Question

The horizontal speed of a jet of water is 100 cm/sec and 50 cm³ of water hits the plate each second. Assume that the water moves parallel to the plate after striking it. The force exerted on the stationary plate if it is held perpendicular to the jet of water is:

A. $5 \times 10^{-2}$ N
B. $5 \times 10^{2}$ N
C. $5 \times 10^{-1}$ N
D. 5 N

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Answer

To find the force exerted on the stationary plate by the water jet, we will use the concept of momentum change. The key steps and calculations involved are as follows:

Given Data:

  • Horizontal speed of water jet ($v$): 100 cm/sec = 1 m/sec
  • Volume of water hitting the plate per second ($\Delta V$): 50 cm³/s = $50 \times 10^{-6}$ m³/s
  • Density of water ($\rho$): 1000 kg/m³

Calculations:

  1. Convert the volume flow rate to mass flow rate:

    Since the density of water is $\rho = 1000$ kg/m³, the mass flow rate $\left(\frac{dm}{dt}\right)$ can be found using: $$ \frac{dm}{dt} = \rho \cdot \Delta V = 1000 \times 50 \times 10^{-6} = 0.05 \text{ kg/s} $$

  2. Calculate the change in momentum:

    Before striking, the water has momentum due to its velocity. After striking the plate, it moves parallel to the plate causing a change in its momentum in the horizontal direction.

    The initial horizontal momentum per second is: $$ \text{Momentum} = \left(\frac{dm}{dt}\right) \cdot v = 0.05 \times 1 = 0.05 \text{ kg m/s} $$

    Note: Since the water moves parallel after hitting the plate, its horizontal momentum goes to zero.

  3. Calculate the force exerted:

    The force exerted by the water on the plate, which is the rate of change of momentum, is given by: $$ F = \frac{\Delta p}{\Delta t} = 0.05 \text{ N} $$

Conclusion:

The force exerted on the stationary plate is therefore: $$ \boxed{5 \times 10^{-2} \text{ N}} $$

Thus, the correct answer is Option A: $5 \times 10^{-2} \text{ N}$.


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