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7 m, a particle of mass 9.60 x 103 kg and charge -8.29 uC is fred with an initial speed of 57.3 m/s directly toward the fixed charge. How far does the particle travel before its speed is zero? Units By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor Question Attempts: 0 of 3 used SAVE FOR LATER
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Answer #1

Given that Q = -3.35*10^-6 C, d = 0.0477 m, m = 9.6*10^-3 kg, q' = -8.29*10^-6 C, v = 57.3 m/s
find the distance the particle travel before its speed is zero.
initial energy = kqq'/d + mv^2/2
final energy = kqq'/x
kqq'/d + mv^2/2 = kqq'/x
1/d + mv^2/(2kqq') = 1/x
so x = 1/[1/d + mv^2/(2kqq')]
x = 1/[1/0.0477 + (9.6*10^-3*(57.3)^2)/(2*9*10^9*3.35*10^-6*8.29*10^-6)] = 0.01056 m

the required distance = d - x
= 0.0477-0.01056
= 0.03714 m

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