Two red blood cells each have a mass of 9.05 times 10^-14 kg and carry a negativ

Question

Two red blood cells each have a mass of 9.05 times 10^-14 kg and carry a negative charge spread uniformly over their surfaces. The repulsion arising from the excess charge prevents the cells from clumping together. One cell carries -2.70 pC and the other -3.50 pC, and each cell can be modeled as a sphere 3.75 times 10^-6 m in radius. If the red blood cells start very far apart and move directly toward each other with the same speed, what initial speed would each need so that they get close enough to just barely touch? Assume that there is no viscous drag from any of the surrounding liquid. What is the maximum acceleration of the cells as they move toward each other and just barely touch?

Explanation / Answer

using conservation of energy

total mechanical energy at the far away seperation = Total mechanical energy at the just barely touch

K1+U1= K2+U2

2*(0.5*m*v^2) + 0 = -K*q1*q2/(2*r)

2*0.5*9.05*10^-14*v^2 = -9*10^9*2.7*3.5*10^-24/(2*2*3.75*10^-8)

v = 2.5 *10^3 m/s

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F = K*q1*q2/r^2

m*a = k*q1*q2/r^2

accelaration a = (k*q1*q2/(m*r^2) = (9*10^9*2.7*3.5*10^-24)/(9.05*10^-14*(2*3.75*10^-8)^2) = 1.67*10^14 m/s^2

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