Squid rely on jet propulsion when a rapid escape is necessary a 1.5 kg squid at rest pulls 0.10 kg of water into its mantle, then ejects this water at a remarkable 45 m/s. right after this ejection, how fast is the squid moving? express your answer with the appropriate units.

The final velocity of the squid after the ejection of the water is 3 m/s backwards.

The given parameters;

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = m₁v₁  +  m₂v₂

1.5(0) + 0.1(0) = 1.5v₁  +  45(0.1)

0 = 1.5v₁  + 4.5

1.5v₁ = -4.5

Thus, the final velocity of the squid after the ejection of the water is 3 m/s backwards.

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-3 m/s

Explanation:

The problem can be solved by using the law of conservation of momentum.

In fact, the system squid+water inside is initially at rest, so the total momentum is zero:

while the final momentum is:

where

is the mass of the squid

is the velocity of the squid

is the mass of water

is the water velocity

Since the total momentum must be conserved,

So we have

where the negative sign means the direction is opposite to that of the water.

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6

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