To solve the problem, we need to analyze the relationships given for velocity, acceleration, and force in two different systems of units and derive the relationships for length, mass, time, and momentum.
### Step-by-Step Solution:
1. **Understanding the given relationships**:
- We have three relationships:
1. \( v' = \frac{\alpha^2}{\beta} v \)
2. \( a' = \alpha \beta a \)
3. \( F' = \frac{1}{\alpha \beta} F \)
- Here, \( v' \), \( a' \), and \( F' \) are quantities in one system, while \( v \), \( a \), and \( F \) are in another system. \( \alpha \) and \( \beta \) are dimensionless constants.
2. **Finding the relationship for time**:
- The dimension of velocity \( v \) is \( [L][T]^{-1} \) and for acceleration \( a \) is \( [L][T]^{-2} \).
- From the first relationship, we can express it in terms of dimensions:
\[
[L][T]^{-1} = \frac{\alpha^2}{\beta} [L][T]^{-1}
\]
- For acceleration:
\[
[L][T]^{-2} = \alpha \beta [L][T]^{-2}
\]
- Dividing the two equations gives:
\[
\frac{[L][T]^{-1}}{[L][T]^{-2}} = \frac{\frac{\alpha^2}{\beta}}{\alpha \beta}
\]
- This simplifies to:
\[
[T] = \frac{\alpha}{\beta^2} [T]
\]
- Thus, we find:
\[
t' = \frac{\alpha}{\beta^2} t
\]
- This confirms **Option C**: \( t' = \frac{\alpha}{\beta^2} t \) is correct.
3. **Finding the relationship for length**:
- Using the first relationship for velocity:
\[
v' = \frac{\alpha^2}{\beta} v \implies [L][T]^{-1} = \frac{\alpha^2}{\beta} [L][T]^{-1}
\]
- From this, we can express length:
\[
l' = \frac{\alpha^3}{\beta^3} l
\]
- This confirms **Option A**: \( l' = \frac{\alpha^3}{\beta^3} l \) is correct.
4. **Finding the relationship for mass**:
- Using the relationship for force:
\[
F' = \frac{1}{\alpha \beta} F
\]
- We know \( F = ma \), so substituting \( a' = \alpha \beta a \):
\[
F' = m' a' = m' (\alpha \beta a)
\]
- Setting the two expressions for force equal:
\[
\frac{1}{\alpha \beta} F = m' (\alpha \beta a)
\]
- Rearranging gives:
\[
m' = \frac{1}{\alpha^2 \beta^2} m
\]
- This confirms **Option B**: \( m' = \frac{1}{\alpha^2 \beta^2} m \) is correct.
5. **Finding the relationship for momentum**:
- Momentum \( p = mv \) gives:
\[
p' = m' v' = \left(\frac{1}{\alpha^2 \beta^2} m\right) \left(\frac{\alpha^2}{\beta} v\right)
\]
- Simplifying:
\[
p' = \frac{1}{\beta^3} p
\]
- This confirms **Option D**: \( p' = \frac{1}{\beta^3} p \) is correct.
### Conclusion:
All four options (A, B, C, D) are correct.
