Angle of releasing a pendulum and the speed of the ball hit by it I have conducted the experiment. My independent variable is the angle of releasing the pendulum and my dependent variable is the speed of the ball that is hit by it. I know the following quantities: masses of the pendulum ball and the grounded ball, different angles with different velocities which I measured in my experiment.
I need help with finding equations that are suitable for my experiment and theories that I can use to make my research more interesting. Please note that I'm a high school student who is studying an advanced course, but certainly not university level.
 A: This question can be answered using high school physics only. Let $1$ be the pendulum and $2$ be the ball, and let $i$ and $f$ be the initial state (just before the impact) and the final state (just after the impact).
The velocity at which the pendulum hits the ball can be found imposing the conservation of energy from the initial position of the pendulum to the instant of the impact.
You should get that
$$
v_{1,i} = \sqrt{ 2gR ( 1-\cos{\theta_i} ) },
$$
where $R$ is the length of the rope, and $\theta_i$ is the initial angle.
After that the impact is governed by conservation of momentum, which relates the initial velocity of the pendulum to the final velocities of the ball and the pendulum:
$$
m_1 v_{1,i} = -m_1 v_{1,f} + m_2 v_{2,f}.
$$
The only unknown variable at this stage is $v_{1,f}$, that can be obtained measuring the maximum angle of the pendulum after the collision and imposing again energy conservation to get
$$
v_{1,f} = \sqrt{ 2gR ( 1-\cos{\theta_f} ) }.
$$
I have left some calculations as an exercise for you, but hopefully I have answered :)
PS: I have neglected any friction force and I have assumed that the pendulum and the ball do not stick together in the impact.
A: If you consider the pendulum, the Earth's gravitational field, and the grounded ball as your system, you could use constant total mechanical energy:
$$K_{1}+U_{g1}=K_{2}+U_{g2},$$
where $K$ is the total kinetic energy, $U$ is the gravitational potential energy of the system (assume a constant field strength, aka, "acceleration due to gravity" of $g$) with an arbitrary fixed zero location, and the subscripts $1$ and $2$ designate two different positions of the pendulum before the collision.
You will need some trigonometry, but the algebra should be fairly simple. Draw pictures and label all your dimensions, coordinates, and distances.
Above all, have phun!
A: Step 1: Form equations from known data
Let $m_1$ and $m_2$ be the masses of the pendulum bob and grounded ball respectively. Let $u_1,u_2,v_1$ and $v_2$ be the initiall and final velocities of both the bodies. Let $R$ be the distance between the  point of suspension of pendulum and centre of mass of the bob.
$\bullet$ Coefficient of restitution $e$ is constant for two given materials over low speeds and is defined as:
$e=\frac{velocity-of-separation}{velocity-of-approach}$
$e=\frac{v_2 - v_1}{u_1-u_2}$
$\bullet$ Using Principle of Conservation of Energy, form the equation for velocity at lowest position:
$m_1gR(1-\cos(\theta))=\frac{1}{2}m_1{v_1}^2$
$\bullet$ From Conversation of Momentum, form:
$m_1u_1+m_2u_2=m_1v_1+m_2v_2$
Step 2: Eliminate $em_1v_1$
Multiply the first equation with $m_1$, and second equation with $e$. Then substract either of the new equations from the other. On rearranging the term you will get a equation for $v_2$ which is independent of $v_1$. Please bear in mind that $u_2$ is 0. You can take care of that in the beginning.
I believe that my equation is better than Matteo's since you don't have to measure $\theta_f$.
Note that in ideal case $e=1$ where both bodies are perfectly elastic.
