Two identical balls A and B collide head on elastically. If velocities of A and B, before the collision are +0.5 m/s and -0.3 m/s respectively, then their velocities, after the collision, are respectively
An elastic collision is one in which both momentum and kinetic energy are conserved. During the collision, the objects may deform momentarily but return to their original shapes without loss of energy. Examples include collisions between billiard balls or gas molecules under ideal conditions, maintaining the system’s total energy and momentum.
Class 11 Physics, Chapter 5: Work, Energy and Power, explores fundamental concepts like work done by a force, various forms of energy including kinetic and potential, conservation of energy and power. It explains the relationships between these quantities through mathematical derivations and examples, preparing students for problem-solving and conceptual understanding essential for CBSE EXAM 2024-25.
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To find the velocities of two identical balls A and B after an elastic collision, we can use the conservation of momentum and the fact that kinetic energy is also conserved.
1. Let the initial velocities be:
Velocity of A before collision (uA) = +0.5 m/s
Velocity of B before collision (uB) = -0.3 m/s
2. For two equal masses in an elastic collision, the final velocities can be determined using the following formulas:
vA = (uA + uB) / 2 + (uA – uB) / 2
vB = (uA + uB) / 2 – (uA – uB) / 2
– First, we find the average velocity:
Average velocity (u_avg) = (uA + uB) / 2 = (0.5 – 0.3) / 2 = 0.1 / 2 = 0.05 m/s.
Now, we compute the change in velocities:
Change in velocity = uA – uB = 0.5 – (-0.3) = 0.5 + 0.3 = 0.8 m/s.
3. Using the equations above:
– Final velocity of A (vA):
vA = 0.05 + 0.4 = 0.45 m/s to be corrected depending on the direction.
– Final velocity of B (vB):
vB = 0.05 – 0.4 = -0.35 m/s to be corrected depending on the direction.
4. In a head-on collision, however, for two identical masses, their velocities will swap:
– After collision, ball A takes B’s initial velocity, and ball B takes A’s initial velocity.
– Thus, the final velocities will be:
– vA = -0.3 m/s (B’s initial velocity)
– vB = +0.5 m/s (A’s initial velocity)
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