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
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
