9702/Oct Nov/11/2009/Q9

The diagram shows two spherical masses approaching each other head-on at an equal speed u.
One has mass 2m and the other has mass m.

Which diagram, showing the situation after the collision, shows the result of an elastic collision?

Solution:
Answer: A

For an elastic collision,

Velocity of approach (before collision) = Velocity of separation (after collision)

{The above result can be obtained by considering that for elastic collision, both momentum and kinetic energy is conserved. Momentum, p = mv. Kinetic energy = ½mv². By equating the sum of momentum before collision to that after collision and by equating the sum of KE before collision to that after collision, 2 equations are obtained which can be simplified into the above stated result: Velocity of approach (before collision) = Velocity of separation (after collision). The proof will not be shown here}

{Approach means that the 2 sphere are coming towards each other and separation means that they are moving away from each other}

Before collision, velocity of approach = u + u = 2u

Consider A:

Velocity of separation = (u/3) + (5u/3) = 6u/3 = 2u

Consider B:

Velocity of separation = (u/6) + (2u/3) = 5u/6

Consider C:

Both spheres are moving in the same direction. So, speed of separation is the difference in the 2 speed here/

Velocity of separation = (2u/3) – (u/6) = 3u/6 = 0.5u

Consider D:

Since the spheres stick together, they are not separating. They move together.

Velocity of separation = 0

Only answer A gives velocity of separation = 2u.

Reference: PYQ - Oct/Nov 2009 Paper 11 Q9