.. A 15.0-kg fish swimming at 1.10 m>s suddenly gobbles up a 4.50-kg fish that is initially stationary. Ignore any drag effects of the water. (a) Find the speed of the large fish just after it eats the small one. (b) How much mechanical energy was dis- sipated during this meal?

(a) 0.846 m/s

(b) 2.097J

Explanation:

Parameters given:

Mass of big fish, M = 15 kg

Mass of small fish, m = 4.5 kg

Initial speed of big fish, U = 1.1 m/s

Initial speed of small fish, u = 0 m/s (it is stationary)

(a) We apply the principle of conservation of momentum:

Total initial momentum = Total final momentum

Since both fish have the same final speed, V, (the small fish is in the mouth of the big fish), we have:

MU + mu = (M + m)*V

(15 * 1.1) + (4.5 * 0) = ( 15 + 4.5) * V

16.5 = 19.5V

=> V = 16.5/19.5

V = 0.846 m/s

The speed of the large fish after the meal is 0.846 m/s.

(b) We need to find the change in Kinetic energy of the entire system to find the total mechanical energy dissipated.

Initial Kinetic energy:

KEini = (½ * M * U²) + (½ * m * u²)

KEini = (½ * 15 * 1.1²) + (½ * 4.5 * 0²)

KEini = 9.075 J

Final Kinetic Energy:

KEfin = (½ * M * V²) + (½ * m * V²)

KEfin = (½ * 15 * 0.846²) + (½ * 4.5 * 0.846²)

KEfin = 5.368 + 1.610 = 6.978 J

Change in kinetic energy will be:

KEfin - KEini = 9.075 - 6.978

ΔKE = 2.097 J

The energy dissipated in eating the meal is 2.097 J

b would be the answer

trig 15.3 cos