A bullet of mass 4 g is fired horizontally with a speed of 300 m/s into 0.8 kg block of wood at rest on a table. If the coefficient of friction between the block and the table is 0.3, how far will the block slide approximately?

First, we have a bullet of mass 4 g fired horizontally at 300 m/s into a wooden block of mass 0.8 kg at rest. The bullet embeds into the block, making this an inelastic collision. We use conservation of momentum to find the velocity of the block-bullet system right after the collision.

Convert the bullet's mass to kilograms: mass of bullet, $$ m_b = 4 $$ g = $$ 0.004 $$ kg. Mass of block, $$ m_{bl} = 0.8 $$ kg. Initial velocity of bullet, $$ u_b = 300 $$ m/s. Initial velocity of block, $$ u_{bl} = 0 $$ m/s. Let $$ v $$ be the common velocity after collision.

Conservation of momentum:

$$ m_b u_b + m_{bl} u_{bl} = (m_b + m_{bl}) v $$

Substitute the values:

$$ (0.004 \times 300) + (0.8 \times 0) = (0.004 + 0.8) v $$

$$ 1.2 + 0 = 0.804 v $$

$$ 1.2 = 0.804 v $$

Solve for $$ v $$:

$$ v = \frac{1.2}{0.804} = \frac{1200}{804} $$

Simplify the fraction by dividing numerator and denominator by 12:

$$ \frac{1200 \div 12}{804 \div 12} = \frac{100}{67} $$

So, $$ v = \frac{100}{67} $$ m/s ≈ 1.4925 m/s.

Now, the block-bullet system slides on the table with initial velocity $$ v = \frac{100}{67} $$ m/s and comes to a stop due to friction. The coefficient of friction is $$ \mu = 0.3 $$. The friction force opposes the motion.

The normal force $$ N $$ equals the weight of the system since the table is horizontal. Total mass, $$ m = m_b + m_{bl} = 0.004 + 0.8 = 0.804 $$ kg. Using $$ g = 9.8 $$ m/s², weight = $$ m g = 0.804 \times 9.8 $$ N.

Friction force, $$ f = \mu N = \mu m g = 0.3 \times 0.804 \times 9.8 $$.

By Newton's second law, friction causes deceleration $$ a $$:

$$ f = m a $$

$$ 0.3 \times 0.804 \times 9.8 = 0.804 a $$

The mass $$ 0.804 $$ cancels out:

$$ a = 0.3 \times 9.8 = 2.94 \text{ m/s}^2 $$

So, deceleration $$ a = -2.94 $$ m/s² (negative indicates opposition to motion).

We need to find the distance $$ s $$ the block slides before stopping. Initial velocity $$ u = v = \frac{100}{67} $$ m/s, final velocity $$ v_f = 0 $$ m/s, acceleration $$ a = -2.94 $$ m/s².

Use the equation of motion:

$$ v_f^2 = u^2 + 2 a s $$

Substitute the values:

$$ 0^2 = \left( \frac{100}{67} \right)^2 + 2 \times (-2.94) \times s $$

$$ 0 = \frac{10000}{4489} - 5.88 s $$

Solve for $$ s $$:

$$ 5.88 s = \frac{10000}{4489} $$

$$ s = \frac{10000}{4489 \times 5.88} $$

Convert $$ 5.88 $$ to a fraction: $$ 5.88 = \frac{588}{100} = \frac{147}{25} $$.

$$ s = \frac{10000}{4489} \times \frac{1}{5.88} = \frac{10000}{4489} \times \frac{25}{147} $$

$$ s = \frac{10000 \times 25}{4489 \times 147} $$

Calculate numerator: $$ 10000 \times 25 = 250000 $$.

Calculate denominator: $$ 4489 \times 147 $$. Break it down:

$$ 4489 \times 100 = 448900 $$

$$ 4489 \times 40 = 179560 $$

$$ 4489 \times 7 = 31423 $$

Add them: $$ 448900 + 179560 = 628460 $$, then $$ 628460 + 31423 = 659883 $$.

So,

$$ s = \frac{250000}{659883} \approx 0.37878 \text{ m} $$

Approximately, $$ s \approx 0.379 $$ m.

Comparing with the options, 0.379 m corresponds to Option B.

Hence, the correct answer is Option B.