Question

Three identical spheres are in a straight line on a table and one moves towards the other two which are at rest and not in contact, if $e=\frac{1}{2}$, then

• A. There are two impacts between them
• B. There are three impacts between them
• C. The ratio of their final velocities is $13:15:63$
• D. The ratio of their final velocities is $13:15:36$

Answer 1

Answer: (B), (D)

The correct options are
B There are three impacts between them
D The ratio of their final velocities is $13:15:36$



First collision:
Velocity of first ball after first collision:
$v_1=\frac{(m-em)}{2m}u+0=\frac{u}{4}$
Velocity of second ball after first collision:
$v_2=\left(\frac{1+e}{2}\right)u=\frac{3}{4}u$

Second collision:
Second ball will undergo inelastic collision with third ball. So,
Velocity of thrid ball after second collision:
$v_3=\left(\frac{1+e}{2}\right)\frac{3}{4}u=\frac{\frac{3}{2}}{2}\times \frac{3}{4}u=\frac{9}{16}u$
Velocity of second ball after second collision
$\therefore v_2^{\prime }=\frac{1}{4}\times \frac{3}{4}u=\frac{3}{16}u$

Third collision:
As $v_2^{\prime }<v_1,$ a collision occurs between first and second balls.
Velocity of first ball after third collision:
$v{"}_1=\frac{1-e}{2}\times \frac{u}{4}+\frac{1+e}{2}\times \frac{3}{16}u=\frac{u}{16}+\frac{9}{64}u=\frac{13}{64}u$
Velocity of second ball after third collision:
$v{"}_2=\frac{1}{4}\times \frac{3u}{16}+\frac{3}{4}\times \frac{u}{4}=\frac{15}{64}u$
Ratio of final velocities of the balls:
$\therefore v{"}_1:v{"}_2:v_3=\frac{13}{64}u:\frac{15}{64}u:\frac{9}{16}u=13:15:36$