Question

Two red blood cells each have a mass of 4.60 x 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. Once cell carries-2.40 pC of charge and the other-3-10 pc, and each cell can be modeled as a sphere 7.60 μrm in diameter. What minimum relative speed o would the red blood cells need when very far away from each other to get close cenough to just touch? Ignore viscous drag from the surrounding liquid. m/s US What is the magnitude of the maximum acceleration amax of each cell?

Answer 1

The mass of each red blood cell is, The charge of one of the red blood cells is, g!--240PC The charge of another red blood cell is., Initially the two red blood cells move with initial velocity u. Initially the red blood cells are far away from each other, then consider the initial or maximum distance between the two red blood cells is infinity The final or minimum distance between the two red blood cells is, radius of red blood cell-)-( radius of red blood cell-2) The two red blood cells come to rest after they close to each other.

According to the law of conservation energy, the total energy of the two red blood cells before and after they come close to each other is, 2 2 2 2 kgn92 The radius of the two red blood cells are equal and the value of the radius of the red blood cell is, 2 7.6um 2 -3.8um -3.8x10 m Calculate the minimum velocity of the red blood cells as follows: ką,q2

9x10 N m'/C (-2.40 pC) (-3.10pC) 4.60x1014 kg (3.8x106 m+3.8x106 m 9x10 N.m2/C2(-2.40x10-12 c)(-3.10x10-12 C 4.60x10-" kg (3.8x10* m +3.8x106 m 437.64 m/s The net force that acts on the red blood cells is, net ma-

Then calculate the acceleration of the red blood cells as follows: kg,92 (4.60x10 " kg) (3.8x10 m+3.8x10 m) ( Эк 109 N. m2/C)(-2.40 10-12 C)(-3.10 10-12 с ) (4.60x10 1 kg)(3.8x10 m+3.8x10 m 2.52x1010 m/s2