Deriving formula for final velocity in 2 repelling bodies

In summary, the problem involves two magnetic masses, m1 and m2, which repel each other for a distance s with a force f. We are asked to predict the final velocities of m1 and m2, assuming the final distance to be v1 and v2 respectively, and that both masses are initially at rest. We are also given that the direction of motion of m1 is positive and the direction of motion of m2 is negative. Using the formula 2as = v^2 - u^2, we can derive two equations with four unknowns. To solve for the final velocities, we need to make some assumptions, such as the force not being a function of distance and the initial separation between the masses being
  • #1
dE_logics
742
0

Homework Statement



Consider 2 magnetic masses, m1 and m2; these 2 repel each other for a distance s with force f (relative to each other)...which applies from m1 to m2.

Predict final velocity of m1 and m2.

Assume final distance to be v1 and v2

Assume the 2 masses at rest initially.

Take any values pointing to the direction of motion of m1 as positive; that is the final velocity of m1 will be positive and m2 will be negative.

So the f by the above definition will be negative (when real world values are taken).



Homework Equations



2as = v^2 - u^2; this might be used...I used it



The Attempt at a Solution



Using standard formula 2as = v^2 – u^2

Its to be noted that the formula is applicable only when the formula from where the normal reaction is derived is stationary with respect to the frame of the observer, so the distance df that the body and mass covers is partially due to the distance traveled by m1 and partially due to the the distance that the body travels (this will be observed by the observer).

So the actual distance that the mass travels (while accelerating) is d1, and that for the body is d2 (new variables assumed, we do not know this value), however the sum of d1 and d2 is equal to the total distance traveled will be equal to the distance for which the force applies (s).

Also the distance traveled by each the mass and the body is an inverse and direct function of mass.

Here a1 (acceleration on m1) = -f/m1

a2 (acceleration on m2) = f/m2

By 2as = v^2 – u^2 -

2*a2*d2 = v2^2...1

2*a1*d1 = v1^2...2

-d2 + d1 = s...3 (this is done cause the real value of d2 will be negative, following the coordinate system)

d1/d2 = m2/m1...4 (as stated before, distance traveled is a direct and inverse function of mass).

There are 4 equation and 4 unknown.

Since the formula needs to be derived, they cannot be solved simultaneously.

Making d2 the subject from equation 4 -

(d1*m1)/m2 = d2

Substituting in equation 3 -

-((d1*m1)/m2) + d1 = s

-d1((m1)/m2) - 1) = s

-d1((m1-m2)/m2) = s

d1 = -s*m2/(m1-m2)

Substituting above value of d1 in equation 2 -

2*a1*(-s*m2/(m1-m2)) = v1^2

From equation 4 making d1 the subject -

d1 = (d2*m2)/m1

Substituting value of d1 in equation 3 -

-d2 + ((d2*m2)/m1) = s

-d2(1 - (m2)/m1) = s

-d2((m1-m2)/m1) = s

d2 = (-s*m1)/(m1-m2)

Substituting value of d2 in equation 1

2*a2*((-s*m1)/(m1-m2)) = v2^2


Final equations -

2*a2*((-s*m1)/(m1-m2)) = v2^2
and
2*a1*(-s*m2/(m1-m2)) = v1^2

Putting real world values – f = -3000
m1 = 610
m2 = 10
s = 2
a2 = -300
a1 = 300/61

v2^2 = 1220....ok
v1^2 = -20/61.....the negativity needs to be removed

Squaring v1^2 = -20/61 (is then step OK?)

v1^4 = 400/3721.....ok
 
Physics news on Phys.org
  • #2
I think this won't be called as a HW problem.
 
  • #3
Predict final velocity under repulsion.

Consider 2 magnetic masses, m1 and m2; these 2 repel each other for a distance s with force f (relative to each other)...which applies from m1 to m2.

Predict final velocity of m1 and m2.

Assume final distance to be v1 and v2

Assume the 2 masses at rest initially.

Take any values pointing to the direction of motion of m1 as positive; that is the final velocity of m1 will be positive and m2 will be negative.

So the f by the above definition will be negative (when real world values are taken).


This is not a HW problem...officially too, cause I think the solution is pretty complex in relation to a HW (no one answered it anyway in that section, so I put it here).
 
  • #4
This formula won't work for both the mass being equal, m1-m2 will churn 0 in that case.
 
  • #5


Assume they are initially at rest.

They repel each other for a distance s with force f = f(d), where d = s + a, a being the initial separation (perhaps zero).

W = S f ds, and W = KE_f - KE_i = KE_f - 0 = (1/2) m v^2, where v is the final velocity.

Note that m1 will have v > 0, and m2 will have v < 0, but |v| is the same for both, by symmetry.

Does that make sense? Am I just totally wrong?
 
  • #6


csprof2000 said:
f = f(d)

Ok...assume force not to be a function of distance...just assume that to make things simple.

d = s + a, a being the initial separation (perhaps zero).

Why has s been added to a?...I mean the distance traveled by both the bodies should be less than s.

Can you pls define the variables? :smile:
 
  • #7
Problem solved....
 

1. What is the formula for calculating the final velocity in a system of two repelling bodies?

The formula for calculating the final velocity in a system of two repelling bodies is v = √(2GM/r), where v is the final velocity, G is the gravitational constant, M is the mass of one body, and r is the distance between the two bodies.

2. How is this formula derived?

This formula is derived from the conservation of energy and the law of universal gravitation. By equating the initial kinetic energy of the bodies to the final gravitational potential energy, we can solve for the final velocity.

3. Can this formula be used for any two repelling bodies, regardless of their mass or distance?

Yes, this formula can be used for any two repelling bodies, as long as the distance between them is large enough to be considered negligible compared to their masses.

4. Is this formula accurate for all scenarios, or are there any limitations?

This formula is accurate for scenarios where the two bodies are moving in a straight line and no other external forces are acting on them. However, it may not be accurate for systems with multiple bodies or when the bodies are moving in a curved trajectory.

5. How can this formula be applied in real-life situations?

This formula can be applied in real-life situations to calculate the final velocity of objects in space, such as planets or satellites, as they orbit around each other. It can also be used to predict the motion of celestial bodies, such as comets or asteroids, as they travel through our solar system.

Similar threads

  • Advanced Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
817
Replies
5
Views
310
Replies
2
Views
971
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
581
Back
Top