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Water leaves a horizontal tube at speed v and strikes a vertical wall placed right at the tube mouth, falling straight down after impact (it loses all horizontal momentum). When the water speed is raised to 2*v, which statement is correct?

1. The thrust the water exerts on the wall is doubled.
2. The thrust the water exerts on the wall stays unchanged.
3. The kinetic energy lost per second by the water hitting the wall becomes four times.
4. The kinetic energy lost per second by the water hitting the wall becomes eight times.

Correct answer: The kinetic energy lost per second by the water hitting the wall becomes eight times.

Solution

Mass per second: dm/dt = rho*A*v. Thrust = (dm/dt)*v = rho*A*v², so doubling v makes thrust four times (not double or unchanged). Energy lost per second = (1/2)*(dm/dt)*v² = (1/2)*rho*A*v³, proportional to v³. Doubling v -> 2³ = 8 times. Hence the energy lost per second becomes eight times.

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