- #1
josephyarwick
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Homework Statement
Two balls, of masses mA=45g and mB=65g, are suspended as shown in figure 9-53. The lighter ball is pulled away to a 66 degree angle with the vertical and released. (a.) What is the velocity of the lighter ball before impact? (b.) What is the velocity of each ball after the elastic collision? (c.) What will be the maximum height of each ball after the elastic collision?
mA=.045kg mB=.065 theta(0)=66 degrees L=.3m
Homework Equations
1/2mv^2+mgh=1/2mv^2+mgh
1/2mv^2+1/2mv^2=1/2mv^2+1/2mv^2
The Attempt at a Solution
I'm very lost, I originally tried to find (a.) by setting v=(sqrroot)(2g(.3)sin(66)) but I got 2.3 m/s which is supposedly incorrect. I then tried replacing sin(theta) with (1-cos(theta)) and somehow got the right answer, 1.9 m/s. For part (b.) I tried to find the velocity immediately after the collision by solving for one of the v(prime) in both the conservation of kinetic energy and the conservation of momentum equations. I then substituted the results but I my answer was .9 m/s for v(prime)A when it was supposed to be -.3 m/s. please help