Two red blood cells each have a mass of 9.05×10−14 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 −2.10 pc and the other −3.30 pc, and each cell can be modeled as a sphere 3.75×10−6 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.

v = 302.923 m/s

Explanation:

We can answer this question using conservation of energy. Since there is no energy loss (e.g. no viscous drag) the energy when they are far apart and the energy when they barely touch must be the same.

The initial energy must be equal to the sum of their kinetic energies, since they are far apart to feel their electrical interaction.

Ei = (1/2)mv1^2 + (1/2)m*v2^2

Let us consider that they move with the same speed:

Ei = mv^2

If we consider the case when they barely touch, there won't be any kinetic energy, just pure electromagnetic energy:

Ef = k q1q2/(r1+r2) = k q1q2/(2r1)

Since Ei = Ef

v^2 = (k/m) q1q2/(2r1)

where

  k = 8.98755 x10^9 Nm^2/C^2

  m = 9.05 x10^-14 kg

  q1 = −2.10 pC

  q2 = −3.30 pC

  r1 = 3.75×10^−6 m

v^2 = 91762.4 m^2/s^2

v = 302.923 m/s

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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