Two vectors A and B have magnitudes such that A + B = 16 N and |A + B| = 8 N. If the resultant is perpendicular to A, what is the magnitude of B?

Let the magnitudes of vectors $\vec{A}$ and $\vec{B}$ be $A$ and $B$ respectively.
The sum of their magnitudes is $16 \, \text{N}$:

$$A + B = 16 \quad \text{(Equation 1)}$$

The magnitude of their resultant $\vec{R} = \vec{A} + \vec{B}$ is $8 \, \text{N}$:

$$R = |\vec{A} + \vec{B}| = 8 \, \text{N}$$

The resultant $\vec{R}$ is perpendicular to $\vec{A}$.

The magnitude of the resultant of two vectors $\vec{A}$ and $\vec{B}$ with an angle $\theta$ between them is given by:

$$R^2 = A^2 + B^2 + 2AB \cos \theta$$

Substitute $R = 8 \, \text{N}$:

$$8^2 = A^2 + B^2 + 2AB \cos \theta$$ $$64 = A^2 + B^2 + 2AB \cos \theta \quad \text{(Equation 2)}$$

If the resultant $\vec{R}$ is perpendicular to $\vec{A}$, the angle $\alpha$ between $\vec{R}$ and $\vec{A}$ is $90^\circ$. The direction of the resultant is given by:

$$\tan \alpha = \frac{B \sin \theta}{A + B \cos \theta}$$

Since $\alpha = 90^\circ$, $\tan 90^\circ$ is undefined. This implies the denominator must be zero:

$$A + B \cos \theta = 0$$ $$B \cos \theta = -A \quad \text{(Equation 3)}$$

Substitute $B \cos \theta = -A$ into Equation 2:

$$64 = A^2 + B^2 + 2A(-A)$$ $$64 = A^2 + B^2 - 2A^2$$ $$64 = B^2 - A^2$$

This is a difference of squares:

$$64 = (B - A)(B + A) \quad \text{(Equation 4)}$$

From Equation 1, we know $A + B = 16$. Substitute this into Equation 4:

$$64 = (B - A)(16)$$

Divide by 16:

$$B - A = \frac{64}{16}$$ $$B - A = 4 \quad \text{(Equation 5)}$$

Now we have a system of two linear equations:

1. $A + B = 16$
2. $-A + B = 4$ Add Equation 1 and Equation 5:

$$(A + B) + (-A + B) = 16 + 4$$ $$2B = 20$$ $$B = 10 \, \text{N}$$

Substitute $B = 10 \, \text{N}$ into Equation 1:

$$A + 10 = 16$$ $$A = 16 - 10$$ $$A = 6 \, \text{N}$$

$$\frac{A}{B} = \frac{6 \, \text{N}}{10 \, \text{N}} = \frac{3}{5}$$

The magnitudes of $\vec{A}$ and $\vec{B}$ are $6 \, \text{N}$ and $10 \, \text{N}$ respectively.
The ratio $A/B$ is $3/5$.

The magnitude of $\vec{A}$ is $\boxed{\text{6 N}}$
The magnitude of $\vec{B}$ is $\boxed{\text{10 N}}$
The ratio $A/B$ is $\boxed{\frac{3}{5}}$