Mighty Mosquito and the Runaway Train

Yes, great problem , how do you think of such things Sir. Here's my way of doing it.

At first sight it may seem that initial velocity of the train is required, but it is not so.

Hello

I have diagram showing the velocities of mosquito and train just before their $(n+1)$ th collision and just after $(n+1)$ th collision.

I am taking ${a}_{n}$ =Velocity of train after $n$ -collisions

Also ${b}_{n}$ = Velocity of mosquito after $n$ -collisions

${m}_{1}$ =Mass of train , ${m}_{2}$ =Mass of mosquito

For the $(n+1)th$ collision we have :

By conservation of linear momentum.

${m}_{1}{a}_{n}-{m}_{2}{b}_{n}={m}_{1}{a}_{n+1}+{m}_{2}{b}_{n+1}$

Also for an elastic collision $e=1$ hence we have :

${a}_{n}+{b}_{n}={b}_{n+1}-{a}_{n+1}$

Solving both the equations we get :

${a}_{n+1}=\frac{({m}_{1}-{m}_{2}){a}_{n}-2{m}_{2}{b}_{n}}{{m}_{1}+{m}_{2}}$

${b}_{n+1}=\frac{({m}_{1}-{m}_{2}){b}_{n}+2{m}_{1}{a}_{n}}{{m}_{1}+{m}_{2}}$

We take $\frac{{m}_{1}}{{m}_{2}}=k$ , The above equations simplify into :

$\large { a }_{ n+1 }=\frac { (k-1){ a }_{ n }-2{ b }_{ n } }{ k+1 }$

$\large { b }_{ n+1 }=\frac { 2k{ a }_{ n }+(k-1){ b }_{ n } }{ k+1 }$

Dividing the equations we get :

$\large \frac { { b }_{ n+1 } }{ { a }_{ n+1 } } =\frac { 2k{ a }_{ n }+(k-1){ b }_{ n } }{ (k-1){ a }_{ n }-2{ b }_{ n } }$

Here comes the main step we take $\large \frac { { { b }_{ n } } }{ { a }_{ n } } ={ r }_{ n }$

Our above equation turns into :

$\large r_{ n+1 }=\frac { 2k+(k-1){ r }_{ n } }{ (k-1)-2{ r }_{ n } }$

So we have formed our recurrence relation.

We know that ${r}_{0}=\frac{{b}_{0}}{{a}_{0}} = 0$ Since initially the velocity of the mosquito was zero.

To solve the recurrence we have to take :

${r}_{n}=\sqrt{k}tan({\theta}_{n})$ and observe that :

$\large r_{ n+1 }=\sqrt { k } (\frac { \frac { 2\sqrt { k } }{ k-1 } +tan({ \theta }_{ n }) }{ 1-\frac { 2\sqrt { k } }{ k-1 } tan({ \theta }_{ n }) } )=\sqrt { k } tan({ \theta }_{ n }+{ tan }^{ -1 }(\frac { 2\sqrt { k } }{ k-1 } ))$

Also since ${r}_{0}=0$ we have our explicit formulae for expressing ${r}_{n}$ as :

$r_{ n }=\sqrt { k } tan(n{ tan }^{ -1 }(\frac { 2\sqrt { k } }{ k-1 } ))$

The mosquito's job would be done if the train's speed is greater that mosquito's both going in same direction.

Mathematically I should say that :

${r}_{n}>-1$ for some $n$ , applying this condition in our formula we get :

$\large \sqrt { k } tan(n{ tan }^{ -1 }(\frac { 2\sqrt { k } }{ k-1 } ))>-1$

$\large \Rightarrow n>\frac { \pi -{ tan }^{ -1 }(\frac { 1 }{ \sqrt { k } } )) }{ { tan }^{ -1 }(\frac { 2\sqrt { k } }{ k-1 } )) }$

Putting the value of $k={10}^{12}$ we get :

$n>1570795.8$

Hence we get :

$n=1570796$ , since $n$ is an integer .

Also for every collision to the train it collides once to the dead end.

Hence total number of collisions is given by :

$\boxed{X=3141592}$

And yes the answer has the first 7 digits of ${\pi}$

Here I have used exact values but I can approximate it to.

We had $n>\frac { \pi -{ tan }^{ -1 }(\frac { 1 }{ \sqrt { k } } )) }{ { tan }^{ -1 }(\frac { 2\sqrt { k } }{ k-1 } )) }$

Approximating it we get :

$n\approx \frac { \pi }{ 2 } \sqrt { k }$

If mass ratio is ${10}^{2x}$ then we have total number of collisions as :

$=\left\lfloor \pi .{ 10 }^{ x } \right\rfloor$

which as michael pointed out is a great thing to observe.

Ronak, thank you very much for your excellent explanation of how it's worked out. This is one of my favorite oddities in math, given a mass ratio of ${ 10 }^{\displaystyle 2n }$ , the number of collisions is exactly the first $n+1$ digits of $\pi$ , beginning with $3$ for a mass ratio of $1$ , i.e., where $n=0$ . Specifically, it works out to

$Floor\left\lfloor \dfrac { \pi }{ ArcTan\left( { 10 }^{\displaystyle -n } \right) } \right\rfloor$

In rare cases, the last digit is off by $1$ , as for example with $n=0$ .

Maybe I should post a note about a full explanation of this oddity, someday.

Something like a million of Pi.

[Original formula solved included a not catch for elastic collision.]

m1 v1 + m2 v2 = m1 u1 + m2 u2 {Conservation of momentum with no external force against.}

m1 v1^2 + m2 v2^2 = m1 u1^2 + m2 u2^2 {Conservation of energy with no lost of energy.}

Generally, with v1 and v2 as only unknowns to be found, via solving quadratic equation:

v1 = (m1 u1 + m2 u2 - | m2 u1 – m2 u2 |)/ (m1 + m2) {Determined answer from quadratic roots.}

v2 = (m1 u1 + m2 u2 + | m1 u1 – m1 u2 |)/ (m1 + m2) {Determined answer from quadratic roots.}

Since | -x | = | x |, and | x – y | = | y – x |, while | x | = x for x >=0 or | x | = - x for x <0, | x – y | = | y – x | is particularly risky to remain as a form of expression.

Therefore, rewrite the above formula using real axis convention for either or both direction(s):

For u1 >= u2, {Benefited from original negative and positive.}

v1 = (m1 u1 + m2 u2 – (m2 u1 – m2 u2))/ (m1 + m2) = [u1 (m1 – m2) + 2 m2 u2)]/ (m1 + m2)

v2 = (m1 u1 + m2 u2 + (m1 u1 – m1 u2))/ (m1 + m2) = [2 m1 u1 - u2 (m1 – m2)]/ (m1 + m2)

For u1 < u2, {Explained why not positive and negative.}

v1 = (m1 u1 + m2 u2 – (- m2 u1 + m2 u2))/ (m1 + m2) = u1

v2 = (m1 u1 + m2 u2 + (- m1 u1 + m1 u2))/ (m1 + m2) = u2

Specifically, overall momentum is not conserved once external force changed u2 before each collision (as from the passage, it does not include impact with the dead end; but it meant on answer.) between train and mosquito. In a context of ratios only, initial speed of train and mass of mosquito are assigned 1. Replace u2 with -u2:

v1 = [u1 (m1 – m2) - 2 m2 u2)]/ (m1 + m2)

v2 = [2 m1 u1 + u2 (m1 – m2)]/ (m1 + m2)

Gradual change of what happen when the train knock onto the deed end is being transferred to the mosquito for a detail understanding about the mechanism occurs for such impact. The mosquito represents impulses of tiny mass with high speeds. Using computing loop to substitute vi to ui repeatedly, we can observe by counting to the 7 digits for an attention. Using Turbo Pascal, V[flip] of EXTENDED type array and flip of BOOLEAN type can access previous value conveniently for a catch of wanted change.

The mosquito gained a speed of nearly twice the speed of train initially after the first collision until less than one time after 1570796th collision, which is bounced back from 1570796th knock onto the dead end. Since then, the train’s speed is greater than the mosquito’s speed both in negative direction finally, with u1 < u2 as greater magnitude in negative means smaller. Against <> exerted. u1 is always at left side of u2. Mosquito posses -0.653587 times of initial train speed eventually while the train posses -0.99999? times, not certain because of accumulation of computing errors.

Finally, collisions here do meant for all, twice as 3141592, confident because of Pi.