3. -/2 points My Notes (a) Two particles which have the same magnitude charge bu

Question

3. -/2 points My Notes (a) Two particles which have the same magnitude charge but opposite sign are held 7.00 nm apart. Particle I is then released while Particle II is held steady; the released particle has a mass of 1.21 x 10-22 kg. Particle I's speed is 131 km/s (b) If the particles are still initially held 7.00 nm apart but both particles are released, when they are 3.57 nm away from each other, how would Particle I's speed compare to the speed used in part (a) above (Assume that Particle II's mass is not the same as Particle I's; you should be able to answer this without performing a detailed calculation.) To answer this at all, Particle II's mass would be needed Particle I would be slower than it was in (a) Particle I would be moving at the same speed that it had in (a) Particle I's speed is unknown but it will definitely be the same as Particle II's Particle I would be faster than it was in (a) O

Explanation / Answer

(a) Initial energy = Potential energy = kq*(-q) / r1 =- kq2/r1
=- 9*109*q2/ (7*10-9)
= -1.286*1018q2 -------------(1)
Now at final position
Energy = Potential energy+ Kinetic energy
= -kq2/r2+ (1/2)m1V2
= -9*109)q2 /(3.57*10-9) + (1//2)(1.21*10-22)*(131*103)2
= -2.521*1018q2 + 1.038*10-12 --------------(2)
By conservation of energy
-2.521*1018q2 + 1.038*10-12   = -1.286*1018q2
1.235*1018q2 = 1.038*10-12
q = 9.167*10-16 C
(b) It will be slower than what in part a because total intial energy and final potential energy is same in second case also.
But in second case both will have kinetic energy therefore the energy will be distributed which makes the partcle I slower compairing to first case.
hence the correct option is 2.

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