Conservation of Momentum and Kinetic Energy in 2D Collisions

In summary, Object A has a mass of 2.0 kg and an initial velocity of 2.5 m/s. It strikes Object B, which has a mass of 2.0 kg and is at rest. After the collision, the objects travel in different directions with Object B traveling at an angle of 44 degrees from its original position.
  • #1
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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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  • #2
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?
 
  • #3
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.
 

FAQ: Conservation of Momentum and Kinetic Energy in 2D Collisions

What is 2D conservation of momentum?

The 2D conservation of momentum is a fundamental law in physics that states that the total momentum of a system remains constant in all directions, as long as there are no external forces acting on the system.

How is 2D conservation of momentum different from 1D conservation of momentum?

In 1D conservation of momentum, the total momentum is only conserved in one direction. However, in 2D conservation of momentum, the total momentum is conserved in both the x and y directions simultaneously.

Why is 2D conservation of momentum important?

2D conservation of momentum is important because it helps us understand and predict the motion of objects in two dimensions. It is also a fundamental principle in many areas of physics, such as mechanics, fluid dynamics, and electromagnetism.

What are some real-world applications of 2D conservation of momentum?

2D conservation of momentum is applied in various fields, such as rocket propulsion, collisions between objects, and motion of fluids. It is also used in sports, such as billiards and ice hockey.

How can the 2D conservation of momentum equation be used to solve problems?

To solve problems involving 2D conservation of momentum, we can use the equation: ΣPinitial = ΣPfinal. This equation states that the sum of the initial momenta of all objects in a system is equal to the sum of the final momenta of all objects in the system. By setting up and solving equations using this principle, we can determine the final velocities or momenta of objects in a given scenario.

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