2 Two shuffleboard disks of equal mass, one orange and the other yellow,
3 are involved in an elastic, glancing collision. The yellow disk is
4 initially at rest and is struck by the orange disk moving with a speed
5 $v_i$. After the collision, the orange disk moves along a direction
6 that makes an angle $\theta$ with its initial direction of motion.
7 The velocities of the two disks are perpendicular after the collision.
8 Determine the final speed of each disk.
12 Let the final speed of the orange disk be $v_o$, the final speed of
13 the yellow disk be $v_y$, and $m$ be the mass of one disk. Calling
14 the initial direction of the orange disk \ihat, and the direction
15 perpendicular to that \jhat\ (such that the final direction of
16 $\vect{v}_o$ has positive components in both directions), we see
18 v_{o\ihat} &= v_o \cos\theta \\
19 v_{o\jhat} &= v_o \sin\theta
21 For the orange puck, and that since the motion of the yellow is
22 perpendicular the the orange, the angle between the final motion of
23 the yellow and the $-\jhat$ direction is also $\theta$, so
25 v_{y\ihat} &= v_y \sin\theta \\
26 v_{y\jhat} &= -v_y \cos\theta
29 Conserving momentum in both directions we have
31 P_{i\jhat} = 0 &= P_{f\jhat} = m v_{y\jhat} + m v_{o\jhat} = m v_o \sin\theta - m v_y\cos\theta \\
32 v_y &= v_o \frac{\sin\theta}{\cos\theta}\\
33 P_{i\ihat} = m v_i &= P_{f\ihat} = m v_{y\ihat} + m v_{o\ihat} = m v_o \cos\theta + m v_y \sin\theta \\
34 v_i \cos\theta &= v_o \cos^2 \theta + \left(v_o \frac{\sin\theta}{\cos\theta}\right)\sin\theta\cos\theta
35 = v_o \cos^2 \theta + v_o \sin^2 \theta
40 \sin^2 \theta + \cos^2 \theta = 1
44 v_o &= \ans{v_i \cos\theta} \\
45 v_y &= v_o \frac{\sin\theta}{\cos\theta} = \ans{v_i \sin\theta}