Electric Projectile - Isaac Physics

Electric Projectile — Isaac Physics - The question in reference

Part A of the question is asking for the value of rho_m, the mass density, that minimises the horizontal displacement, R.

I shall now explain my working out.

I decided to write out suvat for the horizontal and vertical motion. I noticed that the vertical acceleration should be the resultant force of the electrostatic force of attraction towards the upper plate minus the particle's weight divided by the mass of the particle. Note that the mass, m, can be written as rho_m * V, where V is the volume and rho_m is the mass density, and the same applies for the charge, Q, which can be written as rho_q * V, where rho_q is the charge density. This led me to a resultant vertical acceleration of (E*rho_q)/(rho_m))-g.

For the horizontal motion, I know that there is a impulse acting horizontally on the particle and the particle was initially at rest. Hence, I = mv and so I = rho_m V * v.

I finally equated the time of motion for both, leading R being expressed in terms of the other variables and constants in the question (Electric Field Strength and etc.). Since the question is asking for a value of rho_m that minimises R, I thought maybe finding dR/d(rho_m) and equating to zero would give a solution. I have tried this and yet this is incorrect. I think I have maybe messed this step up, but at this point I have no clue whether I am in the right direction. Not to mention that 'hints' don't do justice to this question.

Reply 1

Nitrotoluene

Original post by KettleNoodles

Electric Projectile — Isaac Physics - The question in reference
Part A of the question is asking for the value of rho_m, the mass density, that minimises the horizontal displacement, R.
I shall now explain my working out.
I decided to write out suvat for the horizontal and vertical motion. I noticed that the vertical acceleration should be the resultant force of the electrostatic force of attraction towards the upper plate minus the particle's weight divided by the mass of the particle. Note that the mass, m, can be written as rho_m * V, where V is the volume and rho_m is the mass density, and the same applies for the charge, Q, which can be written as rho_q * V, where rho_q is the charge density. This led me to a resultant vertical acceleration of (E*rho_q)/(rho_m))-g.
For the horizontal motion, I know that there is a impulse acting horizontally on the particle and the particle was initially at rest. Hence, I = mv and so I = rho_m V * v.
I finally equated the time of motion for both, leading R being expressed in terms of the other variables and constants in the question (Electric Field Strength and etc.). Since the question is asking for a value of rho_m that minimises R, I thought maybe finding dR/d(rho_m) and equating to zero would give a solution. I have tried this and yet this is incorrect. I think I have maybe messed this step up, but at this point I have no clue whether I am in the right direction. Not to mention that 'hints' don't do justice to this question.

I did not do the exercise, but I read your solution carefully and came to the following conclusions.

Your approach is well thought out, and you've framed the problem correctly.

Note: I think, however, that the final differentiation is complicated and you might generate errors here. IMHO...of course!

Kind regards,
Sandro

Reply 2

mqb2766

Original post by KettleNoodles

Electric Projectile — Isaac Physics - The question in reference
Part A of the question is asking for the value of rho_m, the mass density, that minimises the horizontal displacement, R.
I shall now explain my working out.
I decided to write out suvat for the horizontal and vertical motion. I noticed that the vertical acceleration should be the resultant force of the electrostatic force of attraction towards the upper plate minus the particle's weight divided by the mass of the particle. Note that the mass, m, can be written as rho_m * V, where V is the volume and rho_m is the mass density, and the same applies for the charge, Q, which can be written as rho_q * V, where rho_q is the charge density. This led me to a resultant vertical acceleration of (E*rho_q)/(rho_m))-g.
For the horizontal motion, I know that there is a impulse acting horizontally on the particle and the particle was initially at rest. Hence, I = mv and so I = rho_m V * v.
I finally equated the time of motion for both, leading R being expressed in terms of the other variables and constants in the question (Electric Field Strength and etc.). Since the question is asking for a value of rho_m that minimises R, I thought maybe finding dR/d(rho_m) and equating to zero would give a solution. I have tried this and yet this is incorrect. I think I have maybe messed this step up, but at this point I have no clue whether I am in the right direction. Not to mention that 'hints' don't do justice to this question.

Youre right with your basic approach, though its hard work to diff and set equal to zero. You should have something like
#/quadratic(rho_q)
which is minimised when the denom is maximum. So by completing the square to maximise the quadratic gives the result. Though really you just need the -b/2a part, so even completing the square is overkill.

(edited 1 month ago)

Reply 3

mqb2766

Original post by TheKnightmare24

I have attached my working above. You were nearly there just some nasty algebra. If you need help with other problems feel free to PM.

Its best to delete and just offer a hint or two if the OP asks. It sounds like they have the right expression for the range, but tried to differentiate = 0 rather than just noting its a quadratic maximisation problem. Your algebra would arguably be a bit simpler / direct to do
t = sqrt(...)
from the s=ut+1/2at^2 vertical motion, then simply sub into R=vt for the horizontal motion. Also, from (1) you could note that one root is at 0 and the first bracket gives the other one, so the max occurs at 1/2 the second root (first bracket).

(edited 1 month ago)

Reply 4

KettleNoodles

Original post by mqb2766

Thank you very much for your hint. I completely forgot that completing the square is one way of finding maximums/minimums.

Reply 5

mqb2766

Original post by KettleNoodles

Thank you very much for your hint. I completely forgot that completing the square is one way of finding maximums/minimums.

There are a fair few questions at this level which are simply quadratics, so even just noting that the stationary point occurs at -b/(2a) is enough, or (from memory as it was a couple of weeks ago) you had something like
-cx(ax + b)
then the max would be at the root average so half the second root as the first root is 0.