Question

Two ivory balls are placed together at rest. A third identical ball moving with velocity u in line of first two balls as shown in the figure collide head on elastically then,

A. Third ball comes to rest with the second ball while the first ball moves with speed u.
B. Third ball comes to rest and other two moves together with speed $\dfrac{u}{2}$
C. All three ball move together with speed $\dfrac{u}{3}$
D. All three balls move in such a manner each makes angle $120{}^\circ$ to each other

Answer 1

Basically, all you have to understand here is that a total of two collisions happen in this situation. One happens between ball 3 and ball 2 and the other between ball 2 and ball 1. You could recall the expression for velocities of two bodies after a head on elastic collision and then substitute the values accordingly for each condition and thus find the answer.

Formula used:
Velocities after head on elastic collision,
${{v}_{1}}=\dfrac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}{{u}_{1}}+\dfrac{2{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}}{{u}_{2}}$
${{v}_{2}}=\dfrac{2{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}}{{u}_{1}}+\dfrac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}{{u}_{2}}$

Complete answer:
In the question, we are given 3 identical ivory balls. We are said that the third ball moving with a velocity u is colliding with the other two balls head on elastically. Using these given conditions, we are asked to find which among the given options is the correct.
Here, in this situation, there are actually two collisions taking place. The ball-3 is colliding with ball 2 and then the ball 2 is colliding with ball 1.
Now let us recall the expressions for velocity after collision of the balls undergoing elastic head on collision,
${{v}_{1}}=\dfrac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}{{u}_{1}}+\dfrac{2{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}}{{u}_{2}}$ ……………………………………. (1)
${{v}_{2}}=\dfrac{2{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}}{{u}_{1}}+\dfrac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}{{u}_{2}}$……………………………………. (2)
In this question,
${{m}_{1}}={{m}_{2}}={{m}_{3}}=m$
In the first collision ball 3 has a velocity u and ball 2 is at rest, so,
${{u}_{1}}=u$
${{u}_{2}}=0$
Substituting all these we get,
${{v}_{1}}=0$
${{v}_{2}}=\dfrac{2m}{m+m}u=u$
Hence, we see that the whole velocity of ball 3 is now transferred to ball 2.
Now the second collision happens for ball 2 moving with velocity u and ball 1 at rest.
For the second collision, just like the previous case,
${{u}_{1}}=u$
${{u}_{2}}=0$
But here ${{u}_{1}}$ and ${{u}_{2}}$ are velocities of ball 2 and ball 1 before collision and their velocities after collision will be again,
${{v}_{1}}=0$
${{v}_{2}}=u$
So, finally the ball 1 moves with a velocity u and the other two balls remain at rest.

Hence, option A is the correct answer.

Note:
At first glance, you may think that on the first collision itself the velocity of ball 3 will be split between the two balls that are at rest and are lying close to each other. But in reality, a second collision happens almost immediately. This causes the result to become very different from our expectation.