balls in a semi-circle

okie...tx.

Ok. I am going to post a somewhat incompletesolution.

Let,initially the ball be coming horizontally with velocity v₀.
Now after first collision, the mass M/N gets deflected at an angle θ with horizintal[in the below side].
consequently to hit the next ball let the ball of mass m move towards second M/N ball.
As N→∞, We may assume that, the mass m goes off at an angle π/N with horizontal[at upper side].
with velocity v₁
So,

mv_0=\frac{MV_1 Cos \theta}{N} + mv_1 cos(\frac{\pi}{2N})
and
mv_1 sin(\frac{\pi}{2N})= \frac{MV_1 sin \theta}{2N}

Putting value of V_1 from second equation in first,
we have
v_1=\frac{v_0}{Sin(\frac{\pi}{2N}cot\theta)+cos(\frac{\pi}{2N})}

clearly, as N→∞, in all collisions, m moves tangentially to "attack" the next.
Thus from above recurrence

replacing v_0 by v_n and v_1 by v_{n+1},
v_n=v_0 (Cosec(\theta+\frac{\pi}{2N})sin(\theta))^n

So, we have v_N=final speed from above by putting n=N.
v_N=v_0 (Cosec(\theta+\frac{\pi}{2N})sin(\theta))^N
As N→∞, this thing goes to
e^{-\0.5 pi cot{\theta}}v_0.
Now as sin^{-1}\frac{M}{Nm} is maximum angle of deviation[as Rohan said],i need to prove min{cot θ}=2, any suggestions...

for deriving the results one has to know a basic thing =

if a ball of mass m collides with another stationary ball of mass M then the maximum angle of deviation of the ball of mass m is sin^-1M/m

now here we have that if the seperation between two adjacent balls is θ then the angle of deviation for the colliding ball is θ/2 from figure
(sorry i am not able to upload my figure)

further we know the maximum angle of deviation of m after colliding with each M/N is sin^-1M/Nm ---- (i)
if the ratio M/m decreases from a particular value then the angle goes less than θ/2 and hence the ball fails to go to the next one and hit it obliquely as hit the previous one

so in the limiting condition θ/2=sin^-1M/Nm

further the total deviation from all collisions=Nθ/2 as it hits each one similarly after the first one

further when it emerges from the upper end it should face left hence the total angle of deviation=Nθ/2=π

putting the values and applying L hospital rule to equation (i)

we get M/m=Ï€

further for the second part ..

for the first collision let the intitial velocity of mass m be v and let us take our x axis along that direction

after deflection the mass m is at an angle θ/2 with horizonal and the mass M/N is at β downwards with horizontal

conserving momentum

mv=mv₂cosθ/2 + M/Nv₃cosβ
and

mv₂sinθ/2=M/Nv₃sinβ
by conservation of energy
mv²/2=mv₂²/2+(M/2N)(v₃)²

solving the above equations we get

v2=v√(1-π/N)/(1+π/N)

after N collisions

vf=v((1-Ï€/N)/(1+Ï€/N))^N/2

we see that it is of the form 1^∞

hence this is equal to e^{(-2Ï€/N*N/2)/(1+Ï€/N)}

=e^-Ï€
as N->∞ 1/N→0

thus the answers

Sky : "bounces (elastically)"

this implies e = 1 naa.....

Or is it dat elastically dusnt mean perfectly elastic

@ritika... yep... i've done it using limits... dunno if any other method exists...!!!

let initial speed of m be 'v'

As 'm' collides wid d first ball, its velocity reduces to v'=(mN-M/mN+M).v

After second collision, it will be v''= (mN-M/mN+M)².v

after Nth collision, it will b v^N=(mN-M/mN+M)^N.v

now as N tends to infinity... V_f = e^-2M/m.v

=> v_f / v = e^-2M/m ............ (ii)

but i couldn't get d first one...!!!

thinking... [12]

cant get the question :( how can the ball of mass m come to the place where the arrow is placed in the figure

:-)