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Question

# A cruise ship with a mass of $1.00 \times 10^{7}\ \mathrm{kg}$ strikes a pier at a speed of 0.750 m/s. It comes to rest 6.00 m later, damaging the ship, the pier, and the tugboat captain's finances. Calculate the average force exerted on the pier using the concept of impulse.

Solution

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Here the mass of the ship

\begin{align*} m = 1.00\times 10^7\,\mathrm{kg} \end{align*}

Speed of the ship at the time of strike with the pier

\begin{align*} v_{\text{i}} = 0.750\,\mathrm{m\,s^{-1}} \end{align*}

The distance covered by the ship before come to rest

\begin{align*} S = 6.00\,\mathrm{m} \end{align*}

Finally the ship comes to rest, so its final speed is

\begin{align*} v_{\text{f}} = 0.0\,\mathrm{m\,s^{-1}} \end{align*}

Now from the kinametics, we get

\begin{align*} v_{\text{f}}^2 = v_{\text{f}}^2 + 2\,a\,S \end{align*}

By solving the above equation for the acceleration (a), we get

\begin{align*} a & = \frac{v_{\text{f}}^2 - v_{\text{i}}^2}{2\,S}\\ & = \frac{\left(0.0\,\mathrm{m\,s^{-1}} \right)^2 - \left(0.750\,\mathrm{m\,s^{-1}} \right)^2}{2\,\times 6.00\,\mathrm{m}}\\ & = 0.047\,\mathrm{m\,s^{-2}} \end{align*}

Now we know the force is given by

\begin{align*} F & = m\,a\\ & = 1.00\times 10^7\,\mathrm{kg} \times 0.047\,\mathrm{m\,s^{-2}}\\ & = 468750\,\mathrm{N}\\ & = 469\,\mathrm{kN} \end{align*}

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