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): \((\frac{4}{3})^{n-1}V_0\) 

We are analyzing a sequence of elastic collisions where each ball is lighter than the previous one.

Let the initial velocity of the first ball be \(v_1 = v\) and the second ball be at rest: \(v_2 = 0\).

Using the velocity formula after elastic collision:

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

Given \(m_2 = \frac{m}{2}\) and \(v_2 = 0\),

\(v_2' = \frac{2m}{m + \frac{m}{2}}v = \frac{2m}{\frac{3m}{2}}v = \frac{4}{3}v\)

Now this ball hits a third ball with mass \(m_3 = \frac{m}{4}\), again initially at rest.

By similar logic, velocity after next collision:

\(v_3' = \frac{2m_2}{m_2 + m_3}v_2' = \frac{2 \cdot \frac{m}{2}}{\frac{m}{2} + \frac{m}{4}} \cdot \frac{4}{3}v = \frac{4}{3} \cdot \frac{4}{3}v = \left(\frac{4}{3}\right)^2 v\)

So, by this pattern, after the \(n^\text{th}\) collision:

\(v_n = \left(\frac{4}{3}\right)^{n-1}v = \left(\frac{4}{3}\right)^{n-1}V_0\)

Therefore, the final velocity of the \(n^\text{th}\) ball is:

\(\boxed{\left(\frac{4}{3}\right)^{n-1}V_0}\)

Answer 2

Suppose collision of two balls is as follows :

 

By conservation of momentum: 

\(mv_0 = mv_1 + \frac{m}{2}v_2\)

\(\Rightarrow v_0 = v_1 + \frac{v_2}{2} \quad \text{...(i)}\)

For elastic collision: \(e = 1 = \frac{v_2 - v_1}{v_0 - 0}\)

\(\Rightarrow v_0 = v_2 - v_1 \quad \text{...(ii)}\)

From equations (i) and (ii):

Add both equations:

\(v_0 + v_0 = v_1 + \frac{v_2}{2} + v_2 - v_1 = \frac{3v_2}{2}\)

\(\Rightarrow 2v_0 = \frac{3v_2}{2}\)

\(\Rightarrow v_2 = \frac{4v_0}{3}\)

Similarly, the velocity of the 3rd ball:

\(v_3 = \left(\frac{4}{3}\right)^2 v_0\)

Therefore, velocity of the \(n^\text{th}\) ball:

\(v_n = \left(\frac{4}{3}\right)^{n-1} v_0\)

So, the correct option is (C): \(\left(\frac{4}{3}\right)^{n-1} V_0\)