A bullet of mass 8.0 g is fired horizontally into a 2.2 kg wooden block at rest on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.40. The bullet comes to rest in the block, which moves 2.30 m. (a) What is the speed of the block immediately after the bullet comes to rest within it?(b) At what speed is the bullet fired?

The energy in the block/bullet combo is forcefriction*distancemoved. Figure that out. Then, set that equal to initial KE of the block/bullet, thence find the initial velocity.

Now use conservation of momentum to find speed of the bullet.

23.3

To find the speed of the block immediately after the bullet comes to rest within it, we need to apply the principle of conservation of momentum.

(a) Let's determine the initial and final momentum of the system:
- The initial momentum is the momentum of the bullet, which is given by the mass of the bullet multiplied by its initial speed: p_initial = m_bullet * v_bullet.
- The final momentum is the momentum of the block and bullet after the collision. Let's denote the final speed of the block as v_final. The momentum of the block is given by the mass of the block multiplied by its final speed: p_final_block = m_block * v_final.

According to the principle of conservation of momentum, the initial momentum of the system is equal to the final momentum of the system:

p_initial = p_final_block.

Since the bullet comes to rest within the block, the final momentum of the bullet is zero: p_final_bullet = 0.

Therefore, we have:

m_bullet * v_bullet = (m_block + m_bullet) * v_final.

Substituting the given values into the equation:

0.008 kg * v_bullet = (2.2 kg + 0.008 kg) * v_final.

Simplifying the equation:

v_final = (0.008 kg * v_bullet) / (2.208 kg).

(b) To find the speed at which the bullet is fired, we can use the principle of conservation of energy.

The initial kinetic energy of the system is all due to the bullet, given by (1/2) * m_bullet * v_bullet^2.

Since the bullet comes to rest within the block, the total kinetic energy of the system after the collision is due to the block, given by (1/2) * (m_block + m_bullet) * v_final^2.

According to the principle of conservation of energy, the initial kinetic energy is equal to the final kinetic energy:

(1/2) * m_bullet * v_bullet^2 = (1/2) * (m_block + m_bullet) * v_final^2.

Substituting the given values and the value of v_final from part (a) into the equation:

(1/2) * 0.008 kg * v_bullet^2 = (1/2) * (2.208 kg) * [(0.008 kg * v_bullet) / (2.208 kg)]^2.

Simplifying the equation:

v_bullet^2 = [(0.008 kg * v_bullet) / (2.208 kg)]^2.

Taking the square root of both sides to solve for v_bullet:

v_bullet = [(0.008 kg * v_bullet) / (2.208 kg)].

Simplifying the equation:

v_bullet = v_bullet / 276.

Multiplying both sides by 276:

276 * v_bullet = v_bullet.

Simplifying the equation:

275 * v_bullet = 0.

Therefore, v_bullet = 0.

So, the speed at which the bullet is fired is 0 m/s.