A particle with mass m and initial speed `V_(0)` is a subject to a velocity-dependent damping force of the form `bV^(n)`.With dimensional analysis determine how the stopping time depends on m, `V_(0)` and b for begin with writing `Deltat=Am^(alpha)b^(beta)V_(0)^(gamma)`, powers `alpha, beta` and `gamma` will be.

Options

A. `alpha=1, beta=-1, gamma=1-n`
B. `alpha=2-n, beta=-1, gamma=2`
C. `alpha=1, beta=1, gamma=1-n`
D. `alpha=-1, beta=-1, gamma=1-n`

Explanation

To determine how the stopping time (\(\Delta t\)) depends on mass (\(m\)), initial speed (\(V_0\)), and the damping coefficient (\(b\)), we can use dimensional analysis. We start by expressing \(\Delta t\) in terms of these variables:

\[ \Delta t = A m^{\alpha} b^{\beta} V_0^{\gamma} \]

where \(A\) is a dimensionless constant, and \(\alpha\), \(\beta\), and \(\gamma\) are the powers we need to find.

### Step 1: Identify the dimensions of each variable
- Mass (\(m\)): The dimension is \([M^1 L^0 T^0]\)
- Damping coefficient (\(b\)): The damping force is given by \(F = bV^n\). The dimensions of force (\(F\)) are \([M^1 L^1 T^{-2}]\), and the dimensions of velocity (\(V\)) are \([L^1 T^{-1}]\). Thus, we can express \(b\) in terms of dimensions:

\[ [b] = \frac{[F]}{[V^n]} = \frac{[M^1 L^1 T^{-2}]}{[L^1 T^{-1}]^n} = [M^1 L^{1-n} T^{-2+n}] \]

- Initial speed (\(V_0\)): The dimension is \([L^1 T^{-1}]\)

### Step 2: Write the dimensions of \(\Delta t\)
The dimension of time (\(\Delta t\)) is \([M^0 L^0 T^1]\).

### Step 3: Substitute the dimensions into the equation
Now we substitute the dimensions into the equation:

\[ [M^0 L^0 T^1] = [M^1 L^0 T^0]^{\alpha} [M^1 L^{1-n} T^{-2+n}]^{\beta} [L^1 T^{-1}]^{\gamma} \]

This expands to:

\[ [M^0 L^0 T^1] = [M^{\alpha + \beta} L^{\beta(1-n) + \gamma} T^{\beta(-2+n) - \gamma}] \]

### Step 4: Set up equations for each dimension
Now we can equate the powers of \(M\), \(L\), and \(T\):

1. For mass (\(M\)):
\[ \alpha + \beta = 0 \quad \text{(1)} \]

2. For length (\(L\)):
\[ \beta(1-n) + \gamma = 0 \quad \text{(2)} \]

3. For time (\(T\)):
\[ \beta(-2+n) - \gamma = 1 \quad \text{(3)} \]

### Step 5: Solve the equations
From equation (1):
\[ \beta = -\alpha \]

Substituting \(\beta = -\alpha\) into equation (2):
\[ (-\alpha)(1-n) + \gamma = 0 \implies \gamma = \alpha(n-1) \]

Substituting \(\beta = -\alpha\) and \(\gamma = \alpha(n-1)\) into equation (3):
\[ (-\alpha)(-2+n) - \alpha(n-1) = 1 \]
\[ \alpha(2-n) - \alpha(n-1) = 1 \]
\[ \alpha(2-n-n+1) = 1 \implies \alpha(3-2n) = 1 \implies \alpha = \frac{1}{3-2n} \]

Now substituting \(\alpha\) back to find \(\beta\) and \(\gamma\):
\[ \beta = -\frac{1}{3-2n}, \quad \gamma = \frac{1}{3-2n}(n-1) \]

### Final Result
Thus, the powers are:
- \(\alpha = \frac{1}{3-2n}\)
- \(\beta = -\frac{1}{3-2n}\)
- \(\gamma = \frac{n-1}{3-2n}\)

The final expression for stopping time is:
\[ \Delta t = A m^{\frac{1}{3-2n}} b^{-\frac{1}{3-2n}} V_0^{\frac{n-1}{3-2n}} \]