Question

4. Two identical ball bearings in contact with each other and resting on a frictionless
table are hit head-on by another ball bearing of the same mass moving initially with a
speed V. If the collision is elastic, which of the following figure is a possible result after
collision?




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Answer 1

Answer:

Two balls are identical , Let each ball of mass = m

initial speed of two balls resting on frictionless table is 0 and another ball is v

before collision :

kinetic energy = 1/2m(0)² + 1/2m(0)² + 1/2mv²

= 1/2mv²

because collision is elastic head on collision so kinetics energy must be conserved . e.g kinetic energy before collision = kinetic energy after collision .

check each of the following case :

______________________________

(a) kinetic energy after collision = 1/2m(0)² + 1/2m(v/2)² + 1/2m(v/2)²

= 0 + mv²/8 + mv²/8

= mv²/4

kinetic energy before collision ≠ kinetic energy after collision .

(b) kinetic energy after collision = 1/2m(0)² + 1/2m(0)² + 1/2m(v)²

= 1/2mv²

here,

kinetic energy before Collision= kinetic energy after collision .

(c) kinetic energy after collision = 1/2m(v/3)² + 1/2m(v/3)² + 1/2m(v/3)²

= 3 × 1/2mv²/9

= mv²/6

kinetic energy before collision ≠ kinetic energy after collision.