Question

Two identical balls marked 2 and 3, in contact with each other and at rest on a horizontal frictionless table, are hit head-on by another identical ball marked 1 moving initially with a speed 'v' as shown in the figure. If the collision is elastic, what is observed?

• A. Ball A comes to rest and balls 2 and 3 roll out with speed each.
• B. Balls 1 and 2 comes to rest and ball 3 rolls out with speed v.
• C. Balls 1, 2 and 3 roll out with speed each.
• D. Balls, 1, 2 and 3 come to rest.

Answer 1

The correct option is B Balls 1 and 2 comes to rest and ball 3 rolls out with speed v.

The system consists of three identical balls marked 1, 2, and 3. Let m be the mass of each ball.
Before the collision,
KE of the system = KE of 1 + KE of 2 + KE of 3 = $\frac{1}{2}m{v}^2+0+0=\frac{1}{2}m{v}^2$
Case (a) After the collision,
KE of the system = KE of 1 + KE of 2 + KE of 3
$=0+\frac{1}{2}m{\left(\frac{v}{2}\right)}^2+\frac{1}{2}m{\left(\frac{v}{2}\right)}^2=\frac{1}{4}m{v}^2$
Case (b)
KE of the system = KE of 1 + KE of 2 + KE of 3
$=0+0+\frac{1}{2}m{v}^2=\frac{1}{2}m{v}^2$
Case (c)
KE of the system = KE of 1 + KE of 2 + KE of 3
$=\frac{1}{2}m{\left(\frac{v}{3}\right)}^2+\frac{1}{2}m{\left(\frac{v}{3}\right)}^2+\frac{1}{2}m{\left(\frac{v}{3}\right)}^2$
$=\frac{1}{6}m{v}^2$
Case (d)
KE of the system = 0
Now, in an elastic collision, the kinetic energy of the system remains unchanged. Hence choice (b) is the only possible result of the collision.