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Newton's second law problem

  • Thread starter Idoubt
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Homework Statement



A drop of water of mass m is falling vertically towards the ground. Due to moisture, the mass of the drop is increasing as given by m=kt2. The equation of motion is

mdv/dt + vdm/dt = mg

find v after 1 second.


Homework Equations



The integrating factor of an imperfect integral

Adx + Bdy = 0

is given by Q = eintegral fydy

where fy = 1/A[ dB/dx - dA/dy]

The Attempt at a Solution



rearranging the equation of motion

dv/dt + (v/m)dm/dt -g = 0

substituting m = kt2 and dm/dt = 2kt

dv/dt + 2v/t - g =0

i.e dv + (2v/t -g )dt = 0 , now it is the form Adv + Bdt, where A=1, B=(2v/t-g)


here ft= 1/1[ 2/t-0] = 2/t, integral ftdt = 2lnt = ln t2

integrating factor Q = eintegral ftdt = elnt2 = t2


multplying the DE by the int factor

t2dv + 2vtdt - gt2dt = 0

the first two terms combine to form a perfect integral

dU where U = vt2

so dU - gt2 = 0

integrating

U - gt3/3 = C , where C is a constant

ie vt2 - gt3/3 = C

when t=0, C=0 so

v = gt/3

I can't find anything wrong with my work, but the answer choices to the question are

1, 0.25 g
2, 0.5 g
3, 0.75 g
4, g

Can some1 tell me where I messed up? thanks in advance.
 

Answers and Replies

  • #2
172
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:frown:
 

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