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. If the collision is elastic, which of the following figure is a possible result after collision?
Solution
The total momentum before and after a collision in each case is constant.
For an elastic collision,
Total kinetic energy of a system remains conserved before and after the collision.
For mass of each ball bearing m, we can write,
Total kinetic energy of the system before collision,
=
mV2 +
(2m) × 02
=
mV2
Case (i)
Total kinetic energy of the system after collision,
=
m × 0 +
(2m) (
)2
=
mV2
Hence, the kinetic energy of the system is not conserved in case (i).
Case (ii)
Total kinetic energy of the system after collision,
=
x(2m) × 0 + 
mV2
=
mV2
Hence, the kinetic energy of the system is conserved in case (ii).
Case (iii)
Total kinetic energy of the system after collision,
= (
)(3m)(
)2
= (
)mV2
Hence, the kinetic energy of the system is not conserved in case (iii).
Hence, Case II is the only possibility.
For an elastic collision,
Total kinetic energy of a system remains conserved before and after the collision.
For mass of each ball bearing m, we can write,
Total kinetic energy of the system before collision,
=
mV2 +(2m) × 02=
mV2Case (i)
Total kinetic energy of the system after collision,
=
m × 0 + (2m) ()2=
mV2Hence, the kinetic energy of the system is not conserved in case (i).
Case (ii)
Total kinetic energy of the system after collision,
=
x(2m) × 0 + mV2 =
mV2Hence, the kinetic energy of the system is conserved in case (ii).
Case (iii)
Total kinetic energy of the system after collision,
= (
)(3m)()2= (
)mV2 Hence, the kinetic energy of the system is not conserved in case (iii).
Hence, Case II is the only possibility.
