• We’re currently investigating an issue related to the forum theme and styling that is impacting page layout and visual formatting. The problem has been identified, and we are actively working on a resolution. There is no impact to user data or functionality, this is strictly a front-end display issue. We’ll post an update once the fix has been deployed. Thanks for your patience while we get this sorted.

Help me on my physics HW

Status
Not open for further replies.
O

octopus41092

Golden Member
An object charge q1 = +2.0 mC and mass m1 = 2 kg is placed at the origin of a coordinate system, and another object of charge q2 = +3.0 mC and mass m2 = 3kg is placed at x = 3.0m.

Diagram

The two objects are simultaneously released from rest.
a) What is the location of mass m2 at the instant m1 is located at x = -3.0m (Answer = 5.0 m)
b) What is the speed of mass m2 at the instant mass m1 is located at x = -3.0m (Answer = 54.7m/s)

I have the answers but I don't know how I'm supposed to get them.
 
IIRC, you can use the charge repulsion to calculate the electric repulsive force. That will give you a force, which you can then calculate the acceleration of each object. With the acceleration, you can use a standard, Newtonian velocity/acceleration/position equation with the numbers they give you in a and b.

It's something like F = K(q1q2)/r^2 (?), where K is that constant for electricity - (the equation looks like the gravitation equation).
 
Originally posted by: Brainonska511
IIRC, you can use the charge repulsion to calculate the electric repulsive force. That will give you a force, which you can then calculate the acceleration of each object. With the acceleration, you can use a standard, Newtonian velocity/acceleration/position equation with the numbers they give you in a and b.
Click to expand...

Isn't the acceleration non uniform so you cant use the standard newtonian velocity/acceleration/position equations?
 
Originally posted by: octopus41092
Originally posted by: Brainonska511
IIRC, you can use the charge repulsion to calculate the electric repulsive force. That will give you a force, which you can then calculate the acceleration of each object. With the acceleration, you can use a standard, Newtonian velocity/acceleration/position equation with the numbers they give you in a and b.
Click to expand...

Isn't the acceleration non uniform so you cant use the standard newtonian velocity/acceleration/position equations?
Click to expand...

Touche. I guess the force would not be constant as they are pushed apart.

I can't really help you then. I don't remember that much E&M from last year and physics isn't exactly my thing.
 
the charges repel, the masses attract gravitationally.
Find the net force in both the X and Y direction.


then use F=ma on each to determine its acceleration, once you have acceleration, you can use motion equations to find the answers to questions 1 and 2.
 
Part a is just Newton's laws of motion.
Part b you just need to calculate the changes in energy between the initial state and the state specified. I.e. how much electrical potential energy is converted to kinetic, and the how that KE is divided between the masses.
 
Originally posted by: sao123
the charges repel, the masses attract gravitationally.
Find the net force in both the X and Y direction.


then use F=ma on each to determine its acceleration, once you have acceleration, you can use motion equations to find the answers to questions 1 and 2.
Click to expand...

You don't have to take into account gravitational attraction, but once again with the acceleration not being uniform you can't just use F=ma can you?
 
Originally posted by: Mark R
Part a is just Newton's laws of motion.
Part b you just need to calculate the changes in energy between the initial state and the state specified. I.e. how much electrical potential energy is converted to kinetic, and the how that KE is divided between the masses.
Click to expand...

Ok, I got be with the provided answer to A, but I still don't know how to get the answer to part A. Everyone is saying to use Newton's laws of motion. However, the acceleration isn't uniform and I have to do this w/o the use of Calculus.
 
Originally posted by: octopus41092
Originally posted by: sao123
the charges repel, the masses attract gravitationally.
Find the net force in both the X and Y direction.


then use F=ma on each to determine its acceleration, once you have acceleration, you can use motion equations to find the answers to questions 1 and 2.
Click to expand...

You don't have to take into account gravitational attraction, but once again with the acceleration not being uniform you can't just use F=ma can you?
Click to expand...



if you were supposed to ignore gravity, would not the problem have said massless particles?

The energy conversion Mark said might be the answer you are searching for...
 
Originally posted by: octopus41092

Ok, I got be with the provided answer to A, but I still don't know how to get the answer to part A. Everyone is saying to use Newton's laws of motion. However, the acceleration isn't uniform and I have to do this w/o the use of Calculus.
Click to expand...

The acceleration doesn't have to be uniform, because both particles experience an acceleration that is proportional to the other (constant mass). You need to use the first law - each action has an equal and opposing reaction. You know that Mass 1 has moved 3 units under influence of the force.

Now use the 2nd law to work out how Mass 2 must have moved under the influence of an equal and opposite force.
 
So, I still don't get how to do A

And for b, here's what I did
Change in potential = Kinetic Energy
Delta U = 1/2mv^2
kqQ/r - kqQ/R = 1/2mv^2
8.99e9(2.0e-3)(3.0e-3)/3 - 8.99e9(2.0e-3)(3.0e-3)/8 = 1/2(3)v^2
v = 86.8m/s but the correct answer should be 54.7m/s.

Where am I going wrong 🙁
 
If the force is non uniform, energy method is much easier.
Let's Ue=Vq
Ke=1/2mv^2
This is all I remember from my physics 2 class.
After another look at the problem this seems to be more complicated than I imagined especially having gravity and electric forces non uniform. Since there's no calculus involved all that's left is the energy methods which are really derived from calc anyway.
For the gravitational attraction
U gravitational potential= mgh
K = 1/2mv^2

For the electric fields
U electrical potential=Vq where V is potential=kq/r^2 for the spherical shape of the charge and q is the 2nd charge involved in the interaction
K potential -same as before
Emech in= Emech final
(Kinitial=0)+Ue initial+Ug initial=Kfinal +Ue final + Ug final
Plug in the formulas and simplify. At least that's how it looks to me.
 
I GOT B!!

For those interested in how I did it...

Change in Potential = Kinetic Energy
kqQ/r - kqQ/R = 1/2m1v1^2 + 1/2m2v2^2
9e9(2e-3)(3e-3)/3 - 9e9(2e-3)(3e-3)/8 = v1^2 + 3/2(v2^2)
1.12e4 = v1^2 + 3v2^2/2

And I used
m1v1 + m2v2 = m1v1o + m2v2o
2v1 = -3v2
v1 = -3/2(v2)

Substitute that in and solve

54.7m/s! 😀

Still need help on A though.
 
Originally posted by: octopus41092
I GOT B!!

For those interested in how I did it...

Change in Potential = Kinetic Energy
kqQ/r - kqQ/R = 1/2m1v1^2 + 1/2m2v2^2
9e9(2e-3)(3e-3)/3 - 9e9(2e-3)(3e-3)/8 = v1^2 + 3/2(v2^2)
1.12e4 = v1^2 + 3v2^2/2

And I used
m1v1 + m2v2 = m1v1o + m2v2o
2v1 = -3v2
v1 = -3/2(v2)

Substitute that in and solve

54.7m/s! 😀

Still need help on A though.
Click to expand...

Nice job on B. I was about to post this after reading your previous posts, but you got it 🙂

A is easier. Instead of equating kinetic energies, equate potential energies. That is, the increase in potential energy of one object equals the increase in potential energy of the other object. You can calculate the change in potential energy of m1, so just use that for m2.

edit: more hints.

Potential energy E is proportional to m*x where m is mass and x is the distance you move (think E = mgh for gravity... use a different coefficient for g due to the force).

So you can use m1x1=-m2x2

-m1x1/m2 = x2
-(2)(-3)/3 = x2
x2 = 2

This is of course the change in the position of m2. It started at x=3, so it is now at x=5.

edit 2: Clarification

delta(E) = c*m*delta(x) where c is some constant. You can change this to:

delta(E) is proportional to m*delta(x)

Equating changes in potential energy just gives you:

delta(E1) = m1*delta(x1) and
delta(E2) = m2*delta(x2)

Thus, adding a negative sign to account for different directions,

m1*delta(x1) = -m2*delta(x2)

and what I had above follows.
 
Originally posted by: silverpig
Originally posted by: octopus41092
I GOT B!!

For those interested in how I did it...

Change in Potential = Kinetic Energy
kqQ/r - kqQ/R = 1/2m1v1^2 + 1/2m2v2^2
9e9(2e-3)(3e-3)/3 - 9e9(2e-3)(3e-3)/8 = v1^2 + 3/2(v2^2)
1.12e4 = v1^2 + 3v2^2/2

And I used
m1v1 + m2v2 = m1v1o + m2v2o
2v1 = -3v2
v1 = -3/2(v2)

Substitute that in and solve

54.7m/s! 😀

Still need help on A though.
Click to expand...

Nice job on B. I was about to post this after reading your previous posts, but you got it 🙂

A is easier. Instead of equating kinetic energies, equate potential energies. That is, the increase in potential energy of one object equals the increase in potential energy of the other object. You can calculate the change in potential energy of m1, so just use that for m2.

edit: more hints.

Potential energy E is proportional to m*x where m is mass and x is the distance you move (think E = mgh for gravity... use a different coefficient for g due to the force).

So you can use m1x1=-m2x2

-m1x1/m2 = x2
-(2)(-3)/3 = x2
x2 = 2

This is of course the change in the position of m2. It started at x=3, so it is now at x=5.

edit 2: Clarification

delta(E) = c*m*delta(x) where c is some constant. You can change this to:

delta(E) is proportional to m*delta(x)

Equating changes in potential energy just gives you:

delta(E1) = m1*delta(x1) and
delta(E2) = m2*delta(x2)

Thus, adding a negative sign to account for different directions,

m1*delta(x1) = -m2*delta(x2)

and what I had above follows.
Click to expand...

Cool, thanks 🙂
 
Status
Not open for further replies.

TRENDING THREADS

Back
Top