Elastic Collision in 2 Dimensions

eschw54

1. Homework Statement
A steel ball with a velocity of 2.7 m/s collides elastically with an identical stationary steel ball, and ends up going at a 40. degree angle with its original path. What is the direction of the other ball? What is the final velocity of each?

2. Homework Equations
Conservation of Momentum along x axis: m1v1i = m1v1fcostheta1 + m2v2fcostheta2
Conservation of Momentum along y axis: 0 = -m1v11fsintheta1 + m2v2fsintheta2
Conservation of Kinetic Energy for Elastic Collisions: .5m1v1^2 = .5m1f1f^2 + .5m2v2f^2

It's confusing with all the subscripts, but in short: p1initial + p2initial = p1final + p2final and Kinitial=Kfinal.

3. The Attempt at a Solution

I recognize that the balls are identical; thus, the masses cancel out.

Using the third equation (Conservation of KE): .5 (2.7^2) = .5v1f^2 + .5v2f^2
Solving for v1f: v1f = sqrt(7.29-v2f^2) This is all I have.

According to the answer key:
The direction of the other ball is 50 degrees.
v1f=1.7 m/s
v2f=2.1 m/s

I realize that you need to combine all three equations and use substitution to solve for the velocities and direction, but I'm not sure how. Any help would be greatly appreciated. Thank you!

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cupid.callin

instead of kinetic energy eqn try using eqn of coefficient of restitution

its : v2 - v1 = e(u1 - u2)

e = 1 for elastic collision

AlexChandler

It is not necessary to do the problem the way I am going to describe, but to me it is much more satisfying and enlightening to do it this way.

It may be helpful to view the system from its center of mass. The center of mass always lies halfway between the ball, and travels at a speed half that of the initial speed of the first ball.

$$v_{cm} = \frac{1}{2} * 2.7 \normaltext{m/s}$$

In the center of mass frame of reference, the final velocities will be pointed in opposite directions with an angle of 180 degrees between them.

You should be able to fully solve this problem by only considering vectors in "velocity space", A 2 dimensional space with axes vx and vy.
Establish two frames of reference, the stationary frame (Lab frame) and the center of mass frame. Draw all of the initial and final vectors in both frames, along with the center of mass velocity vector. Notice that the final angle for each ball with respect to the horizontal will be different in the two different frames. You may need the law of cosines to relate them.

I'm going to assume that the ball that is initially moving in the lab frame is on the left and the stationary ball is on the right. The initial velocity vector of the ball on the left, in the lab frame points along the vx axis with a magnitude of 2.7 m/s. The center of mass velocity points along the vx axis in the same direction with half the magnitude. In the center of mass frame, the ball on the left moves at half the speed as it does in the lab frame, and in the center of mass frame, the ball on the right moves to the left at half of 2.7 m/s.

Give this method a try if you want. And let me know if you need more help. This is a much more powerful and intuitive method than trying to solve algebraically in my opinion.

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