A plate moves normally with the speed `v_(1)` towads a horizontal jet of uniform area of cross-section. The jet discharge water at the rate of volume `V` per second at a speed of `v_(2)`. The density of water is `rho`. Assume that water splashes along the surface of the plate ar right angles to the original motion. The magnitude of the force action on the plate due to the jet of water is

Options

A. (a)`rho Vv_(1)`
B. (b)`rho V (v_(1) + v_(2))`
C. (c)`(rho V)/(v_(1) + v_(2)) V_(1)^(2)`
D. (d)`rho [(V)/(v_(2))] (v_(1) + v_(2))^(2)`

Explanation

To solve the problem, we need to determine the magnitude of the force acting on a plate due to a horizontal jet of water. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the Setup
We have a plate moving towards a horizontal jet of water. The plate moves with a speed \( v_1 \) and the jet discharges water at a rate of volume \( V \) per second with a speed \( v_2 \). The density of water is \( \rho \).

**Hint:** Visualize the scenario and draw a diagram to understand the directions of the velocities.

### Step 2: Determine the Change in Momentum
The force acting on the plate due to the jet of water can be calculated using the principle of momentum change. The change in momentum per unit time (force) is given by:

\[ F = \frac{\Delta P}{\Delta t} \]

where \( \Delta P \) is the change in momentum.

**Hint:** Remember that momentum \( P \) is defined as mass times velocity.

### Step 3: Relate Momentum to Mass and Velocity
The momentum \( P \) can be expressed as:

\[ P = m \cdot v \]

where \( m \) is mass and \( v \) is velocity. The change in momentum when the water hits the plate can be expressed as:

\[ \Delta P = \text{mass} \times \Delta v \]

where \( \Delta v \) is the change in velocity of the water as it strikes the plate.

**Hint:** Consider how the water's velocity changes when it strikes the plate.

### Step 4: Calculate the Change in Velocity
Since the water is moving towards the plate with speed \( v_2 \) and the plate is moving towards the water with speed \( v_1 \), the effective change in velocity \( \Delta v \) when the water hits the plate is:

\[ \Delta v = v_1 + v_2 \]

**Hint:** Think about the direction of the velocities and how they combine.

### Step 5: Find the Mass Flow Rate
The mass flow rate of the water can be expressed as:

\[ \text{mass flow rate} = \rho \cdot V \]

where \( V \) is the volume of water discharged per second.

**Hint:** Use the definition of density to relate mass and volume.

### Step 6: Substitute into the Force Equation
Now, substituting the mass flow rate into the force equation, we have:

\[ F = \text{mass flow rate} \times \Delta v = \rho V (v_1 + v_2) \]

**Hint:** Ensure that you are using the correct units for each variable.

### Step 7: Finalize the Expression for Force
Thus, the magnitude of the force acting on the plate due to the jet of water is:

\[ F = \rho V (v_1 + v_2) \]

**Hint:** Check if this expression matches any of the provided options in the question.

### Conclusion
The final expression gives us the force acting on the plate due to the water jet. If the options are provided, we can identify the correct one based on our derived formula.