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A jet of water with cross section of 6 c...

A jet of water with cross section of `6 cm^(2)` strikes a wall at an angle of `60^(@)` to the normal and rebounds elastically from the wall without losing energy. If the velocity of the water in the jet is `12 m//s`, the force acting on the wall is

A

0.864 N

B

86.4 N

C

72 N

D

7.2 N

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we will follow these steps: ### Step 1: Identify the given data - Cross-sectional area of the jet, \( A = 6 \, \text{cm}^2 = 6 \times 10^{-4} \, \text{m}^2 \) - Velocity of the water jet, \( v = 12 \, \text{m/s} \) - Angle of incidence with the normal, \( \theta = 60^\circ \) ### Step 2: Calculate the components of the velocity The velocity of the water jet can be broken down into two components: - The component perpendicular to the wall (y-component): \[ v_y = v \sin(60^\circ) = 12 \sin(60^\circ) = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3} \, \text{m/s} \] - The component parallel to the wall (x-component): \[ v_x = v \cos(60^\circ) = 12 \cos(60^\circ) = 12 \times \frac{1}{2} = 6 \, \text{m/s} \] ### Step 3: Analyze the momentum change Since the water jet rebounds elastically, the y-component of the velocity remains the same, while the x-component changes direction: - Initial momentum in the x-direction: \( p_{x, \text{initial}} = m v_x = m (6) \) - Final momentum in the x-direction (after rebounding): \( p_{x, \text{final}} = m (-v_x) = m (-6) \) The change in momentum in the x-direction is: \[ \Delta p_x = p_{x, \text{final}} - p_{x, \text{initial}} = m(-6) - m(6) = -12m \] ### Step 4: Calculate the mass flow rate The mass flow rate (\( \dot{m} \)) of the water jet can be calculated using the formula: \[ \dot{m} = \rho A v \] Where: - \( \rho \) (density of water) = \( 1000 \, \text{kg/m}^3 \) - \( A = 6 \times 10^{-4} \, \text{m}^2 \) - \( v = 12 \, \text{m/s} \) Substituting the values: \[ \dot{m} = 1000 \times (6 \times 10^{-4}) \times 12 = 7.2 \, \text{kg/s} \] ### Step 5: Calculate the force acting on the wall The force (\( F \)) acting on the wall is given by the rate of change of momentum: \[ F = \frac{\Delta p_x}{\Delta t} = \dot{m} \cdot \Delta v_x \] Where \( \Delta v_x = 12 \, \text{m/s} \) (since the change in momentum is 12m for each second). Thus: \[ F = \dot{m} \cdot 12 = 7.2 \cdot 12 = 86.4 \, \text{N} \] ### Final Answer The force acting on the wall is \( 86.4 \, \text{N} \). ---
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