Conservation of Momentum and Kinetic Energy in 2D Collisions

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SUMMARY

The discussion focuses on the conservation of momentum and kinetic energy in a two-dimensional collision involving two objects of equal mass (2.0 kg). Object A, initially moving at 2.5 m/s, collides with Object B, which is at rest. The participant successfully applies the equations for momentum and kinetic energy conservation, ultimately deriving the post-collision velocities and angles through systematic substitution and simplification. The final results confirm the conservation principles, validating the calculations and approach taken.

PREREQUISITES
  • Understanding of basic physics concepts: momentum and kinetic energy
  • Familiarity with vector components in two-dimensional motion
  • Proficiency in solving algebraic equations and trigonometric functions
  • Knowledge of conservation laws in physics
NEXT STEPS
  • Study the derivation of momentum conservation equations in two dimensions
  • Learn about the application of trigonometric identities in physics problems
  • Explore advanced collision types, such as elastic and inelastic collisions
  • Investigate the role of angles in vector resolution during collisions
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and collision theory, as well as educators seeking to enhance their teaching of conservation laws in two-dimensional contexts.

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Homework Statement



Object A has a mass of 2.0 kg and an initial velocity of 2.5 m/s. It strikes Object B, which is at rest and has a mass of 2.0 kg as well.

After the collision, the objects travel in different directions, with Object B traveling at an angle of 44 degrees from its original position.

What is the velocity of Object B after the collision and what is the displacement angle of Object A from the point of collision?

Homework Equations



Momentum before collision = momentum after collision

Px = (m1 * vf1 * cosX) + (m2 * vf2 * cosY)

Py = (m1 * vf1 * sinX) + (m2 * vf2 * sinY)

KE before collision = KE after collision

KEf = (.5)(m1)vf12 + (.5)(m2)vf22

The Attempt at a Solution



I first calculated the components of momentum and kinetic energy before impact.

Px = 2 kg * 2.5 m/s = 5.0 kg*m/s
Py = 0 kg*m/s
KE = .5 * 2 kg * 2.5 m/s = 6.25 J

Then, I setup equations relating the objects post-impact to the momentum and energy they should have.

5 = 2*vf1*cosX + 2*vf2*cos44
0 = 2*vf1*sinX + 2*vf2*sin44

6.25 = vf12 + vf22

I've tried using substitution to solve for one of the variables, but each time I end up getting arcsines within cosines equaling sines. And I really don't know how to solve from there.

Am I at least on the right track? Should I solve for the angle first? Does it matter?
 
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This is good so far. Now suppose you took the first two (momentum conservation) equations and you solved for the components of the momentum of mass 1 in terms of the other quantities. In other words, put everything that has subscript 1 on one side and everything that has subscript 2 or is constant on the other side. What do you get?
 
Awesome. Thanks!

I super simplified the problem following your post:

M = m1v1 + m2v2

Mx = m1v1x + m2v2x = 5
My = m1v1y - m2v2y = 0

KE = .5m1v12 + .5m1v22

KE = (.5*m1*v1x2 + v1y2) + (.5*m2*v2x2 + v2y2)

Taking your advice, I basically got:

v1y = v2y

v1x = [tex]\frac{5}{m}[/tex] - v2x

Plugged it all in and got my answers! Thanks again.
 

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