Question

Animal Propulsion. Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracting the cavity to force out the water through an opening. A$\mathbf{\text{6.50}}\mathit{k}\mathit{g}$ squid (including the water in the cavity) at rest suddenly sees a dangerous predator. (a) If the squid has $\mathbf{1}\mathbf{.}\mathbf{75}\mathit{k}\mathit{g}$of water in its cavity, at what speed must it expel this water suddenly to achieve a speed of $\mathbf{2}\mathbf{.}\mathbf{50}\mathit{m}\mathbf{/}\mathit{s}$to escape the predator? Ignore any drag effects of the surrounding water. (b) How much kinetic energy does the squid create by this maneuver?

Short answer

(a) The speed of the water expelled is $9.29m/s$.

(b) The kinetic energy is$75.51\text{J}$

Step-by-step solution

Step 1:Given in the question:

The mass of predator squidis $m=6.50\text{kg}$.

The mass of water to be expelled is$M=1.75\text{kg}$

Change in speed is

$\begin{array}{rcl}∆p& =& m∆v\\ & =& \left(6.5kg\right)\left(2.5m/s\right)\\ & =& 16.25kgm/s\end{array}$

Change in momentum of squid is

Law of conservation of momentum:

When two bodies collide each other then the total momentum before collision is equal to the total momentum after collision.

Step 3:(a) Speed of expelling the water:

Use the conservation of momentum to obtain the speed of the water:

Squids’ change in momentum is equal to expelled water momentum change.

$$\begin{gathered} p=mv \\ 16.25\text{kgm/s=}1.75kg\times v \end{gathered}$$

$\begin{array}{rcl}v& =& \frac{16.25}{1.75}\\ & =& 9.29\text{m/s}\\ & & \end{array}$

Hence,the speed of the water expelled is$9.29\text{m/s}$

(b) Kinetic energy created the squid: 

The energy is calculated as follows:

$\begin{array}{rcl}E& =& \frac{1}{2}\left(m{v}^2\right)\\ & =& \frac{1}{2}\times 1.75kg\times {\left(9.29m/s\right)}^2\\ & =& 75.51\text{J}\end{array}$

Hence, the kinetic energy is $75.51\text{J}$