Question 17.SP.9: A 0.05-lb bullet B is fired with a horizontal velocity of 15......

Principle of Impulse and Momentum. We consider the bullet and the panel as a single system and express that the initial momenta of the bullet and panel and the impulses of the external forces are together equipollent to the final momenta of the system. Since the time interval Δt = 0.0006 s is very short, we neglect all nonimpulsive forces and consider only the external impulses \mathbf{A} _x \Delta t \text { and } \mathbf{A} _y \Delta t.

\text { Syst Momenta }_1+\text { Syst Ext Imp } \text { M }_{1 \rightarrow 2}=\text { Syst Momenta }_2

The centroidal mass moment of inertia of the square panel is

\bar{I}_P=\frac{1}{6} m_P b^2=\frac{1}{6}\left(\frac{20\text{ lb}}{32.2}\right)\left(\frac{18}{12} \text{ ft} \right)^2=0.2329 \text{ lb} \cdot \text{ ft} \cdot s ^2

Substituting this value as well as the given data into (1) and noting that

\bar{v}_2=\left(\frac{9}{12} \text{ ft} \right) \omega_2

we write

\left(\frac{0.05}{32.2}\right)(1500)\left(\frac{14}{12}\right)=0.2329 \omega_2+\left(\frac{20}{32.2}\right)\left(\frac{9}{12} \omega_2\right)\left(\frac{9}{12}\right)

\begin{aligned}& \omega_2=4.67\text{ rad} / s \quad \omega_2=4.67 \text{ rad} / s  {\uparrow} \\& \bar{v}_2=\left(\frac{9}{12} \text{ ft} \right) \omega_2=\left(\frac{9}{12} \text{ ft} \right)(4.67 \text{ rad} / s )=3.50 \text{ ft} / s\end{aligned}

Substituting \bar{v}_2=3.50\text{ ft}/ s , \Delta t=0.0006  s, and the given data into Eq. (2), we have

\left(\frac{0.05}{32.2}\right)(1500)+A_x(0.0006)=\left(\frac{20}{32.2}\right)(3.50)

A_x=-259\text{ lb}\quad\quad\quad A _x=259 \text{ lb} \leftarrow

From Eq. (3), we find        \begin{array}{ll}A_y=0 &\quad\quad\quad\quad\quad\quad\quad A _y=0\end{array}