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# A 3000-kg cannon is mounted so that it can recoil only in the horizontal direction. (a) Calculate its recoil velocity when it fires a 15.0-kg shell at 480 m/s at an angle of $20.0^{\circ}$ above the horizontal. (b) What is the kinetic energy of the cannon? This energy is dissipated as heat transfer in shock absorbers that stop its recoil. (c) What happens to the vertical component of momentum that is imparted to the cannon when it is fired?

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$\textbf{(a)}$ Here mass of the cannon

\begin{align*} m_{\text{C}} = 3000\,\mathrm{kg} \end{align*}

Mass of the shell

\begin{align*} m_{\text{S}} = 15.0\,\mathrm{kg} \end{align*}

Since both cannon and shell are stationary before fire so their initial speed is

\begin{align*} v_{\text{C}} = 0.00\,\mathrm{m\,s^{-1}} \end{align*}

and

\begin{align*} v_{\text{S}} = 0.00\,\mathrm{m\,s^{-1}} \end{align*}

Since the shell is fired at $\theta = 20^{\circ}$ above the horizontal so it horizontal component of the speed is

\begin{align*} v_{\text{S}}' = \left(480\,\mathrm{m\,s^{-1}} \right)\,\cos \theta \end{align*}

Let the horizontal speed of the cannon is $v_{\text{C}}'$ so from the law of conservation of momentum, we get

\begin{align*} p_{\text{i}} = p_{\text{f}} \end{align*}

Hence

\begin{align*} m_{\text{C}} \,v_{\text{C}} + m_{\text{S}} \,v_{\text{S}} = m_{\text{C}} \,v_{\text{C}}' + m_{\text{S}} \,v_{\text{S}}' \end{align*}

OR

\begin{align*} 3000\,\mathrm{kg} \times \,0.00\,\mathrm{m\,s^{-1}} + 15.0\,\mathrm{kg} \times \,0.00\,\mathrm{m\,s^{-1}} = 3000\,\mathrm{kg} \times \,v_{\text{C}}' + 15.0\,\mathrm{kg} \times \,480\,\mathrm{m\,s^{-1}} \,\cos 20^{\circ} \end{align*}

OR

\begin{align*} v_{\text{C}}' & = \frac{15.0\,\mathrm{kg} \times \,480\,\mathrm{m\,s^{-1}} \,\cos 20^{\circ}}{3000\,\mathrm{kg}}\\ & = 2.26\,\mathrm{m\,s^{-1}} \end{align*}

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