Question

FIGURE CP12.89 shows a cube of mass m sliding without frictionat speed v0. It undergoes a perfectly elastic collision with the bottom tip of a rod of length d and mass M = 2m. The rod is pivoted about a frictionless axle through its center, and initially it hangs straight down and is at rest. What is the cube’s velocity— both speed and direction—after the collision?

Short answer

The cube’s velocity after collision is v_f=v_o/5, in right direction.

Step-by-step solution

Given information

Mass of cube =m
Initial Velocity of the cube= v_o
Length of rod =d
Mass of the rod = 2m

Explanation

From the law of energy conservation

$$\begin{gathered} E_0=E_F\dots \dots \dots \dots \dots \dots \dots \dots \left(1\right) \\ \Rightarrow \frac{1}{2}m{v}_0^2=\frac{1}{2}m{v}_f^2+\frac{1}{2}I{\omega }^2\dots \dots \dots \dots \left(2\right) \end{gathered}$$

As the collision is elastic momentum is conserved
so,

momentum before collision= momentum after collision

$$\begin{gathered} L=\overrightarrow{r}\times \overrightarrow{p}=mr{v}_0 \\ {L}_i=L_f \\ \Rightarrow m{v}_0\left(\frac{d}{2}\right)=m{v}_f\left(\frac{d}{2}\right)+I\omega \dots \dots \dots \dots \dots \dots \dots \dots \dots .\left(3\right) \end{gathered}$$

Moment of Inertia of the rod is given as

$$\begin{gathered} I=\frac{1}{12}M{d}^2 \\ I=\frac{1}{12}\left(2m\right)d^2=\frac{1}{6}\left(m\right)d^2 \end{gathered}$$

Substitute the value in equation (3)

$$\begin{gathered} m{v}_0\left(\frac{d}{2}\right)=m{v}_f\left(\frac{d}{2}\right)+\frac{1}{6}\left(m\right)d^2 \\ \omega \Rightarrow \frac{v_0}{2}=\frac{v_f}{2}+\frac{\omega d}{6}\Rightarrow \\ \omega =3\left(\frac{v_0-v_f}{d}\right) \end{gathered}$$

Substitute this into the equation (1) , to get velocity

$$\begin{gathered} \frac{1}{2}m{v}_0^2=\frac{1}{2}m{v}_f^2+\frac{1}{2}\left(\frac{1}{6}\left(m\right)d^2\right){\left(3\left(\frac{v_0-v_f}{d}\right)\right)}^2 \\ \Rightarrow v_0^2=v_f^2+\frac{3}{2}{\left(v_0-v_f\right)}^2 \\ \Rightarrow v_0^2=v_f^2+\frac{3}{2}\left(v_0^2+v_f^2-2v_0{v}_f\right) \\ \Rightarrow v_0^2+5v_f^2-6v_0{v}_f=0 \\ \Rightarrow v_f^2-\frac{6}{5}{v}_0{v}_f+\frac{1}{5}{v}_0^2=0 \end{gathered}$$

Solve the equation

$$\begin{gathered} \Rightarrow v_f=\frac{-\left(-\frac{6}{5}{v}_0\right)\pm \sqrt{{\left(-\frac{6}{5}{v}_0\right)}^2-4\left(1\right)\left(\frac{1}{5}{v}_0^2\right)}}{2\left(1\right)} \\ \Rightarrow v_f=\frac{3}{5}{v}_0\pm \frac{2}{5}{v}_0 \\ \Rightarrow v_f=v_0\text{or}\frac{v_0}{5} \end{gathered}$$

So the rod was hit in right direction with $\frac{v_0}{5}$.