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Question: A series of n elastic balls whose masses are \(m, em, e^2 m,… etc\) are at rest separated by interva...

A series of n elastic balls whose masses are m,em,e2m,…etcm, em, e^2 m,… etc are at rest separated by intervals with their centers on a straight line. Here, e is coefficient of restitution for the collision. The first is made to impinge directly on the second with velocity u. Then –
A) The first (n−1)(n - 1) balls will be moving with the same velocity (1−e)u(1 - e)u
B) The last one ball will move with velocity uu
C) The kinetic energy of the system is 12u2(1−e+en)\dfrac{1}{2}{u^2}(1 - e + {e^n})
D) None of these

Explanation

Solution

This problem is an example of imperfect inelastic collisions as there is a coefficient of restitution, e involved. Coefficient of Restitution is the ratio of velocity of separation to the velocity of approach in the collisions. Also momentum is conserved in all collisions whether elastic or inelastic.

Complete step by step solution:
Given-
The masses of the elastic balls are in a series – m,em,e2m,…,enm.m, em, e^2 m,…, e^n m.
The initial velocity of the first ball is u.
Let the velocity after collision of the first two balls be v1v_1 and v2v_2. To determine these velocities –
(i) We know, Coefficient of restitution,
e=v2−v1u−0e = \dfrac{{{v_2} - {v_1}}}{{u - 0}}
⇒v2−v1=eu\Rightarrow {v_2} - {v_1} = eu
⇒v1=v2−eu\Rightarrow {v_1} = {v_2} - eu...................(1)
Also, applying Conservation of Momentum,
mu=mv1+(em)v2mu = m{v_1} + (em){v_2}
⇒u=v1+ev2\Rightarrow u = {v_1} + e{v_2}...........(2)
Putting v1v_1 from (1) in (2),
u=v2−eu+ev2u = {v_2} - eu + e{v_2}
⇒v2(1+e)=u(1+e)\Rightarrow {v_2}(1 + e) = u(1 + e)
⇒v2=u\Rightarrow {v_2} = u
⇒v1=u−eu\Rightarrow {v_1} = u - eu
⇒v1=(1−e)u\Rightarrow {v_1} = (1 - e)u
Similarly, for the next two balls let the velocities be v1v_1 and v2v_2. To determine these velocities –
(i) We know, Coefficient of restitution,
e=V3−V2u−0e = \dfrac{{{V_3} - {V_2}}}{{u - 0}}
⇒V3−V2=eu\Rightarrow {V_3} - {V_2} = eu
⇒V2=V3−eu\Rightarrow {V_2} = {V_3} - eu................(1)
Also, applying Conservation of Momentum,
(em)u=(em)V2+(e2m)V3(em)u = (em){V_2} + ({e^2}m){V_3}
⇒u=V2+eV3\Rightarrow u = {V_2} + e{V_3} - (2)
Putting v1v_1 from (1) in (2),
u=V3−eu+eV3u = {V_3} - eu + e{V_3}
⇒V3(1+e)=u(1+e)\Rightarrow {V_3}(1 + e) = u(1 + e)
⇒V3=u\Rightarrow {V_3} = u
⇒V2=u−eu\Rightarrow {V_2} = u - eu
⇒V2=(1−e)u\Rightarrow {V_2} = (1 - e)u
Thus, we see that the first (n-1) balls after collision will move with the velocity (1-e)u.

Final answer is, The correct option is A. The first (n−1)(n - 1) balls will be moving with the same velocity (1−e)u(1 - e)u.

Note: (1) Conservation of Momentum is applicable even in Inelastic collisions.
(2) Since the masses of each next ball to the previous ball is in a ratio of e and the velocity of the previous ball before impact is the same(= u). Therefore the velocity of each pair of balls after impact is the same.