Q. Four smooth steel balls of equal mass at rest are free to move along a straight line without friction. The first ball is given a velocity of 0.4 ${ms}^{- 1}$. It collides head on with the second one elastically, the second one similarly with the third and so on. The velocity of the last ball is

A) 0.4${ms}^{- 1}$
B) 0.2${ms}^{- 1}$
C) 0.1${ms}^{- 1}$
D) 0.05${ms}^{- 1}$
Explanation

(a) If after the collision of two bodies, the total kinetic energy of the bodies remains the same as it was before the collision, and also momentum remains same, then it is a case of perfectly elastic collision.
Momentum before collision= Momentum after collision
Kinetic energy before collision
=Kinetic energy after collision
Also,${\ u}_{1} - u_{2} = - \left( v_{1} - v_{2} \right)$
Where ($\ u_{1} - u_{2}$) is the relative velocity before the collision and $(v_{1} - v_{2})$is the relative velocity after the collision. Thus, in a perfectly elastic collision the relative velocity remains unchanged in magnitude, but is reserved in direction. Hence, velocity of the last ball is $- 0.4\ ms^{- 1}$.

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.