Question

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 average force the fish exerts on the drop of water?

(a) 0.00015 N;
(b) 0.00075 N;
(c) 0.075 N;
(d) 0.15 N.

Answer 1

(d) 0.15 N

Step-by-step explanation:

mass of the fish, m₁ = 65 g = 0.065 kg

initial velocity of the fish, u₁ = 0

mass of water expelled by the fish, m₂ = 0.30 g = 0.0003 kg

time during which the water was expelled, t = 5.0 ms = 5.0 x 10⁻³ s

velocity of the water, v = 2.5 m/s

The magnitude of force of the exerted water is equal to the magnitude of force the fish exerted on the water.

The magnitude of force of the exerted water is calculated as follows;

`F = ma = m(v)/(t) \\\\F = (mv)/(t) \\\\F = (m_2 \ * \ v)/(t) \\\\F = (0.0003 \ * \ 2.5)/(5 * 10^(-3)) \\\\F = 0.15 \ N`

The correct option is D.

Therefore, the average force the fish exerts on the drop of water is 0.15 N.