answer is: 321 m/s how do I find this (b) What is the maximum acceleration of th

Question

answer is: 321 m/s how do I find this

(b) What is the maximum acceleration of the cells in part (a)?

answer is: 1.38e+10 m/s2 How did they find this answer

Assuming the viscosity of the blood is negligible, we can use conservation of mechanical energy to calculate the speed the red blood cells would need when they are very far such that they just touch; when the cells just touch, their centers will be separated by a distance of d. The magnitude of the net force acting on each red blood cell is given by Coulomb's law. Plugging this into Newton's second law allows us to calculate the magnitude of the maximum acceleration of the cells.

Explanation / Answer

Mass m = 9*10^-14 kg
Charge q1 = -2.5*10^-12 C
charge q2 = -3.1*10^-12 C
a)
Suppose the speed is v, then when they are close enough the speed will be zero and when they are very far the potential energy will be zero, so the initial kinetic energy will be converted into the potential energy completely.
So,
0.5*m*v^2 = kq1q2/r
2*0.5*( 9*10^-14)*v^2 = 9*10^9*(2.5*10^-12)*(3.1*10^-12)/(7.5*10^-6)
9*10^-14*v^2 = 9.3*10^-9
v = 321 m/s each
b) maximum acceleration will be when they are closest,
the force on any one of them will be,
F = kq2q2/r^2 = 9*10^9*(2.5*10^-12)*(3.1*10^-12)/(7.5*10^-6)^2 = 0.00124 N
so acceleration = F/m = 0.00124/(9*10^-14) = 1.38*10^10 m/s^2.

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