A 70.0 kg astronaut pushes to the left on a spacecraft with a force `vec(F)` in "gravity-free" space. The spacecraft has a total mass of `1.0xx10^(4)` kg. During the push, the astronaut accelerates to the right with an acceleration of `0.36" m"//"s"^(2)`.
Determine the magnitude of the acceleration of the spacecraft.

Options

A. `51.4" m"//"s"^(2)`
B. `0.36" m"//"s"^(2)`
C. `2.5xx10^(-3)" m"//"s"^(2)`
D. `7.0xx10^(-3)" m"//"s"^(2)`

Answer

To solve the problem, we will use Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. Here’s how we can determine the magnitude of the acceleration of the spacecraft step by step:

### Step 1: Identify the forces acting on the astronaut and the spacecraft.
- The astronaut pushes the spacecraft to the left with a force \( F \).
- Due to this push, the astronaut experiences an equal and opposite force \( -F \) that accelerates him to the right.

### Step 2: Calculate the force exerted by the astronaut.
- The astronaut has a mass \( m_a = 70.0 \, \text{kg} \) and accelerates to the right with an acceleration \( a_a = 0.36 \, \text{m/s}^2 \).
- Using Newton's second law, the force exerted by the astronaut can be calculated as:
\[ F = m_a \cdot a_a = 70.0 \, \text{kg} \cdot 0.36 \, \text{m/s}^2 = 25.2 \, \text{N} \]

### Step 3: Apply Newton's third law to find the acceleration of the spacecraft.
- According to Newton's third law, the force exerted by the astronaut on the spacecraft is equal in magnitude but opposite in direction to the force exerted by the spacecraft on the astronaut. Therefore, the force acting on the spacecraft is also \( 25.2 \, \text{N} \) to the left.
- The mass of the spacecraft is \( m_s = 1.0 \times 10^4 \, \text{kg} \).

### Step 4: Calculate the acceleration of the spacecraft.
- Using Newton's second law for the spacecraft:
\[ a_s = \frac{F}{m_s} = \frac{25.2 \, \text{N}}{1.0 \times 10^4 \, \text{kg}} = 2.52 \times 10^{-3} \, \text{m/s}^2 \]

### Conclusion:
- The magnitude of the acceleration of the spacecraft is \( 2.52 \times 10^{-3} \, \text{m/s}^2 \) to the left.