Two red blood cells each have a mass of 9.05 x 104 kg and carry a negative charg

Question

Two red blood cells each have a mass of 9.05 x 104 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 -3.00 pC and the other -3.50 pC, and each cell can be modeled as a sphere 3.75 x 106 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 Number m/s What is the maximum acceleration of the cells as they move toward each other and just barely touch? Number m/s

Explanation / Answer

Part A -

The expression for the potential energy is U = kqq/d.
and at a large distance (r) U = 0 and their initial kinetic energy is K = 2 x ½mv² = mv² (where m is the mass of a single cell).

Further, as they get closer, the slow down due to repulsion: kinetic energy is turned to potential kinetic energy. At the point they just touch, they have momentarily stopped and their potential energy equals whatever their initial kinetic energy was.
mv² = kqq/d
v = [kqq/(md)]

'd' is the distance between their centres when they just touch, this is 2 radii.
so -
v = [kqq/(md)]
= [8.99x10^9 x (-3.0x10^-12) x (-3.50x10^-12) /(9.05x10^-14 x 2 x 3.75x10^-6)]

= [1.3907 x 10^5]
= 373 m/s

Part B -

Now the max. acceleration of a cell occurs when the force is a maximum. F = kqq/d². This is when they are closest, just touching (d=2r, as above).

Since F = ma, a = F/m
a_max = (kqq/(2r)²)/m
= kqq/(4r²m)
= 8.99x10^9 x (-3.0x10^-12) x (-3.5x10^-12) /(9.05x10^-14 x 4 x (3.75x10^-6)²)
= 0.1854 x 10^11 = 1.854 x 10^10 m/s²  

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