A jet of water with a cross section of 6 cm^2 strikes a wall at an angle of 60 degrees 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) 2av^2ρcosθ B) avcosθ C) 2avcosθ D) avcosθ
Question
A jet of water with a cross section of 6 cm$^2$ strikes a wall at an angle of 60$^\circ$ 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) $2av^2\rho\cos\theta$ B) av$\cos\theta$ C) 2av$\cos\theta$ D) $av\cos\theta$
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Answer
To solve the problem of finding the force acting on a wall when a jet of water strikes and rebounds elastically, we follow these steps:

Given Data:
 Crosssectional area of the jet, $a = 6 , \text{cm}^2 = 6 \times 10^{4} , \text{m}^2$.
 Angle of incidence and reflection, $\theta = 60^\circ$.
 Velocity of water jet, $v = 12 , \text{m/s}$.
 Density of water, $\rho = 1000 , \text{kg/m}^3$.

Velocity Components:
 The velocity component perpendicular to the wall (ydirection), $v_y = v \sin(60^\circ) = v \cdot \frac{\sqrt{3}}{2}$.
 The velocity component parallel to the wall (xdirection), $v_x = v \cos(60^\circ) = v \cdot \frac{1}{2}$.

Change in Momentum:
 The ycomponent of velocity does not change when it rebounds, thus no change in momentum in the ydirection.
 In the xdirection, the change in velocity is: $$ \Delta v_x = v \cos(60^\circ)  (v \cos(60^\circ)) = 2v \cos(60^\circ) = 2v \cdot \frac{1}{2} = v $$
 So, the change in momentum per unit volume is $2v \cos(60^\circ)$.

Force Calculation:
 Mass flow rate per second, $\dot{m} = \rho \cdot a \cdot v$, is used to find the rate of change of linear momentum.
 Force, $F$, acting on the wall is: $$ F = \dot{m} \cdot (\text{change in velocity component in the xdirection}) $$ $$ F = \rho \cdot a \cdot v \cdot (2v \cos(60^\circ)) $$ $$ F = \rho \cdot a \cdot v^2 \cdot \cos(60^\circ) $$

Substituting Values:
 Calculate $F$ using given values: $$ F = 1000 , \text{kg/m}^3 \cdot 6 \times 10^{4} , \text{m}^2 \cdot (12 , \text{m/s})^2 \cdot \cos(60^\circ) $$ $$ F = 1000 \cdot 6 \times 10^{4} \cdot 144 \cdot \frac{1}{2} = 1000 \cdot 6 \times 10^{4} \cdot 72 $$ $$ F = 6 \times 72 = 432 \times 10^{4} = 43.2 , \text{N} $$
From the options provided, this doesn't match directly as the given options focus on the generalized expression, not numeric values. Thus, the most appropriate general formula for force is: Option A: $2av^2\rho\cos\theta$
However, since the numeric calculations seem more appropriate: Answer: 86.4 N (none of the options precisely match this general formula as rest option was already not matching initially)
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