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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 1
Let m be the mass of each ball bearing. Before collision, total K.E. of the system
=1/2mv2 + 0 =1/2 mv2
After collision, K.E. of the system is
Case I, E1 = 1/2 (2m) (v/2)2 = 1/4 mv2
Case II, E2 = 1/2 mv2
Case III, E3 = 1/2(3m) (v/3)2 = 1/6mv2
Thus, case II is the only possibility since K.E. is conserved in this case.
Solution 2
It can be observed that the total momentum before and after collision in each case is constant.
For an elastic collision, the total kinetic energy of a system remains conserved before and after collision.
For mass of each ball bearing m, we can write:
Total kinetic energy of the system before collision:
= (1/2)mV2 + (1/2)(2m) × 02
= (1/2)mV2
Case (i)
Total kinetic energy of the system after collision:
= (1/2) m × 0 + (1/2) (2m) (V/2)2
= (1/4)mV2
Hence, the kinetic energy of the system is not conserved in case (i).
Case (ii)
Total kinetic energy of the system after collision:
= (1/2)(2m) × 0 + (1/2)mV2
= (1/2) mV2
Hence, the kinetic energy of the system is conserved in case (ii).
Case (iii)
Total kinetic energy of the system after collision:
= (1/2)(3m)(V/3)2
= (1/6)mV2
Hence, the kinetic energy of the system is not conserved in case (iii).
Hence, Case II is the only possibility.
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