If velocity, force and time are taken to be fundamental quantities find dimensional formula for (a) Mass, and (b) Energy.

Answer

To find the dimensional formulas for mass and energy using velocity, force, and time as fundamental quantities, we can follow these steps:

### Part (a): Finding the Dimensional Formula for Mass

1. **Assume the relationship**: We assume that mass (M) can be expressed as a function of velocity (V), force (F), and time (T). We write:
\[ M = K \cdot V^a \cdot F^b \cdot T^c \]
where \( K \) is a constant and \( a, b, c \) are powers we need to determine.

2. **Write the dimensional formulas**:
- Velocity (V) has the dimensional formula: \( [V] = [L^1 T^{-1}] \)
- Force (F) has the dimensional formula: \( [F] = [M^1 L^1 T^{-2}] \)
- Time (T) has the dimensional formula: \( [T] = [T^1] \)
- Mass (M) has the dimensional formula: \( [M] = [M^1] \)

3. **Substitute the dimensional formulas**:
\[ [M] = K \cdot [L^1 T^{-1}]^a \cdot [M^1 L^1 T^{-2}]^b \cdot [T^1]^c \]
This expands to:
\[ [M] = K \cdot [M^b L^{a+b} T^{-a-2b+c}] \]

4. **Equate the dimensions**: Since the left side is \( [M^1] \), we equate the powers of M, L, and T:
- For M: \( b = 1 \)
- For L: \( a + b = 0 \)
- For T: \( -a - 2b + c = 0 \)

5. **Solve the equations**:
- From \( b = 1 \), substitute into \( a + b = 0 \) to get \( a + 1 = 0 \) → \( a = -1 \).
- Substitute \( a = -1 \) and \( b = 1 \) into \( -a - 2b + c = 0 \):
\[ -(-1) - 2(1) + c = 0 \implies 1 - 2 + c = 0 \implies c = 1 \]

6. **Final expression**: Thus, the dimensional formula for mass is:
\[ [M] = K \cdot V^{-1} \cdot F^{1} \cdot T^{1} \]

### Part (b): Finding the Dimensional Formula for Energy

1. **Assume the relationship**: We assume that energy (E) can be expressed as:
\[ E = K \cdot V^a \cdot F^b \cdot T^c \]

2. **Write the dimensional formulas**:
- Energy (E) has the dimensional formula: \( [E] = [M^1 L^2 T^{-2}] \)

3. **Substitute the dimensional formulas**:
\[ [E] = K \cdot [L^1 T^{-1}]^a \cdot [M^1 L^1 T^{-2}]^b \cdot [T^1]^c \]
This expands to:
\[ [E] = K \cdot [M^b L^{a+b} T^{-a-2b+c}] \]

4. **Equate the dimensions**: Since the left side is \( [M^1 L^2 T^{-2}] \), we equate the powers:
- For M: \( b = 1 \)
- For L: \( a + b = 2 \)
- For T: \( -a - 2b + c = -2 \)

5. **Solve the equations**:
- From \( b = 1 \), substitute into \( a + b = 2 \) to get \( a + 1 = 2 \) → \( a = 1 \).
- Substitute \( a = 1 \) and \( b = 1 \) into \( -a - 2b + c = -2 \):
\[ -(1) - 2(1) + c = -2 \implies -1 - 2 + c = -2 \implies c = 1 \]

6. **Final expression**: Thus, the dimensional formula for energy is:
\[ [E] = K \cdot V^{1} \cdot F^{1} \cdot T^{1} \]

### Summary of Dimensional Formulas
- **Mass**: \( [M] = K \cdot V^{-1} \cdot F^{1} \cdot T^{1} \)
- **Energy**: \( [E] = K \cdot V^{1} \cdot F^{1} \cdot T^{1} \)