- #1
anightlikethis
- 10
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"Simple" part of a hard problem involving momentum/kinetic energy conservation
Two balls, of masses mA = 36 g and mB = 64 g are suspended as shown in Figure 7-44. The lighter ball is pulled away to a 60° angle with the vertical and released.
(a) What is the velocity of the lighter ball before impact? (Take the right to be positive.)
c) What will be the maximum height of each ball (above the collision point) after the elastic collision?
I found the velocity for ball A which I know to be correct -.48 m/s with the kinetic energy and momentum conservation equations, and also the ball B speed 1.23 which I'm not sure about. The part I'm having problems with is the heights. I've been told to use this equation: mgh=1/2mv^2, but this equation gives me .01 for Ball A which is not correct. What am I doing wrong?
Two balls, of masses mA = 36 g and mB = 64 g are suspended as shown in Figure 7-44. The lighter ball is pulled away to a 60° angle with the vertical and released.
(a) What is the velocity of the lighter ball before impact? (Take the right to be positive.)
c) What will be the maximum height of each ball (above the collision point) after the elastic collision?
I found the velocity for ball A which I know to be correct -.48 m/s with the kinetic energy and momentum conservation equations, and also the ball B speed 1.23 which I'm not sure about. The part I'm having problems with is the heights. I've been told to use this equation: mgh=1/2mv^2, but this equation gives me .01 for Ball A which is not correct. What am I doing wrong?