Question

a 0.040 kg bullet was traveling at 150 m/s, inelastically collided with a 5 kg pendulum that was at rest. what is the maximum height h can the pendulum go after the collision ignoring the air resistance

Answer 1

We will have the following:

First, using the conservation of momentum to find the velocity of the system, that is:

$\begin{gathered} (0.040kg)(150m/s)+(5kg)(0m/s)=(0.040kg+5kg)v_f \\ \\ \Rightarrow v_f=\frac{(0.040kg)(150m/s)}{(0.040kg+5kg)}\Rightarrow v_f=\frac{25}{21}m/s \end{gathered}$

Now that we have the initial velocity we will use conservation of energy to determine the height of the pendulum:

$\begin{gathered} \frac{1}{2}(0.040kg+5kg)(\frac{25}{21}m/s)=(0.040kg+5kg)(9.8m/s^2)h \\ \\ \Rightarrow\frac{1}{2}(\frac{25}{21}m/s)=(9.8m/s^2)h\Rightarrow h=\frac{(1/2)(25/21m/s)}{(9.8m/s^2)} \\ \\ \Rightarrow h=\frac{125}{2058}m\Rightarrow h\approx0.06m \end{gathered}$

So, the pendulum mover a height of exactly 125/2058 m, that is approximately 0.06m.