Urgent!! Pendulum problem elastic collision 1. The problem statement, all variables and given/known data An elastic ball (A) with mass m is released from a horizontal position, connected via a massless string to a rod. When the ball reaches the bottom, it collides with another elastic ball (B) of twice the mass, aslo connected via a massless string to the rod. How high can each ball swing, respectively? (The collision is total elastic.) Assume the +x direction is to the right a)What is the velocity of ball A after the collision? (5pt) b)What is the velocity of ball B after the collision? (5pt) c)How high will ball A swing relative to the bottom position? (5pt) d)What is magnitude of the tension force (T) in the string connected to the ball A at the moment right after the collision? (5pt) A picture of the problem can be found at http://www2.fiu.edu/~leguo/Site/PHY2048_files/ExtraCredit2.pdf 2. Relevant equations 3. The attempt at a solution M=Mtotal=3m Va= initial velocity A Vb= initial Velocity B V1= final velocity A V2= final velocity B pi= initial momentum pf= final momentum ki = initial kinetic energy kf= final kinetic energy So what I first did was I wanted to find the initial velocity ( Va) before the collision. To that I did: 1/2MVa^2=MgH Va= √2gh The collision is total elastic which means momentum and energy is conserved Pi=pf MaVa=Mav1+Mbv2 Im going to solve for v2: MaVa-Mav1/Mb=v2 Plugging in known variables m√2gh-mv1/2m=v2 m's cancel and you get √2gh-v1/2=v2 Here is where I get confused my main problem lies with h. So I want to make that h cancel since it is circular motion as the picture shows I try using uniform circular motion g=Va^2/H H= Va^2/g I want to make sure if im doing good so far. This is a problem that will be on my exam tomorrow so much help would be appreciated.