Using impulse momentum\[ p_{1}+\int _{0}^{t}F_{av}\left ( t\right ) dt=p_{2}\] But \(p_{1}=mv_{1}=0\) since starting from rest and \(p_{2}=mv_{2}\), therefore\begin{align*} \int _{0}^{t}F_{av}\left ( t\right ) dt & =\frac{\left ( \frac{181}{7000}\right ) }{32.2}\left ( 3347\right ) \\ & =2.688\text{ lb-sec} \end{align*}
Therefore\begin{align*} F_{av}\left ( 0.0011\right ) & =2.688\\ F_{av} & =\frac{2.688}{0.0011}\\ & =2443.636\text{ lb} \end{align*}
\[ \,p_{1}+\int _{0}^{t}Tdt=p_{2}\] Where \(T\) is the thrust. But \(p_{1}=0\), therefore\begin{align*} Tt & =mv_{2}\\ t & =\frac{mv_{2}}{T}\\ & =\frac{\left ( \frac{45500}{32.2}\right ) \left ( 162\left ( \frac{5280}{3600}\right ) \right ) }{33000}\\ & =10.174\text{ sec} \end{align*}
To find how long a runway is needed\[ x_{f}=x_{1}+v_{1}t+\frac{1}{2}at^{2}\] But \(x_{1}=0\) and \(a=\frac{v_{2}-v_{1}}{t}\), and \(v_{1}=0\) since starting from rest, hence\begin{align*} x_{f} & =\frac{1}{2}at^{2}\\ & =\frac{1}{2}\left ( \frac{v_{2}}{t}\right ) t^{2}\\ & =\frac{1}{2}v_{2}t\\ & =\left ( \frac{1}{2}\right ) 162\left ( \frac{5280}{3600}\right ) \left ( 10.174\right ) \\ & =1208.671\text{ ft} \end{align*}
This is 4 times as long as without the catapults.
\begin{align*} \bar{p}_{1}+\int _{0}^{t}\bar{F}dt & =\bar{p}_{2}\\ -mv_{1}\hat{\imath }+\int _{0}^{t}\left ( F_{x}\hat{\imath }+F_{y}\hat{\jmath }\right ) dt & =mv_{2}\cos \alpha \hat{\imath }+mv_{2}\sin \alpha \hat{\jmath }\\ \hat{\imath }\left ( -mv_{1}+F_{x}t\right ) +\hat{\jmath }\left ( F_{y}t\right ) & =mv_{2}\cos \alpha \hat{\imath }+mv_{2}\sin \alpha \hat{\jmath } \end{align*}
Hence we obtain two equations\begin{align*} -mv_{1}+F_{x}t & =mv_{2}\cos \alpha \\ F_{y}t & =mv_{2}\sin \alpha \end{align*}
Or\begin{align*} F_{x}t & =mv_{2}\cos \alpha +mv_{1}\\ F_{y}t & =mv_{2}\sin \alpha \end{align*}
Now \(m=\frac{\frac{5.125}{16}}{32.2}=0.00994\) slug, and \(v_{1}=89\left ( \frac{5280}{3600}\right ) =130.533\) ft/sec and \(v_{2}=160\left ( \frac{5280}{3600}\right ) =234.667\) ft/sec. Hence\begin{align*} F_{x}t & =\left ( 0.00994\right ) \left ( 234.667\right ) \cos \left ( 32\left ( \frac{\pi }{180}\right ) \right ) +\left ( 0.00994\right ) \left ( 130.5333\right ) \\ F_{y}t & =\left ( 0.00994\right ) \left ( 234.667\right ) \sin \left ( 32\left ( \frac{\pi }{180}\right ) \right ) \end{align*}
Or \begin{align*} F_{x}t & =1.978\,+1.298=3.276\\ F_{y} & =1.236\, \end{align*}
Hence impulse is\[ \bar{I}=3.276\hat{\imath }+1.236\hat{\jmath }\] To find average force, we divide by time\begin{align*} \bar{F}_{av} & =\frac{3.276}{0.001}\hat{\imath }+\frac{1.236}{0.001}\hat{\jmath }\\ & =3276\hat{\imath }+1236\hat{\jmath } \end{align*}
Since there is no external force, then \(p_{1}=p_{2}\) or\begin{equation} m_{B}v_{B}^{-}+m_{A}v_{A}^{-}=m_{B}v_{B}^{+}+m_{A}v_{A}^{+}\tag{1} \end{equation} Where \(+\) means after impact and \(-\) means before impace. Therefore (using positive going to the right)\begin{align*} v_{A}^{-} & =-57\left ( \frac{5280}{3600}\right ) =-83.6\text{ ft/sec}\\ v_{B}^{-} & =32\left ( \frac{5280}{3600}\right ) =46.933\text{ ft/sec}\\ m_{A} & =\frac{8110}{32.2}=251.8634\text{ slug}\\ m_{B} & =\frac{2070}{32.2}=64.2857\text{ slug} \end{align*}
Hence (1) becomes\begin{align} \left ( 64.2857\right ) \left ( 46.933\right ) -\left ( 251.8634\right ) \left ( 83.6\right ) & =\frac{2070}{32.2}v_{B}^{+}+\frac{8110}{32.2}v_{A}^{+}\nonumber \\ -18038.66 & =64.286v_{B}^{+}+251.863\,4v_{A}^{+}\tag{2} \end{align}
And since \(e=0\), then\begin{align} e & =0=\frac{v_{B}^{+}-v_{A}^{+}}{v_{A}^{-}-v_{B}^{-}}\nonumber \\ v_{B}^{+} & =v_{A}^{+}\tag{3} \end{align}
Using (2,3) we solve for \(v_{B}^{+},v_{A}^{+}\). Plug (3) into (2) gives\begin{align*} -18038.66 & =64.286v_{A}^{+}+251.863\,4v_{A}^{+}\\ -18038.66 & =316.\,1494v_{A}^{+}\\ v_{A}^{+} & =\frac{-18038.66}{316.\,1494}\\ & =-57.057\,39\text{ ft/sec} \end{align*}
Hence \[ v_{B}^{+}=-57.057\,39\text{ ft/sec}\]
Let \(v_{B}^{-}\) be speed of bullet befor impact. Assume that after imapct bullet and mass \(A\) are stuck togother with speed \(v^{+}\). Hence\begin{equation} m_{B}v_{B}^{-}=\left ( m_{B}+m_{A}\right ) v^{+}\tag{1} \end{equation} Now we apply work-energy. Hence\begin{equation} \frac{1}{2}\left ( m_{B}+m_{A}\right ) \left ( v^{+}\right ) ^{2}=\left ( m_{B}+m_{A}\right ) g\left ( L-L\cos \theta \right ) \tag{2} \end{equation} Where datum is taken at the horizontal level. From (2) we solve for \(v^{+}\) and use it in (1) to find \(v_{B}^{-}\). (2) becomes\begin{align*} \frac{1}{2}\left ( 0.09+5.8\right ) \left ( v^{+}\right ) ^{2} & =\left ( 0.09+5.8\right ) \left ( 9.81\right ) \left ( 1.7\right ) \left ( 1-\cos \left ( 51\left ( \frac{\pi }{180}\right ) \right ) \right ) \\ 2.945\left ( v^{+}\right ) ^{2} & =36.410\,94\\ v^{+} & =\sqrt{\frac{36.410\,94}{2.945}}\\ & =3.516\,\text{\ m/sec} \end{align*}
Then (1) becomes\begin{align*} 0.09v_{B}^{-} & =\left ( 0.09+5.8\right ) \left ( 3.516\right ) \\ v_{B}^{-} & =\frac{\left ( 0.09+5.8\right ) \left ( 3.516\right ) }{0.09}\\ & =230.103\text{ m/sec} \end{align*}
Applying impulse momentum\[ m_{B}v_{B}^{-}=\left ( m_{B}+m_{A}\right ) v^{+}\] Solving or \(v^{+}\)\begin{align*} v^{+} & =\frac{m_{B}v_{B}^{-}}{m_{B}+m_{A}}\\ & =\frac{\left ( 1860\right ) \left ( 38\left ( \frac{1000}{3600}\right ) \right ) }{1860+1524}\\ & =5.802\text{ m/sec} \end{align*}
Now applying work-energy\begin{align*} T_{1}+U^{12} & =T_{2}\\ \frac{1}{2}\left ( m_{B}+m_{A}\right ) \left ( v^{+}\right ) ^{2}-\int _{0}^{d}\mu \left ( m_{B}+m_{A}\right ) gdx & =0 \end{align*}
We now solve for \(d\)\begin{align*} \frac{1}{2}\left ( 1860+1524\right ) \left ( 5.802\right ) ^{2}-\left ( 0.67\right ) \left ( 1860+1524\right ) \left ( 9.81\right ) d & =0\\ d & =\frac{\frac{1}{2}\left ( 1860+1524\right ) \left ( 5.802\right ) ^{2}}{\left ( 0.67\right ) \left ( 1860+1524\right ) \left ( 9.81\right ) }\\ & =2.561\text{ meter} \end{align*}