Answer:
0.25 m/s
Explanation:
This problem can be solved by using the law of conservation of momentum - the total momentum of the squid-water system must be conserved.
Initially, the squid and the water are at rest, so the total momentum is zero:
[tex]p_i = 0[/tex]
After the squid ejects the water, the total momentum is
[tex]p_f = m_s v_s + m_w v_w[/tex]
where
[tex]m_s = 1.60 kg[/tex] is the mass of the squid
[tex]v_s[/tex] is the velocity of the squid
[tex]m_2 = 0.115 kg[/tex] is the mass of the water
[tex]v_w = 3.50 m/s[/tex] is the velocity of the water
Due to the conservation of momentum,
[tex]p_i = p_f[/tex]
so
[tex]0=m_s v_s + m_w v_w[/tex]
so we can find the final velocity of the squid:
[tex]v_s = -\frac{m_w v_w}{m_s}=-\frac{(0.115 kg)(3.50 m/s)}{1.60 kg}=-0.25 m/s[/tex]
and the negative sign means the direction is opposite to that of the water.