Two body final velocities (check this please)

  • Thread starter Thread starter dean barry
  • Start date Start date
  • Tags Tags
    Body Final
AI Thread Summary
The discussion focuses on calculating the final velocities of two masses, m1 and m2, that are released from rest and move towards each other under gravitational attraction. The total kinetic energy (KE) at the final distance d is derived from the difference in gravitational potential energy at distances D and d. The kinetic energy is then split according to the mass ratio, leading to the final velocities expressed in terms of their respective kinetic energies. Key principles highlighted include the conservation of linear momentum and mechanical energy, emphasizing that the ratio of kinetic energies does not equate to the ratio of masses. The concept of lost mass is also suggested as a potential consideration in the calculations.
dean barry
Messages
311
Reaction score
23

Homework Statement


Two masses m1 & m2 in free space are released from rest at a distance D and draw together under gravity to a distance d, calculate the final velocity of each mass.


Homework Equations


Final total KE of both bodies at d from :
KE (t) = ( ( G * m1 * m2 ) / d ) - ( ( G * m1 * m2 ) / D )
(Joules)



The Attempt at a Solution


Split the KE (t) by mass ratio, so :
KE (m1) = ( m2 / ( m1 + m2 ) ) * KE (t)
KE (m2) = ( m1 / ( m1 + m2 ) ) * KE (t)

The final velocities from :
v (m1) = sqrt ( ( KE ( m1 ) ) / ( ½ * m1 ) )
v (m2) = sqrt ( ( KE ( m2 ) ) / ( ½ * m2 ) )

The final momentums are equal, suggesting a good process, comments please.

 
Physics news on Phys.org
dean barry said:

Homework Statement


Two masses m1 & m2 in free space are released from rest at a distance D and draw together under gravity to a distance d, calculate the final velocity of each mass.


Homework Equations


Final total KE of both bodies at d from :
KE (t) = ( ( G * m1 * m2 ) / d ) - ( ( G * m1 * m2 ) / D )
(Joules)



The Attempt at a Solution


Split the KE (t) by mass ratio, so :
KE (m1) = ( m2 / ( m1 + m2 ) ) * KE (t)
KE (m2) = ( m1 / ( m1 + m2 ) ) * KE (t)

The final velocities from :
v (m1) = sqrt ( ( KE ( m1 ) ) / ( ½ * m1 ) )
v (m2) = sqrt ( ( KE ( m2 ) ) / ( ½ * m2 ) )

The final momentums are equal, suggesting a good process, comments please.

Ratio of kinetic energies of two particles is NOT the same as ratio of their masses.

Use following laws:

1. Conservation of linear momentum
2. Conservation of mechanical energy.

CONCEPT of LOST MASS might also be useful.
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'Collision of a bullet on a rod-string system: query'

Thread 'Collision of a bullet on a rod-string system: query'

In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

How to find the equilibrium position of three masses (two of them fixed)?
Replies
4
Views
2K
Why is my solution to finding the spring constant incorrect?
Replies
3
Views
2K
Vertical spring connected between two bodies with a rope above them
Replies
16
Views
2K
Two dimensional momentum problem
Replies
20
Views
3K
Escape velocity of an iron asteroid
Replies
5
Views
2K
Two manned satellites approaching one another
Replies
4
Views
4K
Finding the Line of Motion of Two Particles
Replies
16
Views
1K
Find second object's velocity in relativity
Replies
2
Views
1K
Kleppner - blocks and a compressed spring
Replies
3
Views
3K
What if check: Am I calculating tension wrong?
Replies
3
Views
2K

Hot Threads

Recent Insights

Back
Top