Print E Calculator tion 17 of 36 Map niversity Physics presented by Sapling Lear

Question

Print E Calculator tion 17 of 36 Map niversity Physics presented by Sapling Learning Two red blood cells each have a mass of 9.05 x 10* kg and carry a negative ch their surfaces. The repulsion arising from arge spread uniformly over the excess charge prevents the cells from clumping together One cell carries -2.70 pC and the other -3.30 pC, and each cell can be modeled as a sphere 3.75 x 10 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 mis maximum acceleration of the cells as they move toward each other and just barely touch? Number m/ s PreviousCheck Answer Next E 27 F4 F7 F8

Explanation / Answer

Their potential energy is U = kqq/d.
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).

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.

I’ll use exponential notation, e.g. 3x10 is written as 3e8.

v = [kqq/(2md)]
. .= [8.99e9 x (-2.70e-12) x (-3.30e-12) /(9.05e-14 x 2 x 3.75e-6)]
. .= 343.5 m/s
__________________________

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)
. . . . . .= 8.99e9 x (-2.70e-12) x (-3.30e-12) x/(9.05e-14 x 4 x (3.75e-6)²)
. . . . . .= 1.57e10 m/s² (= 1.57x10¹ m/s²)

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