Elastic collision between ball and block

[dauto]

$\dfrac{dL}{L} = \dfrac{-2 \cos \lambda n}{\cos \lambda n + 2 \sin \lambda n} dn \\ \lambda = 2 \sqrt{\mu}$

I'm getting
$\dfrac{dL}{L} = \dfrac{-2 \cos \lambda n}{\cos \lambda n + \mu^{-1/2} \sin \lambda n} dn$

 

[utkarshakash]

I'm getting
$\dfrac{dL}{L} = \dfrac{-2 \cos \lambda n}{\cos \lambda n + \mu^{-1/2} \sin \lambda n} dn$

I'm posting my solution step-by-step.

$\dfrac{dL}{L} = \dfrac{-2 Mv_0 \cos \lambda n}{Mv_0 \cos \lambda n + (2 mv_0/ \mu) \sin \lambda n} dn$

If I substitute the value of μ, I get my equation. What's wrong with it?

 

[dauto]

I don't think that's correct either. I'll have to go now. I'll have more time later today

 

[utkarshakash]

I don't think that's correct either. I'll have to go now. I'll have more time later today

OK. Please take your time.

 

[dauto]

OK. Here's what I got

L_i+1=L_i(1−2V_i/(V_ii+v))

(1/L) dL/dn = -2V/(V+v)

(1/L) dL/dn = -2P/(P+p/μ)

(1/L) dL/dn = -2cos(λn)/[cos(λn)+μ^1/2sin(λn)/μ]

(1/L) dL/dn = -2cos(λn)/[cos(λn)+μ^-1/2sin(λn)],

where I used

P=P₀cos(λn)

and

p=μ^1/2P₀sin(λn)

 

[utkarshakash]

OK. Here's what I got

L_i+1=L_i(1−2V_i/(V_ii+v))

(1/L) dL/dn = -2V/(V+v)

(1/L) dL/dn = -2P/(P+p/μ)

(1/L) dL/dn = -2cos(λn)/[cos(λn)+μ^1/2sin(λn)/μ]

(1/L) dL/dn = -2cos(λn)/[cos(λn)+μ^-1/2sin(λn)],

where I used

P=P₀cos(λn)

and

p=μ^1/2P₀sin(λn)

Looks like there was a mistake in my equation for p(n). OK, how do you integrate it then? Substitutions aren't working here. What else should I try?

 

[dauto]

Here's the trick

$$\frac{cos\theta}{cos\theta+A sin\theta} = \frac{cos\theta}{cos\theta+A sin\theta} +\xi -\xi,$$
where $\xi$ is to be chosen later to our own convenience.
$$\frac{cos\theta d\theta}{cos\theta+A sin\theta} = \frac{[(1+\xi)cos\theta+\xi A sin\theta]d\theta}{cos\theta+A sin\theta} -\xi d\theta,$$
We are planning on making the change of variables
$u=cos\theta+A sin\theta,$ and $du=(-sin\theta+A cos\theta)d\theta,$
So we want to chose $\xi$ in such a way that the numerator $[(1+\xi)cos\theta+\xi A sin\theta]d\theta$ is proportional to $du$
$$[(1+\xi)cos\theta+\xi A sin\theta]d\theta = \eta du = \eta (-sin\theta+A cos\theta)d\theta$$ which gives us two equations
$$(1+\xi) = \eta A$$ and
$$\xi A = - \eta$$ The set is easily solved to $\xi = -\frac{1}{1+A^2}$ and $\eta=\frac{A}{1+A^2}$. Now that we know what $\xi$ and $\eta$ are, we have
$$\frac{cos\theta d\theta}{cos\theta+A sin\theta} = \frac{[(1+\xi)cos\theta+\xi A sin\theta]d\theta}{cos\theta+A sin\theta} -\xi d\theta = \frac{\eta du}{u} - \xi d\theta.$$
Integration now is a trivial matter.

 

[utkarshakash]

Here's the trick

$$\frac{cos\theta}{cos\theta+A sin\theta} = \frac{cos\theta}{cos\theta+A sin\theta} +\xi -\xi,$$
where $\xi$ is to be chosen later to our own convenience.
$$\frac{cos\theta d\theta}{cos\theta+A sin\theta} = \frac{[(1+\xi)cos\theta+\xi A sin\theta]d\theta}{cos\theta+A sin\theta} -\xi d\theta,$$
We are planning on making the change of variables
$u=cos\theta+A sin\theta,$ and $du=(-sin\theta+A cos\theta)d\theta,$
So we want to chose $\xi$ in such a way that the numerator $[(1+\xi)cos\theta+\xi A sin\theta]d\theta$ is proportional to $du$
$$[(1+\xi)cos\theta+\xi A sin\theta]d\theta = \eta du = \eta (-sin\theta+A cos\theta)d\theta$$ which gives us two equations
$$(1+\xi) = \eta A$$ and
$$\xi A = - \eta$$ The set is easily solved to $\xi = -\frac{1}{1+A^2}$ and $\eta=\frac{A}{1+A^2}$. Now that we know what $\xi$ and $\eta$ are, we have
$$\frac{cos\theta d\theta}{cos\theta+A sin\theta} = \frac{[(1+\xi)cos\theta+\xi A sin\theta]d\theta}{cos\theta+A sin\theta} -\xi d\theta = \frac{\eta du}{u} - \xi d\theta.$$
Integration now is a trivial matter.

Integrating the equation I get

$\log L = \left( \dfrac{A}{1+A^2} \log (\cos \theta + A \sin \theta) + \dfrac{\theta}{1+A^2} \right) (-2) + \log L_0$

Substituting $n=\frac{\pi}{4 \sqrt{\mu}}$

$\log L = \dfrac{\sqrt{\mu}}{1+\mu} \log \mu - \dfrac{\mu \pi}{1+\mu} + \log L_0$

But this is not even close to the correct answer. Are you getting the same result?

 

[dauto]

Integrating the equation I get

$\log L = \left( \dfrac{A}{1+A^2} \log (\cos \theta + A \sin \theta) + \dfrac{\theta}{1+A^2} \right) (-2) + \log L_0$

Substituting $n=\frac{\pi}{4 \sqrt{\mu}}$

$\log L = \dfrac{\sqrt{\mu}}{1+\mu} \log \mu - \dfrac{\mu \pi}{1+\mu} + \log L_0$

But this is not even close to the correct answer. Are you getting the same result?

I got

$$log (L/L_0) =-\left[\frac{log [\mu^{1/2}cos\theta +sin\theta] + \mu^{1/2}\theta}{1+\mu} \right]_{\theta=0}^{\pi/2}$$

What do you think the correct answer should be?

 

[utkarshakash]

I got

$$log (L/L_0) =-\left[\frac{log [\mu^{1/2}cos\theta +sin\theta] + \mu^{1/2}\theta}{1+\mu} \right]_{\theta=0}^{\pi/2}$$

What do you think the correct answer should be?

The correct answer is L₀ √μ. Your equation seems to give a different answer.

 

[dauto]

The correct answer is L₀ √μ. Your equation seems to give a different answer.

Look closely. There are a couple of terms that are negligible. One of the terms in the numerator may be dropped. And the term μ^1/2 in the numerator is also negligible. It simplifies as

$$log (L/L_0) =-\left[log [\mu^{1/2}cos\theta +sin\theta] \right]_{\theta=0}^{\pi/2}$$

Do you see it now?

 

[TSny]

Here is another approach to part (a). In post #3 you have $$L_{i+1} = \frac{v_i-V_i}{v_i+V_i}L_i$$
So, in particular,

$$L_2 = \frac{v_1-V_1}{v_1+V_1}L_1$$ and $$L_3= \frac{v_2-V_2}{v_2+V_2}L_2 = \frac{v_2-V_2}{v_2+V_2} \frac{v_1-V_1}{v_1+V_1}L_1$$

For an elastic collision, the relative speed of approach before the collision equals the relative speed of separation after the collision.

For example, at the first collision

$V_0 =v1-V_1$, which means $L_2 = \frac{V_0}{v_1+V_1}L_1$

For the second collision

$v_1+V_1=v2-V_2$ which can be used to show $L_3= \frac{V_0}{v_2+V_2}L_1$.

Generalize this to write $L_N$ in terms of $L_1, v_{N-1}, V_{N-1}$ and $V_0$.

See if you can apply this to the case where $\small N -1$ is the index of the collision for which $\small M$ essentially comes to rest.

 

[haruspex]

This is an intriguing question. I wasn't sure what approximation would be reasonable, so I set out to solve it exactly to start with - looks like I did what TSny suggests. I took u_n, v_n to be the speeds, to the right, of M, m respectively, just before each time they collide (so the v_n are negative). I got $\left[\stackrel{u_n}{v_n}\right] = \left(\frac{M-m}{M+m}\right)^nPD^nP^{-1}\left[^u_0\right]$, where u is the initial speed of M, and PDP^-1 is the diagonalisation of $A = \left[\begin{array}{cc} 1 & b \\ a & 1 \end{array} \right]$, $a = \frac{2m}{M-m}$, $b = -\frac{2M}{M-m}$. For u_n = 0, this gave
$(1+\sqrt{ab})^n = -(1-\sqrt{ab})^n$. (Note that b is negative.) I.e. $n = \frac{i \pi}{\ln(\frac{1+i\mu}{1-i\mu})}$, where $\mu = \frac{2\sqrt{Mm}}{M-m}$. For M>>m this reduces to $n = \frac{\pi}{4}\sqrt{\frac Mm}$.
I'm sure this was all a lot harder than picking the right approximation up front, but it was interesting.
Applying the same approach to the distance, I got this dreadful formula, where x_n is the remaining distance at the nth collision:
$x_{n+1} = x_n\frac{(1+i\rho)^n(1-i\alpha)-(1-i\rho)^n(1+i\alpha)}{(1+i\rho)^n(1+i\alpha)-(1-i\rho)^n(1-i\alpha)}$, where $\rho = 2\frac{\sqrt{Mm}}{M-m}, \alpha = \sqrt{\frac mM}$.
For M>>m, $\rho ≈ 2\alpha << 1$, reducing it to $\frac{\tan(2n\alpha)-\alpha}{\tan(2n\alpha)+\alpha}$, further approximating to $\frac{2n-1}{2n+1}$. Thus x_n ≈ x₁/(2n-1). Does that give the right answer?

 

[utkarshakash]

Look closely. There are a couple of terms that are negligible. One of the terms in the numerator may be dropped. And the term μ^1/2 in the numerator is also negligible. It simplifies as

$$log (L/L_0) =-\left[log [\mu^{1/2}cos\theta +sin\theta] \right]_{\theta=0}^{\pi/2}$$

Do you see it now?

Yes. I got the correct answer. Thanks for helping !

 

[utkarshakash]

This is an intriguing question. I wasn't sure what approximation would be reasonable, so I set out to solve it exactly to start with - looks like I did what TSny suggests. I took u_n, v_n to be the speeds, to the right, of M, m respectively, just before each time they collide (so the v_n are negative). I got $\left[\stackrel{u_n}{v_n}\right] = \left(\frac{M-m}{M+m}\right)^nPD^nP^{-1}\left[^u_0\right]$, where u is the initial speed of M, and PDP^-1 is the diagonalisation of $A = \left[\begin{array}{cc} 1 & b \\ a & 1 \end{array} \right]$, $a = \frac{2m}{M-m}$, $b = -\frac{2M}{M-m}$. For u_n = 0, this gave
$(1+\sqrt{ab})^n = -(1-\sqrt{ab})^n$. (Note that b is negative.) I.e. $n = \frac{i \pi}{\ln(\frac{1+i\mu}{1-i\mu})}$, where $\mu = \frac{2\sqrt{Mm}}{M-m}$. For M>>m this reduces to $n = \frac{\pi}{4}\sqrt{\frac Mm}$.
I'm sure this was all a lot harder than picking the right approximation up front, but it was interesting.
Applying the same approach to the distance, I got this dreadful formula, where x_n is the remaining distance at the nth collision:
$x_{n+1} = x_n\frac{(1+i\rho)^n(1-i\alpha)-(1-i\rho)^n(1+i\alpha)}{(1+i\rho)^n(1+i\alpha)-(1-i\rho)^n(1-i\alpha)}$, where $\rho = 2\frac{\sqrt{Mm}}{M-m}, \alpha = \sqrt{\frac mM}$.
For M>>m, $\rho ≈ 2\alpha << 1$, reducing it to $\frac{\tan(2n\alpha)-\alpha}{\tan(2n\alpha)+\alpha}$, further approximating to $\frac{2n-1}{2n+1}$. Thus x_n ≈ x₁/(2n-1). Does that give the right answer?

This was indeed a difficult problem, much difficult than what I assumed. Lots of approximations are required. Your method seems complicated to me. But, anyways, thanks for posting your solution. Knowing different methods to solve a problem is certainly beneficial.

 

[TSny]

Using the fact that the relative speed of approach equals the relative speed of separation at each collision, you can iterate $L_{i+1} = \frac{v_i-V_i}{v_i+V_i}L_i$ to get

$L_{N+1} = \frac{V_0}{v_N+V_N}L_1$.

Suppose $\small M$ essentially comes to rest at the $\small N^{th}$ collision. Then to a good approximation

$L_{N+1} = L_{N} = L_{closest}$ and $V_{N}=0$. So,

$L_{closest} = \frac{V_0}{v_N}L_1$. Conservation of energy implies $\frac{1}{2}MV_0^2 = \frac{1}{2}mv_N^2$. So, $L_{closest} = \sqrt{\frac{m}{M}}L_1$

 

[haruspex]

Using the fact that the relative speed of approach equals the relative speed of separation at each collision, you can iterate $L_{i+1} = \frac{v_i-V_i}{v_i+V_i}L_i$ to get

$L_{N+1} = \frac{V_0}{v_N+V_N}L_1$.

Suppose $\small M$ essentially comes to rest at the $\small N^{th}$ collision. Then to a good approximation

$L_{N+1} = L_{N} = L_{closest}$ and $V_{N}=0$. So,

$L_{closest} = \frac{V_0}{v_N}L_1$. Conservation of energy implies $\frac{1}{2}MV_0^2 = \frac{1}{2}mv_N^2$. So, $L_{closest} = \sqrt{\frac{m}{M}}L_1$

Brilliant.

 

[utkarshakash]

Using the fact that the relative speed of approach equals the relative speed of separation at each collision, you can iterate $L_{i+1} = \frac{v_i-V_i}{v_i+V_i}L_i$ to get

$L_{N+1} = \frac{V_0}{v_N+V_N}L_1$.

Suppose $\small M$ essentially comes to rest at the $\small N^{th}$ collision. Then to a good approximation

$L_{N+1} = L_{N} = L_{closest}$ and $V_{N}=0$. So,

$L_{closest} = \frac{V_0}{v_N}L_1$. Conservation of energy implies $\frac{1}{2}MV_0^2 = \frac{1}{2}mv_N^2$. So, $L_{closest} = \sqrt{\frac{m}{M}}L_1$

Nice.

 

[bobie]

$L_{closest} = \frac{V_0}{v_N}L_1$. Conservation of energy implies $\frac{1}{2}MV_0^2 = \frac{1}{2}mv_N^2$. So, $L_{closest} = \sqrt{\frac{m}{M}}L_1$

Can you tell me if that formula allows for a case when M crashes into the wall?
- isn't V/L a relevant factor?

Thanks .

 

[TSny]

Can you tell me if that formula allows for a case when M crashes into the wall?
- isn't V/L a relevant factor?

Thanks .

It's interesting that M never crashes into the wall no matter how large the initial speed of M or how small the mass of the ball!

The distance of closest approach depends only on the initial distance and the ratio of the masses.

Changing the initial speed V of the block just changes the time at which the block reaches the distance of closest approach. If you took a video of the motion, then doubling the initial speed would be equivalent to speeding up the video by a factor of 2.

(Here I'm assuming Newtonian mechanics. If the mass of the ball is extremely small compared to M and if the initial speed of the block is large enough, then the ball could attain relativistic speeds even if V is not relativistic. It might be interesting to think about how that would modify things.)

 

[bobie]

Changing the initial speed V of the block just changes the time at which the block reaches the distance of closest approach.

Thanks for your response,TSny.

. So, $L_{closest} = \sqrt{\frac{m}{M}}L_1$

If M/m = 10⁶, should M stop at 1/1000 L?
suppose V=L= 1m, and the size of the ball is >1 mm, won't the block crash into the ball and the wall? but does really M reach that far?

 

[TSny]

If M/m = 10⁶, should M stop at 1/1000 L?
suppose V=L= 1m, and the size of the ball is >1 mm, won't the block crash into the ball and the wall?

Yes, you are right. If the ball has a finite size greater than the formula's minimum distance of approach then things get complicated. For that situation I guess you would need to add assumptions about the elastic properties of the ball. Ugh. If the ball doesn't "give" at all, then of course the closest the block could make it to the wall would be the diameter of the ball.

I always had in mind a point mass for the ball - an unrealistic assumption to go along with the perfect elastic collisions! Even under these ideal assumptions, the problem is interesting to me.

 

[bobie]

, the problem is interesting to me.

It is really an intriguing problem, but probably the formula needs to be refined.
Please check if what I found is correct :

when M/m = 3 , L_closest = 1/√4 = 1/2 L
when M/m ≈ ∞ L_closest = 1/√49 = , in our example: ≈ .14m
ather the collision N₀ (at L = 1) V₁ = 1 ( and stays practically the same) and

Thanks for your attention

 

[TSny]

It is really an intriguing problem, but probably the formula needs to be refined.
Please check if what I found is correct :

when M/m = 3 , L_closest = 1/√4 = 1/2 L
when M/m ≈ ∞ L_closest = 1/√49 = , in our example: ≈ .14m[/QUOUTE]

ather the collision N₀ (at L = 1) V₁ = 1 ( and stays practically the same) and v₁ = 2 (and grows by 2v+1)
N

Thanks

Can you show how you got these results?

 

[bobie]

Can you show how you got these results?

v₁ = 2 (and grows by 2v+1)
N₁ takes place at L₁ = 1/3 .333m
N₂ at L = .2222m (L₁- 1/3)
N₃ at L = .185 (L2-1/6) and then it decreases very slowly .1697, .1626,.159,.157,.1568,.1564,.1562,.15609 etc ...
probably it won't even be less than .1559 ≈ 1/√41 ,
if M/m varies from
3 to ∞, L should vary from
1/2 to 1/6.4

on my pocket calculator I managed by: 1/3, and then: ANS-ANS/3*2ⁿ up to n=30

you surely can work it out by your computer program

 

[TSny]

I'm not following your statement that v grows by 2v+1. If m << M, then the values of v for the first few collisions are approximately 2, 4, 6, 8, 10, etc. if the block was initially moving at 1 unit of speed. And I don't understand how you are getting some of your other numbers.

The following table shows what I get for the exact closest distance of approach and the approximate distance of closest approach according to L_closest ≈ √(m/M), assuming the first collision occurs at 1 unit of distance from the wall:

$$ \frac{M}{m} = 3 \;\;\;\;\;\;\; L^{closest}_{exact} = 0.50000 \;\;\;\;\;\; L^{closest}_{approx} = 0.577$$ $$ \frac{M}{m} = 25 \;\;\;\;\;\; L^{closest}_{exact} = 0.19967 \;\;\;\;\;\; L^{closest}_{approx} = 0.200$$ $$ \frac{M}{m} = 100 \;\;\;\; L^{closest}_{exact}= 0.09979 \;\;\;\;\;\; L^{closest}_{approx} = 0.100$$ $$ \frac{M}{m} = 10000 \;\;\; L^{closest}_{exact}= 0.0099995 \;\;\; L^{closest}_{approx} = 0.010$$

 

[bobie]

I'm not following your statement,$$ \frac{M}{m} = 3 \;\;\;\;\;\;\; L^{closest}_{exact} = 0.50000 \;\;\;\;\;\; L^{closest}_{approx} = 0.577$$

The KE parameter is not reliable as it's unlikely that M ever stops, when M reaches its closest distance, if still has a lot of positive Ke that at next clash is turned into negative
If M/m =3 the closest L is 1/2, L_closest = .5 m after the initial collision, even if M has still 1/2 V_{0 and therefore} 1/4 of its KE. At the first real clash M bounces back.
Isn' that so?
as to 2v+1 you are right, it's v+2!

So , we must conclude that if a 10-ton-steel-slab is running at 100 Km/h and meets a 1-gr/3cm-pingpongball on its way at 10 m from a wall , the ball will not be crushed but it will stop the slab

 

[TSny]

The KE parameter is not reliable as it's unlikely that M ever stops, when M reaches its closest distance, if still has a lot of positive Ke that at next clash is turned into negative
If M/m =3 the closest L is 1/2, L_closest = .5 m after the initial collision, even if M has still 1/2 V_{0 and therefore} 1/4 of its KE. At the first real clash M bounces back.
Isn' that so?

Yes, that's right. But when m << M, the block will be approximately at rest at the point of closest approach and essentially all of the KE will be in the ball.

So , we must conclude that if a 10-ton-steel-slab is running at 100 Km/h and meets a 1-gr/3cm-pingpongball on its way at 10 m from a wall , the ball will not be crushed but it will stop the slab

Let's assume it's 10 metric tons, so M/m = 10⁷. If the first collision takes place 100 m from the wall, then the closest the slab will make it to the wall is about 3.16 cm from the wall. The initial speed of the slab does not affect this result. (If the initial distance is only 10 m, then the ball will get crushed.)

If the initial speed of the slab is 100 km/h, then at the distance of closest approach, the ball will have a speed of about 316,000 km/h.

 

[bobie]

Thanks for your attention, TSny.