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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?
- The thrust the water exerts on the wall is doubled.
- The thrust the water exerts on the wall stays unchanged.
- The kinetic energy lost per second by the water hitting the wall becomes four times.
- 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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