91Ó°ÊÓ

Conservation of momentum, general law of physics according to which the quantity called momentum that characterizes motion never changes in an isolated collection of objects.

Use the law of conservation of momentum and the formula for the moment of inertia of the rod. The moment of inertia of the rod with respect to its center is as follows:

$I=\frac{1}{21}M{L}^2$

Here, $M$is the mass of the rod and$L$is the length of the rod.

The motion of the meteorite is shown in the following figure:

In the figure, $m$is the mass of the meteorite, $d$is the distance from the center of the rod to the point of application of velocity, $L$is the length of the rod, and$M$is the mass of the rod.

Apply the law of conservation of momentum along the horizontal as follows:

$m{v}_i\cos {\mathrm{θ}}_i=-m{v}_f\cos {\mathrm{θ}}_f+M{V}_{x,cm}$

Here$V_{x,cm}$is the center of mass of the rod along$x-$direction.

Rearrange the equation for the velocity of the center of mass of the rod.

$V_{x,cm}=\frac{m{v}_i\cos {\mathrm{θ}}_i+m{v}_f\cos {\mathrm{θ}}_f}{M}$

Substitute$200m/s$for$v_i,60m/s$for$v_f,0.06kg$for $m,1.3kg$for$M,26°$for ${\mathrm{θ}}_i,$and $82°$for${\mathrm{θ}}_f$ in the equation $V_{x,cm}=\frac{m{v}_i\cos {\mathrm{θ}}_i+m{v}_f\cos {\mathrm{θ}}_f}{M}$and solve for $V_{x,cm}.$

$V_{cm}=\frac{(0.06kg)(200m/s)\cos 26°+(0.06kg)(60m/s)\cos 82°}{1.3kg}$

$=8.68m/s$

Therefore, the velocity of the center of mass of the rod along horizontal is 8.68m/s.

Apply the law of conservation of momentum along the vertical as follows:

$m{v}_i\sin {\mathrm{θ}}_i=m{v}_f+M{V}_f\sin {\mathrm{θ}}_f+M{V}_{y,cm}$

Here$V_{y,cm}$is the center of mass of the rod along$y-$direction.

Rearrange the equation for the velocity of the center of mass of the rod.

$V_{y,cm}=\frac{m{v}_i\sin {\mathrm{θ}}_i-m{v}_f\sin {\mathrm{θ}}_f}{M}$

Substitute $200m/s$for$v_i,60m/s$ for$v_f,0.06kg$for$m,1.3kg$for$M,26°$for ${\mathrm{θ}}_i,$and$82°$for${\mathrm{θ}}_f$in the equation $V_{y,cm}=\frac{m{v}_i\sin {\mathrm{θ}}_i-m{v}_f\sin {\mathrm{θ}}_f}{M}$and solve for $V_{y,cm}$

$V_{y,cm}=\frac{(0.06kg)(200m/s)\sin 26°-(0.06kg)(60m/s)\sin 82°}{1.3kg}$

$=1.3m/s$

Therefore, the velocity of the center of mass of the rod along vertical is $1.3m/s$.

Therefore, the center of mass of the rod is

The magnitude of center of mass of the rod is as follows:

$\begin{array}{rcl}{V}_{cm}& =& \sqrt{{(8.68m/s)}^2+{(1.3m/s)}^2+{(0m/s)}^2}\\ & =& 8.78m/s\end{array}$

Therefore, the magnitude of the center of mass of the rod is 8.78m/s.

$L_i=L_f$

$\left(m{v}_i\cos {\mathrm{θ}}_i\right)d=-\left(m{v}_f\cos {\mathrm{θ}}_f\right)d+I\mathrm{Ó\neg }$$\left(m{v}_i\cos {\mathrm{θ}}_i\right)d=-\left(m{v}_f\cos {\mathrm{θ}}_f\right)d+I\mathrm{Ó\neg }$

Here$I$is the moment of inertia of the rod and is the angular velocity.

Substitute$\frac{1}{12}M{L}^2$for$\left(m{v}_i\cos {\mathrm{θ}}_i\right)d=-\left(m{v}_f\cos {\mathrm{θ}}_f\right)d+I\mathrm{Ó\neg }$in the equation

$m{v}_i\cos {\mathrm{θ}}_{i}d=-md{v}_f\cos {\mathrm{θ}}_f+\left(\frac{1}{12}M{L}^2\right)\mathrm{Ó\neg }$$m{v}_i\cos {\mathrm{θ}}_{i}d=-md{v}_f\cos {\mathrm{θ}}_f+\left(\frac{1}{12}M{L}^2\right)\mathrm{Ó\neg }$

Rearrange the equation for the angular velocity.

$\mathrm{Ó\neg }=\frac{md(v_i\cos {\mathrm{θ}}_i+v_f\cos {\mathrm{θ}}_f)}{\left(\frac{1}{12}M{L}^2\right)}$

Substitute $0.4m$for$L,0.2m$for$d,200m/s$for $v_i,60m/s$for$v_f,0.06kg$for$m,1.3kg$for $M,26°$for ${\mathrm{θ}}_i,$and $82°$for ${\mathrm{θ}}_f.$

$\mathrm{Ó\neg }=\frac{(0.06kg)(0.2m)\left(\right(200m/s)\cos 26°+(60m/s)\cos 82°)}{\left(\frac{1}{12}(1.3kg){(0.4m)}^2\right)}$

$\mathrm{Ó\neg }=\frac{(0.06kg)(0.2m)\left(\right(200m/s)\cos 26°+(60m/s)\cos 82°)}{\left(\frac{1}{12}(1.3kg){(0.4m)}^2\right)}$

$=130rad/s$

Therefore, the angular velocity of the rod is$130rad/s.$

(c) Calculate the kinetic energy is of the system before collision as follows:

Initially the rod is in rest. So, initial kinetic of the rod is zero.

$K_i=\frac{1}{2}m{v_i}^2$

Calculate the kinetic energy is of the system before collision as follows:

$K_f=\frac{1}{2}m{v_f}^2+\frac{1}{2}I{\mathrm{Ó\neg }}^2+\frac{1}{2}M{V_{cm}}^2$

Calculate the increase in internal energy as follows:

The expression for the change in internal energy is as follows:

$$\begin{gathered} \mathrm{Δ}E=K_i-K_f \\ \mathrm{Δ}E=\frac{1}{2}m{v_i}^2-\frac{1}{2}m{v_f}^2-\frac{1}{2}I{\mathrm{Ó\neg }}^2-\frac{1}{2}M{V_{cm}}^2 \end{gathered}$$

Substitute $\frac{1}{12}M{L}^2$for in the equation $\mathrm{Δ}E=\frac{1}{2}m{v_i}^2-\frac{1}{2}m{v_f}^2-\frac{1}{2}I{\mathrm{Ó\neg }}^2-\frac{1}{2}M{V_{cm}}^2$and solve for $\mathrm{Δ}E.$

$\mathrm{Δ}E=\frac{1}{2}m({v_i}^2-{v_f}^2)-\frac{1}{2}\left(\frac{1}{12}M{L}^2\right){\mathrm{Ó\neg }}^2-\frac{1}{2}M{V_{cm}}^2$

Substitute $0.4m$for$L,200m/s$for$v_1,60m/s$for$v_f,0.06kg$for $m,1.3kg$for $M,8.78m/s$for$V_{cm},$and 130rad/s for$\mathrm{Ó\neg }$in the equation

$\mathrm{Δ}E=\frac{1}{2}m({v_i}^2-{v_f}^2)-\frac{1}{2}\left(\frac{1}{12}M{L}^2\right){\mathrm{Ó\neg }}^2-\frac{1}{2}M{V_{cm}}^2$and solve for$\mathrm{Δ}E$.

$$\begin{gathered} \mathrm{Δ}E=\left\{\left(\frac{0.06kg}{2}{\left((200m/s\right)}^2-{(60m/s)}^2\right)-\frac{1}{2}\left(\frac{1}{12}(1.3kg){(0.4m)}^2\right)(130rad/s{)}^2 \\ -\left[\frac{1}{2}(1.3kg){\left((8.68m/s\right)}^2+{(1.3m/s)}^2\right] \\ \right\} \end{gathered}$$

$$\begin{gathered} =895.46J \\ ≈895.5J \end{gathered}$$

Therefore, the increase in internal kinetic energy of the system is $803J$.