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?
Answers
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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