The lead female character in the movie Diamonds Are Forever is standing at the edge of an offshore oil rig. As she fires a gun, she is driven back over the edge and into the sea. Suppose the mass of a bullet is 0.005 kg, and its velocity is +730 m/s. Her mass (including the gun) is 58 kg.

(a) What recoil velocity does she acquire in response to a single shot from a stationary position, assuming that no external force keeps her in place?
1 m/s

(b) Under the same assumption, what would be her recoil velocity if, instead, she shoots a blank cartridge that ejects a mass of 5.0 10-4 kg at a velocity of +730 m/s?
2 m/s

To find the recoil velocity in both cases, we need to apply the principle of conservation of momentum. According to this principle, the total momentum before the shot is equal to the total momentum after the shot.

(a) When she fires a gun with a bullet, the total initial momentum is zero because she is stationary. After firing the gun, the bullet moves forward with a velocity of +730 m/s (assuming in the positive direction), and she acquires an equal but opposite recoil velocity, denoted as V.

Using the conservation of momentum, we can set up the following equation:

m_bullet * v_bullet + m_character * 0 = m_bullet * v_bullet + m_character * V

Here, m_bullet is the mass of the bullet, v_bullet is the velocity of the bullet (+730 m/s), m_character is the mass of the character (including the gun), and V is the recoil velocity we want to find.

Substituting the given values, we have:

0.005 kg * 730 m/s + 58 kg * 0 = 0.005 kg * 730 m/s + 58 kg * V

Simplifying the equation:

3.65 kg m/s = 3.65 kg m/s + 58 kg * V

To solve for V, we subtract 3.65 kg m/s from both sides:

0 kg m/s = 58 kg * V

This means that V = 0 m/s, which implies that the recoil velocity the character acquires is 0 m/s.

Therefore, the answer to (a) is 0 m/s.

(b) When she fires a blank cartridge, the total initial momentum is still zero because she is stationary. The ejected mass of the cartridge is 5.0 * 10^(-4) kg, and it moves with a velocity of +730 m/s (assuming in the positive direction). She acquires an equal but opposite recoil velocity, denoted as V.

Using the conservation of momentum again, we can set up the following equation:

m_cartridge * v_cartridge + m_character * 0 = m_cartridge * v_cartridge + m_character * V

Here, m_cartridge is the mass of the ejected cartridge (5.0 * 10^(-4) kg), v_cartridge is the velocity of the cartridge (+730 m/s), m_character is the mass of the character (including the gun), and V is the recoil velocity we want to find.

Substituting the given values, we have:

5.0 * 10^(-4) kg * 730 m/s + 58 kg * 0 = 5.0 * 10^(-4) kg * 730 m/s + 58 kg * V

Simplifying the equation:

0.365 kg m/s = 0.365 kg m/s + 58 kg * V

To solve for V, we subtract 0.365 kg m/s from both sides:

0 kg m/s = 58 kg * V - 0.365 kg m/s

This means that V = 0.365 kg m/s / 58 kg ≈ 0.006 m/s.

Therefore, the answer to (b) is approximately 0.006 m/s or 2 m/s.