How Far Does m1 Travel Before Colliding with m2?

AI Thread Summary
The discussion focuses on calculating the distance that mass m1 travels before colliding with mass m2, considering the inverse cube relationship of their attraction force. The initial assumption that the center of mass (C.O.M.) would be the collision point is challenged, as m1 has an initial velocity while m2 is at rest, meaning the C.O.M. is moving. This movement affects the distance m1 travels before collision, necessitating a reevaluation of the approach. Participants emphasize the importance of using dynamical equations or conservation of energy to accurately determine the collision time and distance. The conversation highlights the need to account for the motion of both particles rather than solely relying on the C.O.M. concept.
postfan
Messages
259
Reaction score
0

Homework Statement



Two small particles of mass m1 and mass m2 attract each other with a force that varies with the inverse cube of their separation. At time t0, m1 has velocity v directed towards m2, which is at rest a distance d away. At time t1, the particles collide.
How far does m1 travel in the time interval (t0 and t1)? Note: you may use t1 and t0 in your answer. Enter m1, m2, t1 and t0 for masses and times.

Homework Equations





The Attempt at a Solution



I used the center of mass as the collision point, so I calculated (taking x=0 at m1) it to be m2*d/(m1+m2). What did I do wrong?
 

Attachments

  • twodots.png
    twodots.png
    1.9 KB · Views: 612
Physics news on Phys.org
Why did you assume that the C.O.M. would be the collision point? If you jumped off the Brooklyn Bridge, is it reasonable to assume that you would travel to almost the center of the earth?
 
The title of the question literally was 'Center of Mass', in that case what should I do?
 
postfan said:
I used the center of mass as the collision point, so I calculated (taking x=0 at m1) it to be m2*d/(m1+m2). What did I do wrong?
You are overlooking that in the reference frame you are given m1 has an initial velocity but m2 does not. The COM of the system is therefore moving. This increases the distance m1 moves to the collision point.
Even without that complication, I don't believe your formula is right. What is the ratio of the two accelerations? If both started at rest, what would that mean about the ratio of the distances moved?
SteamKing said:
Why did you assume that the C.O.M. would be the collision point?
Because it tells you they are small particles, not bothering to give the actual sizes.
 
The particles will collide at the CM, but the CM moves with a constant velocity, and you need to figure out the time when they collide. It depends on the force of interaction between them, so you need to consider how the particles move, either solving the the dynamical equations F=m1a1 and -F=m2a2, or using conservation of energy. The force of interaction is conservative.

ehild
 

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 '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

Center of mass for two small particles
Replies
4
Views
5K
Solve Pulley-Mass System: m1=4.00 kg, m2=1.00 kg
Replies
2
Views
1K
A mass m1 moving with u collides with mas m2...
Replies
5
Views
2K
You're welcome, Rémi! Good luck with your exercise.
Replies
2
Views
2K
Maximum Angle for 2D Collision: Does θmax Exist if m1 > m2?
Replies
3
Views
3K
How do I solve for an elastic collision going up a ramp?
Replies
10
Views
3K
What Is the Maximum Height Achieved by Mass M1 After Elastic Collision?
Replies
3
Views
2K
Conservation of momentum - with understanding
Replies
2
Views
2K
Center of mass using Mach's restatement of Newtonian Mechanics
Replies
8
Views
2K
Center of Mass/Moment of Inertia Question
Replies
5
Views
4K

Hot Threads

Recent Insights

Back
Top