Two red blood cells each have a mass of 9051054 kg and carry

Two red blood cells each have a mass of 9.05*10^-54 kg and carry a negative charge spread uniformly over their surfaces. The repulsion arising from the excess charge prevents the cells from clumpinq together One cell carries -2.70pC and the other -3.10pC, and each cell can be modeled as a sphere 3.75*10^6min 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? Incorrect The maximum acceleration will occur when the force between the two cells is also at a maximum. The force between the two cells will be the Coulomb force. The magnitude if the Coulomb force increases the distance between the two objects decreases

Solution

When the two cells just touch each other they are 2x3.75e-6 cm apart

The potential enrgy of the system when they just touch each other

= (2.7e-12+3.1e-12)/4₀*3.75e-6

= 1.39e+4 J

For the two cells to come close together they must have an initial KE equal to the PE, their KE will be 0 when they touch each other.

let v be the intial velocity

initial KE = 2*mv²/2 = 9.05e-14v²

                                 = 1.39e+4 J

v² = 1.39e+4/9.05e-14

v= 39.19 e+8m/s

The cells will experience maximum acceleration when they just touch each other.

They are 2*3.75e-6 = 7.5e-6 m

Force on each F = 2.7e-12*3.1e-12/4₀*(7.5e-6)²

                         = 1.34e-3 N

acceleration a = F/m = 1.34e-3/9.05e-14 = 1.48e+10 m/s²