A 11kg big fish moving at 3.5 m/s swallows a stationary fish which with a mass of 2kg. What is the velocity of the big fish after dinner? Only put in the numerical value. Answer to 1 decimal place and do not write the unit in your answer.

Question

A 11kg big fish moving at 3.5 m/s swallows a stationary fish which with a mass of 2kg. What is the velocity of the big fish after dinner? Only put in the numerical value. Answer to 1 decimal place and do not write the unit in your answer.

BlackTom AI · Accepted Answer

To solve this problem, start by recognizing that momentum is conserved in the collision-like event when the big fish swallows the stationary fish. The initial momentum is the momentum of the moving big fish plus the stationary small fish, which is (11 kg)(3.5 m/s) + (2 kg)(0 m/s) = 38.5 kg·m/s. After swallowing, the two fish move together as a single object with total mass (11 kg + 2 kg) = 13 kg and some common velocity v. Apply conservation of momentum: total initial momentum = total final momentum, so 38.5 = (13 kg)·v. Solve for v by dividing both sides by 13: v = 38.5 / 13 ≈ 2.9615 m/s. Rounding to 1 decimal place gives 2.9615 ≈ 3.0 m/s. Since the instruction in this context asks for a numerical value without a unit, we present the numeric result as 3.0.

Key checks:
- The assumption of inelastic collision is appropriate because the fishes stick together after the bite, so momentum is conserved while kinetic energy is not.
- Using only the pre-collision momentum (moving big fish) and zero momentum from the stationary small fish is correct for the initial state.
- Adding masses before applying momentum conservation ensures correct total mass for the final state.

A 11kg big fish moving at 3.5 m/s swallows a stationary fish which with a mass of 2kg. What is the velocity of the big fish after dinner? Only put in the numerical value. Answer to 1 decimal place and do not write the unit in your answer.简答题

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