*66. Go For the situation depicted in the figure, use momentum conserva- tion to determine the magnitude and direction of the final velocity of ball I after the collision. 401 = 0.900 m/s m = 0.150 kg 50.0 +x 35.0 102 = 0.540 m/s m, = 0.260 kg (b) %3D (a) Top view of two balls colliding on a horizontal surface. (b) This part of the drawing shows the x and y components of the velocity of hall I after the collision. 42 = 0.700 m/s (a) Problem 66

College Physics
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Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
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*66. Go For the situation depicted in the figure, use momentum conserva-
tion to determine the magnitude and direction of the final velocity of ball
I after the collision.
401 = 0.900 m/s
m = 0.150 kg
50.0
+x
35.0
102 = 0.540 m/s
m, = 0.260 kg
(b)
%3D
(a) Top view of two balls
colliding on a horizontal surface.
(b) This part of the drawing
shows the x and y components
of the velocity of hall I after the
collision.
42 = 0.700 m/s
(a)
Problem 66
Transcribed Image Text:*66. Go For the situation depicted in the figure, use momentum conserva- tion to determine the magnitude and direction of the final velocity of ball I after the collision. 401 = 0.900 m/s m = 0.150 kg 50.0 +x 35.0 102 = 0.540 m/s m, = 0.260 kg (b) %3D (a) Top view of two balls colliding on a horizontal surface. (b) This part of the drawing shows the x and y components of the velocity of hall I after the collision. 42 = 0.700 m/s (a) Problem 66
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