Collisions in Two Dimensions (Perfectly Elastic)

  • Thread starter calvert11
  • Start date
  • #1
32
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Homework Statement


Assume an elastic collision (ignoring friction
and rotational motion).
A queue ball initially moving at 4.3 m/s
strikes a stationary eight ball of the same size
and mass. After the collision, the queue ball’s
final speed is 2.3 m/s .

Find the queue ball’s angle with respect
to its original line of motion.

v1i = 4.3
v1f = 2.3

Homework Equations


p1i + p2i = p1f + p2f
KEi = KEf

The Attempt at a Solution



sqrt(4.3^2 - 2.3^2) = v2f = 3.6 m/s

4.3 = 2.3*cos θ + 3.6*cos φ (for x direction)

0 = 2.3*sin θ - 3.6*sin φ (for y direction)

2.3*sin θ = 3.6*sin φ

Not quite sure how to proceed from here.
 

Answers and Replies

  • #2
152
0

Homework Statement


Assume an elastic collision (ignoring friction
and rotational motion).
A queue ball initially moving at 4.3 m/s
strikes a stationary eight ball of the same size
and mass. After the collision, the queue ball’s
final speed is 2.3 m/s .

Find the queue ball’s angle with respect
to its original line of motion.

v1i = 4.3
v1f = 2.3

Homework Equations


p1i + p2i = p1f + p2f
KEi = KEf

The Attempt at a Solution



sqrt(4.3^2 - 2.3^2) = v2f = 3.6 m/s

4.3 = 2.3*cos θ + 3.6*cos φ (for x direction)

0 = 2.3*sin θ - 3.6*sin φ (for y direction)

2.3*sin θ = 3.6*sin φ

Not quite sure how to proceed from here.

[tex]m_1u_1+m_2u_2=m_1_v_1+m_2v_2\rightarrow m(u_1+0)=m(v_1+v_2)[/tex].
 
  • #3
32
0
[tex]m_1u_1+m_2u_2=m_1_v_1+m_2v_2\rightarrow m(u_1+0)=m(v_1+v_2)[/tex].

I'm not sure how that helps . The masses are equal so they drop from the equations and the velocities are known.
 
  • #4
152
0
i'm not sure how that helps . The masses are equal so they drop from the equations and the velocities are known.

4.3 = 2.3 + v...
 
  • #5
32
0
4.3 = 2.3 + v...
Still don't know what you're getting at :).

But I managed to solve it using pythagoras: (4.3-2.3 cos θ)^2 + (2.3 sin θ)^2 = 3.6^2

With that I get around 57 degrees which is correct.

Just out of curiosity, would there have been a simpler way of solving this? My solution required using power-reducing formulas for sin/cos...which seems too complicated considering the level of the course.
 

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