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 nth 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

Answer: (C)

$(\frac{4}{3})^{n-1}V_0$

Answer 2

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,

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