Question

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 6.50 kg squid (including the water in the cavity) at rest suddenly sees a dangerous predator. A. If the squid has 1.55 kg of water in its cavity, at what speed must it expel this water to suddenly achieve a speed of 2.60 m/s to escape the predator? Neglect any drag effects of the surrounding water.
B. How much kinetic energy does the squid create by this maneuver?

Answer 1

Answer:

A) The speed of the water must be 8.30 m/s.

B) Total kinetic energy created by this maneuver is 70.12 Joules.

Explanation:

A) Mass of squid with water = 6.50 kg

Mass of water in squid cavuty = 1.55 kg

Mass of squid = $m_1=6.50 kg- 1.55 kg=4.95 kg$

Velocity achieved by squid = $v_1=2.60 m/s$

Momentum gained by squid = $P=m_1v_1$

Mass of water = $m_2=1.55 kg$

Velocity by which water was released by squid = $v_2$

Momentum gained by water but in opposite direction = $P'=m_2v_2$

P = P'

$m_1v_1=m_2v_2$

$v_2=\frac{m_1v_1}{m_2}=\frac{4.95 kg\times 2.60 m/s}{1.55 kg}=8.30 m/s$

B) Kinetic energy does the squid create by this maneuver:

Kinetic energy of squid = K.E  =$\frac{1}{2}m_1v_1^{2}$

Kinetic energy of water = K.E' = $\frac{1}{2}m_2v_2^{2}$

Total kinetic energy created by this maneuver:

$K.E+K.E'=\frac{1}{2}m_1v_1^{2}+\frac{1}{2}m_2v_2^{2}$

$=\frac{1}{2}\times 4.95 kg\times (2.60 m/s)^2+\frac{1}{2}\times 1.55 kg\times (8.30 m/s)^2=70.12 Joules$