For some reason I\'ve had a hard time getting the answers. Two shuffleboard disk
ID: 2105222 • Letter: F
Question
For some reason I've had a hard time getting the answers.
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 5.20 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.
m/s (orange disk)
m/s (yellow disk)
Explanation / Answer
In this center-of-mass reference frame, the orange disk (black arrow in the figure) also has a velocity u=2.60 m/s, the yellow one (blue arrow in the figure) has exactly the same speed, but is in the opposite direction. This is because in the center-of-mass frame, the total momentum must zero, so the momenta of the two particles must add up to zero. And the two disks are of equal mass, so their velocities must be equally large and in opposite directions.
Due to the conservation of energy and momentum the same thing must be true after the collision. So the two disks must remain the same speed after the collision, and they must move in two opposite directions, as shown in the figure |AO|=|BO|=|CO|=|DO|. Note, this is only true in the center-of-mass frame, not the viewer's frame. However they can change a direction of their velocities, as also shown in the figure.
Now to go back to the viewer's frame, we need to add back the velocity of the center of the mass, which is u. Note this velocity never changes, because collision is due to the internal force. The addition of velocities follows the rules of adding vectors. We further exploit the fact that the velocity of the center of the mass is the same as the velocity of the orange disk in the center-of-mass frame, both are u , so we can just do it on the figure. In the figure, the velocities in the viewer's frame are shown by dashed arrows.
We now use a little geometry and trigonometry to calculate the final velocities. First notice |OA|=|OB|, so ∠BAO=∠ABO=40 °, and ∠AOB=100 °.
The speed of the orange disk is the length of the |AB|, which can be calculated from the cosines law,
|AB| ² = |OA| ² + |OB| ² − 2 |OA| |OB| cos(100 °)
==> |AB| = 4.14 m/s.
The speed of the yellow disk is the length of the |AC|, and it can be calculated similarly,
|AC| ² = |OA| ² + |OC| ² − 2 |OA| |OC| cos(80 °)
==> |AC| = 3.47 m/s.
In the viewer's frame, the velocities of the two disks are always perpendicular after the collision, as evidenced in the figure AB ⊥AC. This is because the ABO and ACO are equilateral triangles. So ∠BAO+∠CAO = ∠ABO+∠ACO = (1/2) 180 ° = 90 °.