Question

There are n elastic balls placed on a smooth horizontal plane. The masses of the balls are m,\(\frac{m}{2}\),\(\frac{m}{2^2}\)........\(\frac{m}{2^{n-1}}\) respectively. If the first ball hits the second ball with velocity V₀, then the velocity of the n^th ball will be,

• A. \(\frac{4}{3}V_0\)
• B. \((\frac{4}{3})^{n}V_0\)
• C. \((\frac{4}{3})^{n-1}V_0\)
• D. V₀

Answer 1

The Correct Option is C

The correct answer is option (C): \((\frac{4}{3})^{n-1}V_0\)

\(v_1'=\frac{2m_1v_1}{m_1m_2}+\frac{m_1-m_2}{m_1+m_2}v_2\)

The velocity of the 2nd ball \(v_2=0\)

Let the velocity of the first ball \(v_1=v\)

The collision between the first and second ball \(v_2'=\frac{2mv}{m+\frac{m}{2}}=\frac{4}{3}v\)

\(v_3'=\frac{4}{3}v_3=(\frac{4}{3})^2v\)

so, for n ball,

\(v_n'=\frac{4}{3}v_n=\frac{4}{3}^{(n-1)}v\)

Answer 2

Suppose collision of two balls is as follows :

Now, By conservation of momentum
\(mv_0=mv_1+\frac{m}{2}v_2\)
∴ \(v_0=v_1+\frac{v_2}{2}\ \ ....(i)\)
For elastic collision, \(e=1=\frac{v_2-v_1}{v_0-0}\)
Therefore, \(v_0=v_2-v_1\ \ ....(ii)\)
So, from equation (i) and (ii), we get
\(\frac{3v_2}{2}=2v_0\)
\(⇒v_2=\frac{4v_0}{3}\)
In the same way, the velocity of 3rd ball,
\(v_3=(\frac{4}{3})^2v_0\)
∴ Velocity of n^th ball, \(v_n=(\frac{4}{3})^{n-1}v_0\)
So, the correct option is (C) : \((\frac{4}{3})^{n-1}V_0\)