One particle has a mass of 4.14 x 10 -3 kg and a charge of +8.23 ?C. A second pa
ID: 2125550 • Letter: O
Question
One particle has a mass of 4.14 x 10-3 kg and a charge of +8.23 ?C. A second particle has a mass of 8.32 x 10-3 kg and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is 0.139 m, the speed of the 4.14 x 10-3 kg-particle is 182 m/s. Find the initial separation between the particles.
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One particle has a mass of 4.14 x 10-3 kg and a charge of +8.23 ?C. A second particle has a mass of 8.32 x 10-3 kg and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is 0.139 m, the speed of the 4.14 x 10-3 kg-particle is 182 m/s. Find the initial separation between the particles.Explanation / Answer
As there is no external force ,
momentum will remain conserved
So let velocity of second particle be v
then 8.32 *v = 4.14 *182
v = 90.65 m/s
Now the increase in velocity is due to potentialenergy
Here electrostatic potential energy is converted intokinetic energy
Kinitial = 0
Kfinal = (8.32*10^-3*90.65^2 +4.14*10^-3*182^2) /2
Uinitial =K* (8.23*10^-6)^2 / r
(where r = initial distance)
Ufinal = K* (8.23 *10^-6)^2/ 0.139
Find r by
Ufinal -Uinitail = Kfinal - Kinitial
K *8.23*10^-3 ^2(1 /r - 1/0.139) = 136.96
1/0.139 -1/r = 2.24 *10^-4
r = 0.139 m
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