Question

Squids rely on jet propulsion to move around. A $1.50$ kg squid (including the mass of water inside the squid) drifting at 0.40 m/s suddenly ejects $0.10$ kg of water to get itself moving at $2.50$ m/s. If drag is ignored over the small interval of time needed to expel the water (the impulse approximation), what is the water’s ejection speed relative to the squid?

Short answer

The water's ejection speed relative to the squid$-29.40m/s$

Step-by-step solution

Step 1. Given Information

Mass of squid,$m_{squid+water}=1.50kg$

Speed of squid,$v_{squid+water}=0.40m/s$

Mass of water ejected,$m_w=0.10kg$

Speed of the water ejection,$v_{water}=2.5m/s$

Step 2. Applying conservation principle

As the frictional drag is ignored, applying the law of conservation of momentum,

$$\begin{gathered} p_i=p_f \\ {m}_{\text{squid+water}}{\overrightarrow{v}}_{i_{\text{water}+\text{squid}}}==m_{\text{squid}}{\overrightarrow{v}}_{f_{\text{squid}}}+m_{\text{water}}{\overrightarrow{v}}_{f_{\text{water}}} \\ 1.50kg\left(0.40\frac{m}{s}\right)=1.40kg\left(2.50\frac{m}{s}\right)+0.10kg\left({\overrightarrow{v}}_{f_{\text{water}}}\right) \\ \left({\overrightarrow{v}}_{f_{\text{water}}}\right)=-29m/s \end{gathered}$$

Now the relative speed,$v=-29-0.4=-29.40m/s$