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Archerfish are tropical fish that hunt by shooting drops of water from their mouths at insects above the water’s surface to knock them into the water, where the fish can eat them. A 65-g fish at rest just at the surface of the water can expel a 0.30-g drop of water in a short burst of 5.0 ms. High-speed measurements show that the water has a speed of 2.5 m/s just after the archerfish expels it.

What is the speed of the archerfish immediately after it expels the drop of water? (a) 0.0025 m/s; (b) 0.012 m/s; (c) 0.75 m/s; (d) 2.5 m/s.

Short Answer

Expert verified

The correct answer is (b) 0.012 m/s.

Step by step solution

01

Identification of given data

The velocity of water is vw=2.5 m/s.

The mass of water is mw=0.3×10-3 kg.

The mass of fish is mfish=65×10-3kg.

The velocity of insect is vi=2 m/s.

The time is t=5×10-3 s.

The momentum of the fish is the same for the drop of water ispwater=7.5×10-4kg⋅m/s

02

Concept/Significance of change in Conservation of Momentum

The final total momentum is zero, so apply the law of conservation of momentum

m1v1-m2v2=0m1v1=m2v2

The expression to calculate the required velocity is given by,

pwater=mfishvfish       ......(1)

03

Determine the speed of the archerfish immediately after it expels the drop of water

Substitute mfish=65×10-3kg, andpwater=7.5×10-4kg⋅m/s in the equation (1) to find the speed of the archerfish.

7.5×10-4kg.m/s=vfish65×10-3kgvfish=7.5×10-4kg.m/s65×10-3kg=0.012m/s

Therefore, the correct answer is (b) 0.012 m/s.

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