Chapter 1: Q106PP (page 271)

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 momentum of one drop of water immediately after it leaves the fish’s mouth? (a) $7.5 \times 10^{- 4} kg \cdot m / s$ ; (b) role="math" localid="1667981772098" $1.5 \times 10^{- 4} kg \cdot m / s$ ; (c) $7.5 \times 10^{- 3} kg \cdot m / s$ ; (d) $1.5 \times 10^{- 3} kg \cdot m / s$ .

Expert verified

The correct answer is (a) $7.5 \times 10^{- 4} kg \cdot m / s$ .

Step by step solution

The velocity of water is $v_{w} = 2.5 m / s$ .

The mass of water is $m_{w} = 0.3 \times 10^{- 3} k g$ .

The mass of fish is $m_{f i s h} = 65 \times 10^{- 3} k g$ .

The velocity of insect is $v_{i} = 2 m / s$ .

The time is . $t = 5 \times 10^{- 3} s$

The momentum of the fish is the same for the drop of water is $p_{w a t e r} = 7.5 \times 10^{- 4} k g \cdot m / s$

The momentum of any object can be obtained by multiplying the mass of that object with speed of that object.

The expression to calculate the momentum of the droplet is given by,

$p_{d} = m_{w} v_{w} . . . . . . (1)$

Substitute $m_{w} = 0.3 \times 10^{- 3} k g$ , and $v_{w} = 2.5 m / s$ in the equation (1) to find the momentum of one drop of water.

role="math" localid="1667986703621" $p_{d} = (0.3 \times 10^{- 3} k g) (2.5 m / s) = 7.5 \times 10^{- 4} k g . m / s$

Therefore, the correct answer is (a) $7.5 \times 10^{- 4} kg . m / s$ .

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