Analyzing Two-Dimensional Collisions: Solving for Final Velocities

In summary, the pucks fly apart with the angles shown in the drawing and the final speed of puck A is 5.5m/s and the final speed of puck B is 0m/s.
  • #1
chmilne
10
0
The Problem:

The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of 0.025 kg and is moving along the x-axis with a velocity of +5.5 m/s. It makes a collision with puck B, which has a mass of 0.050 kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with the angles shown in the drawing. Find the final speed of (a) puck A and (b) puck B.

Work Done Thus Far:

I know that before the collision, puck A has momentum in its direction of motion, and all the energy (kinetic). After the collision the momentum of the two pucks must equal the original momentum of the first puck and the sums of their energies must equal the initial energy of puck A. So, here's what I've come up with.

[tex]P_{O1}[/tex] = [tex]M_1[/tex][tex]V_{O1}[/tex]
[tex]P_{O1}[/tex] = .025kg * 5.5m/s
[tex]P_{O1}[/tex] = .1375 kg*m/s

X-Component
[tex]M_{f1x}[/tex][tex]V_{f1x}[/tex] + [tex]M_{f2x}[/tex][tex]V_{f2x}[/tex] = [tex]M_{o1x}[/tex][tex]V_{o1x}[/tex] + [tex]M_{o2x}[/tex][tex]V_{o2x}[/tex]
.025kg([tex]V_{f1x}[/tex]) + .05kg([tex]V_{f2x}[/tex]) = .025kg(Vo1x) + .05kg(Vo2x)
.025kg([tex]V_{f1x}[/tex]) + .05kg([tex]V_{f2x}[/tex]) = .025kg(5.5m/s) + .05kg(0m/s)
.025kg([tex]V_{f1x}[/tex]) + .05kg([tex]V_{f2x}[/tex]) = .025kg(5.5m/s) + 0m/s
.025kg([tex]V_{f1x}[/tex]) + .05kg([tex]V_{f2x}[/tex]) = .1375kg*m/s
.025kg([tex]V_{f1}[/tex]cos65) + .05kg([tex]V_{f2}[/tex]cos37) = .1375kg*m/s

Y-Component
[tex]M_{f1y}[/tex][tex]V_{f1y}[/tex] + [tex]M_{f2y}[/tex][tex]V_{f2y}[/tex] = [tex]M_{o1y}[/tex][tex]V_{o1y}[/tex] + [tex]M_{o2y}[/tex][tex]V_{o2y}[/tex]
.025kg([tex]V_{f1y}[/tex]) + .05kg([tex]V_{f2y}[/tex]) = .025kg(0m/s) + .05kg(0m/s)
.025kg([tex]V_{f1y}[/tex]) + .05kg([tex]V_{f2y}[/tex]) = 0
.025kg([tex]V_{f1y}[/tex]) + .05kg([tex]V_{f2y}[/tex]) = 0
.025kg([tex]V_{f1y}[/tex]) + .05kg([tex]V_{f2y}[/tex]) = 0
.025kg([tex]V_{f1}[/tex]sin65) + .05kg([tex]V_{f2}[/tex]sin37) = 0



So,

.1375 kg*m/s = (.025kg * [tex]V_{f1}[/tex]cos(65)) + (.05kg([tex]V_{f2}[/tex]cos(37))

I'm not sure what I missing to solve for the velocity of both the pucks.
 
Last edited:
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  • #2
You seem to have ignored your y-axis conservation equation. You have two equations and two unknowns, so a solution can be found fairly easily through substitution.
 
  • #3
I'm not following you. Will you please elaborate and be more specific?
 

What is a collision in two dimensions?

A collision in two dimensions occurs when two objects come into contact with each other and exchange energy and momentum. This can happen in any direction within a two-dimensional space.

What factors affect the outcome of a collision in two dimensions?

The outcome of a collision in two dimensions depends on several factors, including the masses and velocities of the objects involved, the angle of impact, and the elasticity of the objects.

How is momentum conserved in a collision in two dimensions?

According to the law of conservation of momentum, the total momentum of a system remains constant before and after a collision. This means that the total momentum of both objects in a collision in two dimensions will remain the same, even if the individual momenta change.

What is the difference between an elastic and an inelastic collision in two dimensions?

In an elastic collision, both the total kinetic energy and the total momentum of the system are conserved. In an inelastic collision, some of the kinetic energy is lost and the objects may stick together after the collision.

How can the equations for a collision in one dimension be applied to a collision in two dimensions?

The equations for a collision in one dimension can be applied to a collision in two dimensions by breaking down the motion in each direction separately. This allows for the conservation of momentum and energy to be examined in each direction, and then combined to determine the overall outcome of the collision.

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