How Do You Solve Momentum and Elastic Collision Problems in Physics?

  • Thread starter Thread starter krypt0nite
  • Start date Start date
  • Tags Tags
    2d Momentum
AI Thread Summary
To solve momentum and elastic collision problems, it's essential to apply the conservation of momentum and kinetic energy principles. In the given scenario, the steel ball's velocity changes direction after colliding with the rubber ball, indicating a need to analyze both x and y components of momentum. When considering elastic collisions, the two objects cannot stick together, and the equations derived from conservation laws must account for all variables. The discussion highlights the complexity of solving for unknown velocities, particularly when multiple solutions arise, suggesting the need for careful consideration of the collision dynamics. Understanding these principles is crucial for accurately determining post-collision speeds.
krypt0nite
Messages
31
Reaction score
0
A steel ball of mass 10kg moves due east at 5.0m/s. It collides with a rubber ball of mass 5.0 kg moving at 10m/s due north. After the collision the steel ball moves at an angle of 60 degrees East of North with a speed of 4.0 m/s.
If the collision were perfectly elastic what would be the speeds of the two balls after the collision?

Where do I start. I know total momentum before = total momentum after. But there seems to be 2 variables. vafter for steel ball and v after for rubber ball.

Can an elastic collision have the 2 objects even stick together?
If so then i got 1.92m/s.
 
Physics news on Phys.org
If the collision is elastic the 2 objects won't stisk together.

Try thinking of momentum as a vector. You can use triginometry for the angel, but I don't think it's really necesry.

P.S. don't forget that the mass of the 2 balls is constant.
 
You have two unknowns so you need two equations. Yes, one of the equations is "momentum before= momentum after", conservation of momenum. Since this collision is "elastic", the other is "kinetic energy before= kinetic energy after": conservation of energy.

Momentum is a vector quantity so you will need to consider x and y components separately. Kinetic energy is a "scalar" quantity so you really have 3 equations for 4 unknowns (x and y components of both velocities) but fortunately you are given the angles.
 
I need help on another question
A ball of mass 3.0kg moving at a speed of 3.0ms has a head-on collision with a stationary ball of mass 4.0kg.

If the collision were perfectly elastic what would be the speeds of the two balls after the collision?

I used the momentum before=momentum after equation and the KE before= Ke after equation.

I get weird answers. I get 2 answers. I get the velocity of the 3kg ball at -.42m/s and 3m/s. So what's the answer? If i know the velocity of the of this ball I can find the velocity of the other.
 
krypt0nite said:
I need help on another question
A ball of mass 3.0kg moving at a speed of 3.0ms has a head-on collision with a stationary ball of mass 4.0kg.

If the collision were perfectly elastic what would be the speeds of the two balls after the collision?

I used the momentum before=momentum after equation and the KE before= Ke after equation.

I get weird answers. I get 2 answers. I get the velocity of the 3kg ball at -.42m/s and 3m/s. So what's the answer? If i know the velocity of the of this ball I can find the velocity of the other.
Using only considerations of energy&momentum conservation before and after collision will always yield two sets of possible answers (solutions):
1) There hasn't been any interacting force between the two objects in the collision period.
That is, there wasn't any collision after all.
2) There has been a net momentum transfer (through a non-zero force) during collision.

Your 3m/s possibility is clearly of the trivial 1)-type.
 
Last edited:
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...
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...
Back
Top