# Engineering Dynamics - Conservation of Energy

1. Sep 23, 2014

### ConnorM

1. The problem statement, all variables and given/known data
Two spheres are hanging from cords as shown. The distance from the ceiling to the center of each sphere is 2m, and the coefficient of restitution is 0.75. If sphere "A" $(m_a = 2kg)$ is drawn back $60^o$ and released from rest, determine:

a) The maximum angle, $θ_B$ , that sphere "B" $(m_B = 3kg)$
b) The angle, $θ_B$, that sphere "A" will rebound as a result of the impact.

2. Relevant equations
$T=1/2 *mv^2$

$V=mgh$

$\sum T_1 + \sum V_1 = \sum T_2 + \sum V_2$

$C_r = \frac{v_B_2 - v_A_2 }{v_A_ - v_B_}$

3. The attempt at a solution
I started by finding the height "A" was released from, since it is pulled back $60^o$,
$sin60^o = \frac{2}{h}$

$2sin60^o= \sqrt{3}$

Next I determined the $V_1$ that "A" had at that height and set it equal to it's velocity just as it reaches sphere "B",

$m_A g h_A = 1/2 m_A v_A ^2$

$2kg * 9.81 m/s^2 * \sqrt{3} = 1/2 * 2kg * v_a ^2$

$v_a = 5.829 m/s$

Next I used my equation for restitution,

$C_r = v_B_2 - v_A_2 / v_A_ - v_B_$

$0.75 = v_B_2 - v_A_2 / 5.829 m/s - 0$

$v_B_2 = 4.372 + v_A_2$

from here I subbed into my conservation of energy equation for when the balls first make contact to when they have separated,

$1/2 m_A v_A ^2 =1/2 m_A v_A_2 ^2 + 1/2 m_B v_B_2 ^2 + m_A g h_A_2 + m_B g h_B$

I'm not quite sure what to do now, I don't think I can solve this since I have too many unknowns.

Last edited: Sep 23, 2014
2. Sep 24, 2014

### Staff: Mentor

In your formula for Cr you substituted for the VA term in the denominator but not for VA in the numerator. Is there a reason for your doing this?