10. 2 points OSUniPhys1 9.5.WA.05 My Notes Ask Your Two ice pucks (one orange an
ID: 2034815 • Letter: 1
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10. 2 points OSUniPhys1 9.5.WA.05 My Notes Ask Your Two ice pucks (one orange and one blue) of equal mass are involved in a perfectly elastic glancing collision as shown in the figure below. The orange puck is initially moving to the right at vor = 7.60 m/s, strikes the initially stationary blue puck, and moves off in a direction that makes an angle of ? = 39.0° with the horizontal axis while the blue puck makes an angle of ? = 51.0° with this axis as in figure (b). Note that for an elastic collision of two equal masses, the separation angle? + ? = 90.0°. Determine the speed of each puck after the collision. Vor m/s m/s Before collision After collision Additional Materials eBookExplanation / Answer
m1v1 = m1v1' + m2v2 (conservation of momentum); however since m1 = m2
v1 = v1' + v2 (where v are vectors, not scalars)
Therefore
Vo = Vo'*cos(th) + Vb*cos(ph), (x direction) and
Vo'*sin(th) = Vb*sin (ph) or Vb = Vo'* (sin(th)/sin(ph)) (y direction; basically the two vertical vectors must cancel one another out since the original vector had no vertical vector component)
Substituting
Vo = Vo'*cos(th) + Vo'* (sin(th)/sin(ph))*cos(ph)
7.60m/s = Vo'*(cos39° + sin39°/tan51°), or
Vo' = 5.90m/s
Vb = 5.90m/s*(sin39°/sin51°) = 4.78m/s
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