How Is Delta_h Calculated in Physics Problems?

[annamal]

Homework Statement

A small diamond of mass 10.0 g drops from a swimmer’s earring and falls through the water, reaching a terminal velocity of 2.0 m/s. a) Assuming the frictional force on the diamond obeys what is b? (b) How far does the diamond fall before it reaches 90 percent of its terminal speed?

Relevant Equations

v_t = terminal velocity
a) -m*g + b*v = m*a = 0 for terminal velocity
b = m*g/v_t
b) My question is here:
0.9v_t = v0 + a*t = a*t
t = 0.9v_t/a
delta_h = v0*t + 0.5*a*t^2 = 0.5*a*t^2 = 0.5*((0.9*v_t)^2)/a =
where a = (-m*g + b*v)/m
delta_h = (0.5*((0.9*v_t)^2)*m)/(-m*g + b*v)
plugging in everything delta_h = 1.65 m which is not what the answer is. Please help

b) My question is here!
0.9v_t = v0 + a*t = a*t
t = 0.9v_t/a
delta_h = v0*t + 0.5*a*t^2 = 0.5*a*t^2 = 0.5*((0.9*v_t)^2)/a =
where a = (-m*g + b*v)/m
delta_h = (0.5*((0.9*v_t)^2)*m)/(-m*g + b*v)
plugging in everything delta_h = 1.65 m which is not what the answer is. Please help

 

[Lnewqban]

Could you clarify this part?
"...Assuming the frictional force on the diamond obeys what is b? (b) How far..."

 

[Orodruin]

0.9v_t = v0 + a*t = a*t

This is true if acceleration is constant. Is acceleration constant?

 

[DaveC426913]

A small diamond of mass 10.0 g

That's a 50 carat diamond! It'd be as big as an acorn!

 

[Orodruin]

That's a 50 carat diamond! It'd be as big as an acorn!

Good observation. This also brings to mind another point though, the total force is not mg-bv ...

 

[kuruman]

As an aside, a 10-g diamond is not exactly "small". Considering that there are 5 carats to a gram, a 10-gram diamond would be 50 carats.

"A 50-carat white diamond has fetched 6.5 million Swiss francs (dollars) at a Christie's auction of jewelry, with the auctioneer saying an anonymous trader snapped up the rare stone."
Source: https://www.dailysabah.com/life/2018/05/17/50-carat-diamond-sold-for-65m-at-christies-auction

How stupid can one be to go swimming with of these hanging from one's ear? I believe that physics problems ought to be reasonably realistic.

On edit: I see @DaveC426913 preempted me.

 

[annamal]

This is true if acceleration is constant. Is acceleration constant?

a = (-m*g + b*v)/m --> acceleration is dependent on velocity. why does acceleration have to be constant?

 

[annamal]

Good observation. This also brings to mind another point though, the total force is not mg-bv ...

No, it is. The solution said so.

 

[annamal]

Homework Statement:: A small diamond of mass 10.0 g drops from a swimmer’s earring and falls through the water, reaching a terminal velocity of 2.0 m/s. a) Assuming the frictional force on the diamond obeys what is b? (b) How far does the diamond fall before it reaches 90 percent of its terminal speed?
Relevant Equations:: v_t = terminal velocity
a) -m*g + b*v = m*a = 0 for terminal velocity
b = m*g/v_t
b) My question is here:
0.9v_t = v0 + a*t = a*t
t = 0.9v_t/a
delta_h = v0*t + 0.5*a*t^2 = 0.5*a*t^2 = 0.5*((0.9*v_t)^2)/a =
where a = (-m*g + b*v)/m
delta_h = (0.5*((0.9*v_t)^2)*m)/(-m*g + b*v)
plugging in everything delta_h = 1.65 m which is not what the answer is. Please help

b) My question is here!
0.9v_t = v0 + a*t = a*t
t = 0.9v_t/a
delta_h = v0*t + 0.5*a*t^2 = 0.5*a*t^2 = 0.5*((0.9*v_t)^2)/a =
where a = (-m*g + b*v)/m
delta_h = (0.5*((0.9*v_t)^2)*m)/(-m*g + b*v)
plugging in everything delta_h = 1.65 m which is not what the answer is. Please help

Assuming the frictional force on the diamond obeys -bv what is b?

 

[annamal]

Could you clarify this part?
"...Assuming the frictional force on the diamond obeys what is b? (b) How far..."

Assuming the frictional force on the diamond obeys -bv what is b?

 

[Orodruin]

No, it is. The solution said so.

That it is in the solution does not make it correct. In this case, the solution is wrong. It would be correct if the water density was very small compared to the diamond density, but that is not the case. The specific gravity of diamond is ca 3.5, which is not that much larger than 1.

a = (-m*g + b*v)/m --> acceleration is dependent on velocity. why does acceleration have to be constant?

The formula you quoted:
$$
v = v_0 + at
$$
assumes that $a$ is constant. It is not valid if $a$ is not constant.

 

[annamal]

That it is in the solution does not make it correct. In this case, the solution is wrong. It would be correct if the water density was very small compared to the diamond density, but that is not the case. The specific gravity of diamond is ca 3.5, which is not that much larger than 1.

The formula you quoted:
$$
v = v_0 + at
$$
assumes that $a$ is constant. It is not valid if $a$ is not constant.

Ok, I found the formula v = v0*e^(-bt/m) but I am confused because isn't v0 the initial velocity 0?

 

[Orodruin]

Ok, I found the formula v = v0*e^(-bt/m) but I am confused because isn't v0 the initial velocity 0?

That is the formula without gravitation or buoyancy. It is not applicable here. You can easily see this from the fact that it goes to zero (and not the terminal velocity) as time goes to infinity.

 

[annamal]

That is the formula without gravitation or buoyancy. It is not applicable here. You can easily see this from the fact that it goes to zero (and not the terminal velocity) as time goes to infinity.

Late, but to solve part b, I did, $0.9v = \int_0^t(mg-bv)dt$ where b= 0.049, but I don't know how to solve that equation

 

[haruspex]

Late, but to solve part b, I did, $0.9v = \int_0^t(mg-bv)dt$ where b= 0.049, but I don't know how to solve that equation

You can't solve it in that form because of the unknown function v(t) inside the integral. You need to get all the occurrences of v together.
Taking it back the the differentIal form, $0.9\dot v = mg-bv$. Now get both v's on the same side. There are two ways to do that. Both work, but the way forward may be more obvious with one than with the other.

 

[annamal]

You can't solve it in that form because of the unknown function v(t) inside the integral. You need to get all the occurrences of v together.
Taking it back the the differentIal form, $0.9\dot v = mg-bv$. Now get both v's on the same side. There are two ways to do that. Both work, but the way forward may be more obvious with one than with the other.

ok, I am rusty with my diff eq, I get $0.9\frac{dv}{dt} + bv = mg$... and then what?

 

[Orodruin]

Then that is a linear ODE with constant coefficients.

 

[haruspex]

ok, I am rusty with my diff eq, I get $0.9\frac{dv}{dt} + bv = mg$... and then what?

Of course, you have picked the harder one. Can you see the other way to get all the v's on the same side?

 

[annamal]

Of course, you have picked the harder one. Can you see the other way to get all the v's on the same side?

Oh nvm. I found the derivation in the book and got the correct answer. Thanks anyways.

 

[haruspex]

Oh nvm. I found the derivation in the book and got the correct answer. Thanks anyways.

Fwiw, the other way is to divide both sides by mg-bv. Then both sides are integrable.

 

[Orodruin]

Oh nvm. I found the derivation in the book and got the correct answer. Thanks anyways.

Getting the right answer is not the end goal. The end goal is understanding how you got it and why it is the right answer. Just looking up the solution is generally not a good approach.