Question

1) 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 −3.00 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.

initial speed=

b) What is the maximum acceleration of the cells as they move toward each other and just barely touch?

maximum acceleration=

Answer 1

Solution : To Their potential energy ve a charge on cell to di distance between K2, 92 d them to at large distance reo U=0 and intial kinete Energy ů ķ2x Imv2 easib Inital 2. = myr mt mass of single cell As they get closer will slow down due to repulsion. kinetic energy converted to potential energy At the point they just touch o u '= Kinitial 161 kq,92 m2 = [K2,22 kq, 92 md Vinitial = [ 8.99 x 10 ** (-3x102) + (-3.30x102) 1 9.05X164 X 3.75 X166 Vinitial = 512.10 m/s

max acceleration of e force acceleration of tell occurs when the is a maximum, ie when they just touching (d=28) F: kº . dag word F = main ploit Wita az E up . amax = kq, 92 amax = md2 s kq, 92 m 422. - 8.99x10°x (+3.0 x1o'a )x (-3-30 2104) 4*9.0510*** (3:45X166) amax = 1.75x100 m/ 22