Two shuffleboard disks of equal mass, one orange and the other green, are involv
ID: 2058356 • Letter: T
Question
Two shuffleboard disks of equal mass, one orange and the other green, are involved in a perfectly elastic glancing collision. The green disk is initially at rest and is struck by the orange disk moving initially to the right at varrowboldoi = 7.60 m/s as in Figure (a) shown below. After the collision, the orange disk moves in a direction that makes an angle of ? = 34.0° with the horizontal axis while the green disk makes an angle of phi = 56.0° with this axis as in figure (b). Determine the speed of each disk after the collision.vof = m/s
vgf = m/s
Explanation / Answer
Momentum must be conserved. The initial momentum of the system is only from the orange disk moving to the right
p = mv = (7.6 m) This momentum is in the positive x direction only. That will be have to be the final total as well to conserve momentum overall
Since the figure is not given, I will assume that the 34o is an upward angle, and 56o is downward
The orange disk, after the collision will have an x and y component of momentum
The green disk, after the collision will have an x and y component of momentum
The y components will have to cancel out to leave a net of zero
The x components will have to add to give a momentum of 7.6m in the positive x direction
In the y direction,
m(v sin 34) orange = m (v sin 56) green Note m cancels
so
(vorangesin 34) = (vgreensin 56)
.559 vorange = .829 vgreen
vorange = 1.48vgreen
In the x direction
m(v cos 34) orange + m(v cos 56) green = 7.6m (again m cancels)
.829 vorange + .559vgreen = 7.6
Substitue from the y direction where we found vorange = 1.48vgreen
.829(1.48vgreen) + .559vgreen = 7.6
1.23vgreen + .559vgreen = 7.6
1.79vgreen = 7.6
vgreen = 4.26 m/s
Then, since vorange = 1.48vgreen
vorange = (1.48)(4.26)
vorange = 6.30 m/s
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