How Does Mass Affect Velocity in a Two-Ball Pendulum Collision?

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SUMMARY

The discussion centers on calculating the velocity of a lighter ball (mass mA = 32 g) in a two-ball pendulum collision scenario. The user initially applied the formula V = (M + m) / m * √(2gl(1 - cos(θ))) but arrived at an incorrect result of 6 m/s. The correct approach emphasizes using the formula [2gl(1 - cos(θ))]^(1/2) directly to find the velocity before impact, which simplifies the calculation and avoids unnecessary complexity.

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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 ??
 
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You solved too far. The answer is in your question.(modified slightly with brackets).
chazgurl4life said:
a) What is the velocity of the lighter ball before impact? (Take the right to be positive.)

[2gl(1-cos of the angle)]^1/2
 

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