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=((0.040kg)(150m/s))/((0.040kg+5kg))\Rightarrow v_f=(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} (1)/(2)(0.040kg+5kg)((25)/(21)m/s)=(0.040kg+5kg)(9.8m/s^2)h \\ \\ \Rightarrow(1)/(2)((25)/(21)m/s)=(9.8m/s^2)h\Rightarrow h=((1/2)(25/21m/s))/((9.8m/s^2)) \\ \\ \Rightarrow h=(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.