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N identical balls lie equally spaced in a semicircle on a frictionless horizontal table, as shown. The total mass of these balls is M. Another ball of mass m approaches the semicircle from the left, with the proper initial conditions so that it bounces (elastically) off all N balls and finally leaves the semicircle, heading directly to the left.
i) In the limit N → ∞ (so the mass of each ball in the semicircle, M/N,goes to zero), find the minimum value of M/m that allows the incoming ball to come out heading directly to the left.
ii) In the minimum M/m case found in part (i), find the ratio of m’s final speed to initial speed.
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26 Answers
1yes sky ..
actually this can also be easily proved if you work at the centre of mass frame of reference and conserve momentum and energy ..
then just the two mass will divert oppoite to each other and then you will get the result ..
1if 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
this shud be remembered always ?
1Ok. 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...
1for 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
1please wait giving sir actually i wanted to confirm my answer first ...
66@rohan,
the answers given by you are correct. What about the intermediate steps?
21Sky : "bounces (elastically)"
this implies e = 1 naa.....
Or is it dat elastically dusnt mean perfectly elastic
1well this one is easier than the earlier one
but sill there may be calculation mistakes
whatever here are the two answers i got
M/m=Ï€
ratio of final and initial velocity=e-Ï€
can i get a conformation
24what a question sirji![11][11][12].........i wish i could try it but can't....have to prepare for boards....[2][2]
66@:-)
What you have done is not correct, since the final velocity you have obtained is only for case of head-on elastic collision.
1@ritika... yep... i've done it using limits... dunno if any other method exists...!!!
1let 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 vN=(mN-M/mN+M)N.v
now as N tends to infinity... Vf = e-2M/m.v
=> vf / v = e-2M/m ............ (ii)
but i couldn't get d first one...!!!
thinking... [12]
66Yes, its the ball of mass m which is colliding with all others. The initial condition has been set up in such a manner that when m collides with the lowest ball of mass M/N, then this ball (with mass M/N) fly somewhere to the right-side (but not necessarily opposite that of m) while m goes on to collide with next ball of M/N. This process is repeated until finally when all collisions are over, m is moving directly towards the left. Now the phenomena should be clear, I hope.
11cant get the question :( how can the ball of mass m come to the place where the arrow is placed in the figure
1no ith power we are talking of the bigger ball not one of the N balls....
but that i only guessed from the language..
1Sorry sir, but i didnot get the part
"so that it bounces (elastically) off all N balls and finally leaves the semicircle, heading directly to the left"
actually what is said here, is one of the N balls leaves as shown in figure, or the incoming ball bounces(then how can it go to the arrowed position on the upper side) or both?
62oh
now I can say
"what an idea sirji"
but then let the users try this one first...
This was another classic from you...
So simple .. yet so intimidating :)
13
great Q!!
very sad i also ve to prepare 4 boards
62What a question Sirji..
But now, who is going to say
"What an Idea Sirji?" ;)
For a moment, I seem to be blank !