- #1

Georgepowell

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I found this question on a past paper (it is from an old syllabus and I was not asked to do this question, so it isn't homework or school work. I'm just curious):

A raindrop falls from rest at time t = 0 and moves through still air. At time t its speed is v and its mass is Me^(kt), where M and k are positive constants. Given that the only force acting on the raindrop as it falls is its weight, show that:

[tex]\frac{dv}{dt} + kv = g[/tex]

But if I just applied F = ma to the situation I get:

[tex]Me^{kt}g = Me^{kt}\frac{dv}{dt}[/tex]

[tex]g = \frac{dv}{dt}[/tex]

Which makes perfect sense because if the weight is the only force then why would the acceleration change? (The acceleration of a hammer is the same as that of a feather) So why would it be different for something gradually changing weight?

I asked my teacher but he just said "For variable mass we don't use F = ma" and he didn't tell me why.

Thanks!

A raindrop falls from rest at time t = 0 and moves through still air. At time t its speed is v and its mass is Me^(kt), where M and k are positive constants. Given that the only force acting on the raindrop as it falls is its weight, show that:

[tex]\frac{dv}{dt} + kv = g[/tex]

But if I just applied F = ma to the situation I get:

[tex]Me^{kt}g = Me^{kt}\frac{dv}{dt}[/tex]

[tex]g = \frac{dv}{dt}[/tex]

Which makes perfect sense because if the weight is the only force then why would the acceleration change? (The acceleration of a hammer is the same as that of a feather) So why would it be different for something gradually changing weight?

I asked my teacher but he just said "For variable mass we don't use F = ma" and he didn't tell me why.

Thanks!

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