- #1
chazgurl4life
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Two balls, of masses mA = 32 g and mB = 80 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.)
a) What is the velocity of the lighter ball before impact? (Take the right to be positive.)
i have been using the formula V=M+m
---- X (2gl(1-cos of the angle)^1/2
m
so when i put it all together i get :
V= .08kg+.032Kg
---------- X (2*9.8*.3m*1-cos60 degrees)^1/2
.032 kg
V=3.5 X(2.94)^1/2
V=6 m/s
but apparently this answer is wrong ..can anyone help ??
a) What is the velocity of the lighter ball before impact? (Take the right to be positive.)
i have been using the formula V=M+m
---- X (2gl(1-cos of the angle)^1/2
m
so when i put it all together i get :
V= .08kg+.032Kg
---------- X (2*9.8*.3m*1-cos60 degrees)^1/2
.032 kg
V=3.5 X(2.94)^1/2
V=6 m/s
but apparently this answer is wrong ..can anyone help ??