrt{2}E_{kin}}{A}
\end{align*}

This is not exactly what I was hoping to get out of this little juggling session. The expression I'm looking for is $\frac{-4AE_{kin}}{(1+A)^{2}}$. Maybe this is because $\varphi$ being $45^\circ$ gets me the minimum of the function, not the maximum. The sine in the denominator being equal to $1$ seems to indicate this.

 

[PeroK]

There's my problem.

Looking at momentum, you have $m \vec{v} + M \vec{V} = m \vec{u}$.

From this, can you see that $v$ is minimised when $\theta = 0$?

KE is minimised when $v$ is mininised.

 

[TheSodesa]

Looking at momentum, you have $m \vec{v} + M \vec{V} = m \vec{u}$.

From this, can you see that $v$ is minimised when $\theta = 0$?

KE is minimised when $v$ is mininised.

So the particle $m$ would have to run into the other one head on, so that it comes to a halt?

Another thing I just realized is that I haven't even taken relativity into account in the energy equations. Would I need to do that?

 

[PeroK]

So the particle $m$ would have to run into the other one head on, so that it comes to a halt?

Intuitively, if you want to lose KE, you would want a head on collision. If you want to prove this algebraically, I would start by using the law of cosines for momentum:

$M^2V^2 = \dots$

And use conservation of energy to give a quadratic in $v$ involving $\theta$ but without $V$.

Then, use implicit differentiation.

Another thing I just realized is that I haven't even taken relativity into account in the energy equations. Would I need to do that?

Maybe stick to classical mechanics for the time being!

 

[TheSodesa]

Intuitively, if you want to lose KE, you would want a head on collision. If you want to prove this algebraically, I would start by using the law of cosines for momentum:

$M^2V^2 = \dots$

And use conservation of energy to give a quadratic in $v$ involving $\theta$ but without $V$.

Then, use implicit differentiation.
Maybe stick to classical mechanics for the time being!

In this case would $(MV)^2 = (mv)^2 + (mv-MV)^2 - 2(mv)(MV)\cos(\theta + \varphi)$?

 

[PeroK]

In this case would $(MV)^2 = (mv)^2 + (mv-MV)^2 - 2(mv)(MV)\cos(\theta + \varphi)$?

That's not right. Draw a vector diagram where $m \vec{v} + M \vec{V} = m \vec{u}$ and apply the law of cosines to that.

For the record, another approach was to follow your original idea and show that:

$V = \frac{2mu \cos(\phi)}{m+M}$

Then, this is maximised when $\phi = 0$, and this minimises $v$.

(The trick you missed was to analyse the collision in a frame rotated by $\phi$ - but that's not easy to spot.)

But, the alternative using the law of cosines correctly should be easier.

 

[TheSodesa]

That's not right. Draw a vector diagram where $m \vec{v} + M \vec{V} = m \vec{u}$ and apply the law of cosines to that.

For the record, another approach was to follow your original idea and show that:

$V = \frac{2mu \cos(\phi)}{m+M}$

Then, this is maximised when $\phi = 0$, and this minimises $v$.

(The trick you missed was to analyse the collision in a frame rotated by $\phi$ - but that's not easy to spot.)

But, the alternative using the law of cosines correctly should be easier.

Alright.

If the picture I drew is correct:
$$M^2 V^2 = (mu)^2 + (mv)^2 - 2(mu)(mv)cos(\theta)\\ \iff\\ V = \sqrt{ \frac{m^2 (u^2 + v^2 - 2uv\cos(\theta))}{M^2} } \stackrel{\frac{M}{m} = A}{=} \sqrt{ \frac{ (u^2 + v^2 - 2uv\cos(\theta)) }{ A^2 } }\\$$

If we then plug this into the (classical) energy equation:
\begin{align*}
\frac{1}{2}mu^2 &=\frac{1}{2}mv^2 + \frac{1}{2}MV^2\\
&\iff\\
mu^2
&= mv^2 + M \left( \sqrt{ \frac{ (u^2 + v^2 - 2uv\cos\theta) }{ A^2 } } \right)^2\\
&= mv^2 + M \frac{ (u^2 + v^2 - 2uv\cos\theta) }{ A^2 }\\

\end{align*}

Solving this for $v$ nets me:
\begin{align*}
(\frac{M}{A^2} + m)v^2 - \frac{2Mu cos\theta }{A^2}v + (\frac{M}{A^2} - m)u^2 = 0
\end{align*}

Now presumably $\theta$ is a function of $v$, $\theta (v)$. If we differentiate with respect to $v$:
\begin{align*}
\frac{d}{dv} \left( (\frac{M}{A^2} + m)v^2 - \frac{2Mu\cos\theta}{A^2}v + (\frac{M}{A^2} - m)u^2 \right)\\
= 2(\frac{M}{A^2} + m)v - \frac{2Mu}{A^2}(\cos\theta -v \theta ' (v) \sin\theta) = 0
\end{align*}

This looks a bit messy. Just to make sure, I want to solve for $\theta$ in order to find out the minimum of $v$, right?

 

[PeroK]

Alright.

If the picture I drew is correct:
$$M^2 V^2 = (mu)^2 + (mv)^2 - 2(mu)(mv)cos(\theta)\\$$

Yes, but the next step is to get rid of $v$ or $V$. These are dependent on each other; whereas $m, M, u$ are constants. It looks easier to get rid of $V$, using your equation, plus conservation of energy:

$mu^2 = mv^2 + MV^2$

Also, you don't need to calculate KE until the end, as the following are equivalent:

Minimising $v$
Maximising $V$
Maximising the loss in KE of $m$

Do you see that?

 

[TheSodesa]

Yes, but the next step is to get rid of $v$ or $V$. These are dependent on each other; whereas $m, M, u$ are constants. It looks easier to get rid of $V$, using your equation, plus conservation of energy:

$mu^2 = mv^2 + MV^2$

Also, you don't need to calculate KE until the end, as the following are equivalent:

Minimising $v$
Maximising $V$
Maximising the loss in KE of $m$

Do you see that?

I think I see intuitively why those are the same. $v$ getting minimized means most of the momentum must have been transferred to $M$. However, I did get rid of $V$, didn't I? I then made a quadratic in $v$ equal zero and differentiated with respect to $v$ to find the extrema. That got me the ugly expression above:
$$2(\frac{M}{A^2} + m)v - \frac{2Mu}{A^2} \left( \cos\theta -v \theta ' (v) \sin\theta \right) = 0$$

I'm not entirely sure what to do with it. I could try solving for $v$ to get the angle, but there that pesky $\theta '(v)$ in there as well...

 

[PeroK]

I think I see intuitively why those are the same. $v$ getting minimized means most of the momentum must have been transferred to $M$. However, I did get rid of $V$, didn't I? I then made a quadratic in $v$ equal zero and differentiated with respect to $v$ to find the extrema. That got me the ugly expression above:
$$2(\frac{M}{A^2} + m)v - \frac{2Mu}{A^2} \left( \cos\theta -v \theta ' (v) \sin\theta \right) = 0$$

I'm not entirely sure what to do with it. I could try solving for $v$ to get the angle, but there that pesky $\theta '(v)$ in there as well...

That's all too ugly for me! The trick is to back up and use implicit differentiation. In any case, you have $\frac{dv}{d \theta} = 0$ when $v$ is a minimum.

On the other point:

Min KE (m) coincides with min $v$ follows directly from $KE = \frac{1}{2} mv^2$

And min $v$ coincides with max $V$ follows directly from $MV^2 + mv^2 =$ constant.

Or, min KE (m) coincides with max KE (M) directly from energy conservation. This is another reason it must be head-on: that's got to be the way to impart maximum KE to M.

Finally, you need these tricks as ploughing straight ahead without any mathematical finesse will often lead to messy and possibly unsolvable equations.

 

[PeroK]

PS that said, I'm not convinced you weren't supposed to assume a 1D collision in the first place!

 

[TheSodesa]

PS that said, I'm not convinced you weren't supposed to assume a 1D collision in the first place!

Could be, although it is not said in the assignment. However I'm still a bit confused about what I should be differentiating. Wasn't the trick I used in post #10 implicit differentiation? You have two unknown variables, one the function of the other, so you differentiate with respect to the other one, taking into account the chain rule.

I'm confused, because I solved for $V$ in the momentum equation as you told me to (Or was I supposed to do it with the energy equation? Trying not to put words in your mouth.), and then proceeded to construct the quadratic in $v$.

In any case, I need to take a small break. I've been at this for some time now and I'm hungry as well.

 

[PeroK]

Could be, although it is not said in the assignment. However I'm still a bit confused about what I should be differentiating. Wasn't the trick I used in post #10 implicit differentiation? You have two unknown variables, one the function of the other, so you differentiate with respect to the other one, taking into account the chain rule.

I'm confused, because I solved for $V$ in the momentum equation as you told me to (Or was I supposed to do it with the energy equation? Trying not to put words in your mouth.), and then proceeded to construct the quadratic in $v$.

In any case, I need to take a small break. I've been at this for some time now and I'm hungry as well.

Okay. But you need to differentiate with respect to $\theta$.

Also, I'm starting to favour the alternative approach: consider the motion/momentum along the path of M. The velocity of M must depend only on the initial velocity of $m$ along this axis, and that is maximised when $m$ approaches directly along that axis, with no other component. Therefore, the velocity of M is maximised by a direct 1D collision.

No need for any calculus!

 

[kuruman]

PS that said, I'm not convinced you weren't supposed to assume a 1D collision in the first place!

I agree with PeroK. There is no need to consider a 2d collision. Just use the head-on elastic collision equation found here
http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html
You get the desired answer.

 

[TheSodesa]

Okay. But you need to differentiate with respect to $\theta$.

Also, I'm starting to favour the alternative approach: consider the motion/momentum along the path of M. The velocity of M must depend only on the initial velocity of $m$ along this axis, and that is maximised when $m$ approaches directly along that axis, with no other component. Therefore, the velocity of M is maximised by a direct 1D collision.

No need for any calculus!

Yeah, differentiating the energy expression that had $V$ eliminated by $\theta$ didn't do much to make it simpler. What I got was:

$$2 \left( \frac{M}{A^2} + m \right) v \frac{dv}{d\theta} - \frac{2Mu}{A^2} \frac{dv}{d\theta}\cos\theta + \frac{2Mu}{A^2}v\sin\theta = 0$$

No way am I solving for either $\theta$ or $v$ out of that.

 

[TheSodesa]

I agree with PeroK. There is no need to consider a 2d collision. Just use the head-on elastic collision equation found here
http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html
You get the desired answer.

I'll give it a shot. Alhough I was kind of exited to try to derive the 2D expression. This might actually be fun if I didn't have deadlines to adhere to.

 

[kuruman]

You could have saved yourself some time if you showed first that the 1d case, i.e. the case when the "after" velocities of the two particles are collinear, provides a larger ΔE_kin than the non-collinear case.

 

[TheSodesa]

You could have saved yourself some time if you showed first that the 1d case, i.e. the case when the "after" velocities of the two particles are collinear, provides a larger ΔE_kin than the non-collinear case.

Yeah, well... The instructor is probably going to be expecting that tomorrow morning, but I've spent enough time as it is on this. And yeah, the answer came pretty much automatically as soon as you assumed a head-on collision.

Thanks for the help.

 

[PeroK]

Yeah, differentiating the energy expression that had $V$ eliminated by $\theta$ didn't do much to make it simpler. What I got was:

$$2 \left( \frac{M}{A^2} + m \right) v \frac{dv}{d\theta} - \frac{2Mu}{A^2} \frac{dv}{d\theta}\cos\theta + \frac{2Mu}{A^2}v\sin\theta = 0$$

No way am I solving for either $\theta$ or $v$ out of that.

What you missed was that if $\frac{dv}{d \theta} = 0$ then you get $\sin \theta = 0$ and $\theta = 0, \pi$

 

[TheSodesa]

What you missed was that if $\frac{dv}{d \theta} = 0$ then you get $\sin \theta = 0$ and $\theta = 0, \pi$

D'oh!

 

[Ray Vickson]

Homework Statement

Show that the largest possible change in the kinetic energy , $\Delta E_{kin}$, of a particle of mass $m$ running into another particle of mass $M$ at rest in the lab coordinate system is
$$\Delta E_{kin} = \frac{-4AE_{kin}}{(1+A)^{2}}$$, where $A = \frac{M}{m}$.

Calculate the limit $$\lim_{M \to m} \Delta E_{kin}$$ and plot $\Delta E_{kin}(A)$.

Homework Equations

Conservation of energy.
Conservation of momentum.
Rutherford's experiment?

The Attempt at a Solution

I'm going to assume, that both energy and momentum are conserved.

Energy:
$$E_{mi} + \stackrel{0}{E_{Mi}} = E_{mf} + E_{Mf}\\ \iff \\ |\Delta E_{kin}| = |E_{mf} - E_{mi}| = |-E_{Mf}| = \frac{1}{2}Mv^{2}$$
View attachment 106518

***** stuff trimmed****

I feel like I've gone down a wrong path. Help, please?

Such problems are almost always more easily analyzed by first transforming to the so-called CM (center of momentum) frame; this is the moving frame in which the initial total momentum equals zero. For a perfectly elastic 2-particle collision in such a frame the final kinetic energies of each particle separately remain unchanged; that is, $\text{KE}_{\text{intial}} = \text{KE}_{\text{final}}$ holds separately for particles 1 and 2. If $(V,0)$ is the initial lab-frame velocity of the moving particle (particle 1), its CM-frame velocity is $(V-v, 0)$ for some easily-computed $v$. Since its speed remains the same in the CM frame after the collision, its post-collision velocity in the CM-frame is $((V-v) c, (V-v) s)$, where $c = \cos \theta$, $s = \sin \theta$ and $\theta$ is the CM-frame deflection angle in the collision. The final lab-frame velocity of particle 1 is $((V-v)c + v, (V-v)s)$, so you can work out its post-collision lab-frame KE fairly easily in terms of $V,v,c$ and $s= \sqrt{1-c^2}$. Finding the maximum change in lab-frame KE as a function of $c$ in the interval $-1 \leq c \leq 1$ is quite straightforward.

For proof of unchanged kinetic energies in the CM frame, see, eg.,
http://www.sfu.ca/phys/120/1131/Solutions/2D_collision_cm.html