Question

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 as
shown in figure.
If the collision is elastic, which of the following (figure) is a
possible result after collision?

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 . so, (a) is wrong

(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 .

hence, option (B) is correct .

(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.

hence (c) is wrong .

hence, only option (B) is correct or suitable .

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