Question

Two shuffleboard disks of equal mass, one orange and the other yellow, are involved in an elastic, glancing collision. The yellow disk is initially at rest and is struck by the orange disk moving with a speed of 5.60 m/s. After the collision, the orange disk moves along a direction that makes an angle of 40.0° with its initial direction of motion. The velocities of the two disks are perpendicular after the collision. Determine the final speed of each disk.
orange disk m/s
yellow disk m/s

Answer 1

I'm going to work this out in steps, so you might see a few comments, or maybe just one.

First thing you need to realize is that the impact will result in the two particles (disks) to move perpendicular to each other.

From here, think of conservation laws for y and for x. since the mass is equal, you can leave it out. I will include it in the general conservation equation for reference.

\[m _{1i}v _{1i} + m_{1i}v_{1i} = m_{1f}v_{1f} + m_{2f}v_{2f}\]

Since we are working with two dimensions, and vectors, we need to break this equation into its components. Remember, there is no initial velocity for disk 2 and no initial velocity for the y component at all. Thus our three basic equations will be as follows.
1: \[v _{1\xi} = v_{1xf} {\cos \theta}+ v_{2xf}{\cos \phi}\]
2: \[0 = v_{1xf} {\sin \theta} - v_{2xf}{\sin \phi}\]
3: \[v_{1i}^2 = v_{1f}^2 + v_{2f}^2 \]

next, solve equations 1 and 2 for v2f

4:\[v_{2xf}{\cos \phi} = v _{1\xi} - v_{1xf} {\cos \theta} \]
5. \[v_{2xf}{\sin \phi} = v_{1xf} {\sin \theta} \]

Now, you'll need to square both equations (4 and 5) and add them together.

\[v_{2f}^2({\cos^2 \phi +\sin^2 \phi})= v _{1 x i}^2 - 2v_{x i}v_{1xf} {\cos \theta} + v_{1f}^2({\cos^2 \theta +\sin^2 \theta})\]

using the Pythagorean identity this becomes:
6. \[v_{2f}^2= v _{1 x i}^2 - 2v_{x i}v_{1xf} {\cos \theta} + v_{1f}^2\]
Substitute this solution in into equation 3:
\[v_{1i}^2 = v_{1f}^2 + v _{1 x i}^2 - 2v_{x i}v_{1xf} {\cos \theta} + v_{1f}^2\]
=> \[(v_{1i}^2 - v _{1 i}^2) = 2v_{1f}^2 - 2v_{ i}v_{1f} {\cos \theta} \rightarrow 0 = v_{1f}^2 - v_{ i}v_{1f} {\cos \theta}\] Set the resulting equation equal to v1f and solve.

7.\[ v_{1f} = v_{ i} {\cos \theta} \] Insert values => 5.6 cos 40 = 4.29m/s
8. set eq 3 = \[v_{2f} = \sqrt{v_{1f}^2 - v_{1i}^2 }\]
insert values and solve: v2f = 3.60m/s

Also, if the mass were included and they were different, the discs wouldn't diverge at 90º. To get the angles above and below horizontal, merely take equation 2 and solve for whichever angle you want.