Question

$n$ elastic balls are placed at rest on a smooth horizontal plane which is circular at the ends with radius $r$ as shown in the figure. The masses of the balls are $m,\frac{m}{2},\frac{m}{2^2}......\frac{m}{2^{n-1}}$ respectively. What is the minimum velocity which should be imparted to the first ball of mass $m$ such that $n^{th}$ ball completes the vertical circle:

• A. $(\frac{3}{4}{)}^{n-1}\sqrt{5gr}$
• B. $(\frac{4}{3}{)}^{n-1}\sqrt{5gr}$
• C. $(\frac{3}{2}{)}^{n-1}\sqrt{5gr}$
• D. $(\frac{2}{3}{)}^{n-1}\sqrt{5gr}$

Answer 1

The correct option is A $(\frac{3}{4}{)}^{n-1}\sqrt{5gr}$

Using the expressions for velocities of the two particles in one dimensional elastic collision,

${v_2}^{\prime }=\frac{2m_1{v}_1}{m_1+m_2}+\frac{m_1-m_2}{m_1+m_2}{v}_2$;

Velocity of the second ball $v_2=0$;
Let velocity of the first ball $v_1=v$;
For collision between the first and second ball,

${v_2}^{\prime }=\frac{2mv}{m+\frac{m}{2}}=\frac{4}{3}v$;

The second ball will have a elastic collision with the third ball.
Since the ratio of consecutive masses of the balls is same, the velocity ratio of the ball will be same.

${v_3}^{\prime }=\frac{4}{3}{v}_3=(\frac{4}{3}{)}^{2}v$;

Continuing this for the n-th ball, we get

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

From motion of a particle in a vertical circle, the minimum velocity required for the particle to reach the highest point on the vertical circle is $(5gR{)}^{\frac{1}{2}}$;

${v_n}^{\prime }=(5gR{)}^{\frac{1}{2}}$;

$\Rightarrow (5gR{)}^{\frac{1}{2}}=(\frac{4}{3}{)}^{n-1}v$;

$\Rightarrow v=\left(\frac{3}{4}{)}^{n-1}\right(5gR{)}^{\frac{1}{2}}$.

$\Rightarrow v=(\frac{3}{4}{)}^{n-1}\sqrt{5gr}$