Answer:
(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
