On an interplanetary mission, a 58.5-kg astronaut is floating toward the front of her ship at 0.15 m/s, relative to the ship. She wishes to stop moving, relative to the ship. She decides to throw away the 2.50-kg book she's carrying. What should the speed and direction of the book be to achieve her goal?

Options

A. 0.15 m/s, toward the front of the ship.
B. 3.5 m/s, toward the back of the ship.
C. 3.7 m/s, toward the front of the ship.
D. 0.30 m/s, toward the back of the ship.

Answer

To solve the problem of the astronaut wanting to stop moving relative to the ship after throwing the book, we can use the principle of conservation of momentum. Here’s a step-by-step solution:

### Step 1: Understand the initial conditions
The astronaut has a mass of \( m_a = 58.5 \, \text{kg} \) and is moving towards the front of the ship at a speed of \( v_a = 0.15 \, \text{m/s} \). The mass of the book is \( m_b = 2.5 \, \text{kg} \).

### Step 2: Set up the momentum conservation equation
Before the astronaut throws the book, the total momentum of the system (astronaut + book) can be expressed as:
\[ p_{\text{initial}} = (m_a + m_b) v_a \]
Since the book is initially moving with the astronaut, its initial speed is also \( v_a = 0.15 \, \text{m/s} \).

### Step 3: Determine the final conditions
After throwing the book, the astronaut wants to be at rest relative to the ship, which means her final velocity \( v_a' = 0 \). Let \( v_b \) be the speed of the book after it is thrown. The direction of \( v_b \) will be opposite to the direction of the astronaut's initial motion to stop her.

### Step 4: Write the final momentum equation
After the book is thrown, the total momentum of the system will be:
\[ p_{\text{final}} = m_a v_a' + m_b v_b \]
Substituting \( v_a' = 0 \):
\[ p_{\text{final}} = 0 + m_b v_b = m_b v_b \]

### Step 5: Apply conservation of momentum
According to the conservation of momentum:
\[ p_{\text{initial}} = p_{\text{final}} \]
Thus,
\[ (m_a + m_b) v_a = m_b v_b \]

### Step 6: Solve for \( v_b \)
Substituting the values:
\[ (58.5 \, \text{kg} + 2.5 \, \text{kg}) (0.15 \, \text{m/s}) = 2.5 \, \text{kg} \cdot v_b \]
Calculating the left side:
\[ 61 \, \text{kg} \cdot 0.15 \, \text{m/s} = 9.15 \, \text{kg m/s} \]
Now, we can solve for \( v_b \):
\[ 9.15 \, \text{kg m/s} = 2.5 \, \text{kg} \cdot v_b \]
\[ v_b = \frac{9.15 \, \text{kg m/s}}{2.5 \, \text{kg}} = 3.66 \, \text{m/s} \]

### Step 7: Determine the direction of \( v_b \)
Since the astronaut is moving towards the front of the ship and wants to stop, she must throw the book in the opposite direction. Therefore, the speed of the book should be \( 3.66 \, \text{m/s} \) in the opposite direction.

### Final Answer
The speed of the book should be \( 3.66 \, \text{m/s} \) in the direction opposite to the astronaut's initial motion.

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