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 (in figure below) is a possible result after collision?
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From the figure, it is observed that the linear momentum is conserved in all the cases. For an elastic collision, kinetic energy is also conserved.
let us assume that mass of each ball is m.
case 1 :- initial kinetic energy = 1/2 mV² + 1/2m(0)² + 1/2m(0)²
= 1/2mV²
final kinetic energy = 1/2m(0)² + 1/2m(V/2)² + 1/2m(V/2)²
= 1/4mV²
here initial kinetic ≠ final kinetic energy
so, case 1 does not show an elastic collision.
case2 :- initial kinetic energy = 1/2mV²
final kinetic energy = 1/2 m(0)² + 1/2m(0)² + 1/2mV²
= 1/2mV²
initial kinetic energy = final kinetic energy
so, case 2 shows an elastic collision.
case 3 :- initial kinetic energy = 1/2mV²
final Kinetic energy = 1/2m(V/3)² + 1/2m(V/3)² + 1/2m(V/3)²
= 1/6mV²
here, initial kinetic energy ≠ final kinetic energy
so, case 3 doesn't show an elastic collision.
hence, option (ii) is correct.
let us assume that mass of each ball is m.
case 1 :- initial kinetic energy = 1/2 mV² + 1/2m(0)² + 1/2m(0)²
= 1/2mV²
final kinetic energy = 1/2m(0)² + 1/2m(V/2)² + 1/2m(V/2)²
= 1/4mV²
here initial kinetic ≠ final kinetic energy
so, case 1 does not show an elastic collision.
case2 :- initial kinetic energy = 1/2mV²
final kinetic energy = 1/2 m(0)² + 1/2m(0)² + 1/2mV²
= 1/2mV²
initial kinetic energy = final kinetic energy
so, case 2 shows an elastic collision.
case 3 :- initial kinetic energy = 1/2mV²
final Kinetic energy = 1/2m(V/3)² + 1/2m(V/3)² + 1/2m(V/3)²
= 1/6mV²
here, initial kinetic energy ≠ final kinetic energy
so, case 3 doesn't show an elastic collision.
hence, option (ii) is correct.
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