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.
http://www.webassign.net/giancoli/7-44fig.gif ( link for Figure 7-44)
(a) What is the velocity of the lighter ball before impact? (Take the right to be positive.) =1.7155
(b) What is the velocity of each ball after the elastic collision?
ball A = -.471
ball B = 1.238
(c) What will be the maximum height of each ball (above the collision point) after the elastic collision?
ball A = ? m
ball B=0.78m
Can anyone please help me to get the height of BALL A ? ( The right answer is not 0.01m for it )
http://www.webassign.net/giancoli/7-44fig.gif ( link for Figure 7-44)
(a) What is the velocity of the lighter ball before impact? (Take the right to be positive.) =1.7155
(b) What is the velocity of each ball after the elastic collision?
ball A = -.471
ball B = 1.238
(c) What will be the maximum height of each ball (above the collision point) after the elastic collision?
ball A = ? m
ball B=0.78m
Can anyone please help me to get the height of BALL A ? ( The right answer is not 0.01m for it )
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a) Vai = √(2gh) = √(2g*.15) = 1.715 m/s
b)
Vcm = Vai*mA/(mA+mB) = .6173 m/s
uA = Vai - Vcm = 1.098 m/s
uB = Vcm = .6173 m/s
vA = Vcm - uA = -.4807 m/s
vB = Vcm + uB = 1.2346 m/s
Ha = vA²/2g = .0118 m
Hb = vB²/2g = .0778 m
These ARE the correct answers; if you use them to find the final PE, you will see that their sum equals the initial PE of the 36 g ball.......
b)
Vcm = Vai*mA/(mA+mB) = .6173 m/s
uA = Vai - Vcm = 1.098 m/s
uB = Vcm = .6173 m/s
vA = Vcm - uA = -.4807 m/s
vB = Vcm + uB = 1.2346 m/s
Ha = vA²/2g = .0118 m
Hb = vB²/2g = .0778 m
These ARE the correct answers; if you use them to find the final PE, you will see that their sum equals the initial PE of the 36 g ball.......