I don't understand how a varied mass changes the acceleration

[Georgepowell]

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:

$$\frac{dv}{dt} + kv = g$$

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

$$Me^{kt}g = Me^{kt}\frac{dv}{dt}$$

$$g = \frac{dv}{dt}$$

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!

 

[berkeman]

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:

$$\frac{dv}{dt} + kv = g$$

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

$$Me^{kt}g = Me^{kt}\frac{dv}{dt}$$

$$g = \frac{dv}{dt}$$

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!

Varying the mass does not change the acceleration, as long as you ignore air resistance. If you have a mass that breaks in half during free fall, both pieces still accelerate the same as a single mass would have. g is a constant.

 

[Georgepowell]

So if I wrote the question correctly (I'm pretty sure I did) then does that mean there is something wrong with the question?

 

[Bob S]

If you have a spherical raindrop falling through a uniform fog and sweeping out mist, it is picking up extra mass with velocity v=0 as it falls. The raindrop acceleration is independent of the density of the fog. The raindrop problem and answer is in problem 8-22 on page 189 of Becker Introduction to Theoretical Mechanics. Your problem appears to be very similar to this one.

Bob S

 

[Georgepowell]

If you have a spherical raindrop falling through a uniform fog and sweeping out mist, it is picking up extra mass with v=0 as it falls. The raindrop acceleration is independent of the density of the fog. The raindrop problem and answer is in problem 8-22 on page 189 of Becker Introduction to Theoretical Mechanics. Your problem appears to be very similar to this one.

Bob S

I don't understand. Are you saying that if an object (not subject to air resistance or any other type of resistance to motion) is falling under gravity and has an increasing mass, its acceleration will not be at a constant g?

 

[Nabeshin]

$$F=\frac{d p} {dt}=\frac{ d (mv)}{dt}=v\frac{dm}{dt} + m\frac{dv}{dt}$$

Cheers!

 

[Bob S]

I don't understand. Are you saying that if an object (not subject to air resistance or any other type of resistance to motion) is falling under gravity and has an increasing mass, its acceleration will not be at a constant g?

Note that I said mass with v=0, meaning that the raindrop has to accelerate all the mass it picks up from v=0 to the raindrop's velocity. This reduces the overall raindrop acceleration.

Bob S

 

[Georgepowell]

Note that I said mass with v=0, meaning that the raindrop has to accelerate all the mass it picks up from v=0 to the raindrop's velocity. This reduces the overall raindrop acceleration.

Bob S

Thanks I understand that. So the mass that is being added in this scenario starts from stationary and that is why the acceleration is not simply g.

 

[Georgepowell]

$$F=\frac{d p} {dt}=\frac{ d (mv)}{dt}=v\frac{dm}{dt} + m\frac{dv}{dt}$$

Cheers!

Thanks, if m is presumed to be constant then F=ma, so I understand why F=ma shouldn't be used with variable mass. I'll explain this to my teacher if he doesn't seem to understand tomorrow.