To solve this problem, we can use the principle of conservation of momentum. When the astronaut throws the camera, she will move in the opposite direction due to the conservation of momentum.
-
Initial conditions:
- Mass of the astronaut \( m_a = 67.5 \) kg
- Mass of the camera \( m_c = 0.875 \) kg
- Speed of the camera when thrown \( v_c = 12 \) m/s
-
Applying conservation of momentum: Since the astronaut and the camera initially have zero momentum before the throw, we can set up the equation:
\[ m_a \cdot v_a + m_c \cdot v_c = 0 \]
where \( v_a \) is the velocity of the astronaut after throwing the camera. Rearranging the equation gives us:
\[ v_a = -\frac{m_c}{m_a} \cdot v_c \]
-
Calculating the astronaut's velocity: Plugging in the values:
\[ v_a = -\frac{0.875 , \text{kg}}{67.5 , \text{kg}} \cdot 12 , \text{m/s} = -\frac{0.875 \times 12}{67.5} \approx -0.1556 , \text{m/s} \]
Here, the negative sign indicates that the astronaut is moving towards the shuttle.
-
Finding the time taken to reach the shuttle: The astronaut is initially 70.9 meters away from the shuttle, and she moves towards it at a speed of \( 0.1556 \) m/s. The time \( t \) taken to reach the shuttle can be calculated using the formula:
\[ t = \frac{\text{distance}}{\text{speed}} = \frac{70.9 , \text{m}}{0.1556 , \text{m/s}} \approx 455.5 , \text{s} \]
-
Converting seconds to minutes: To convert seconds to minutes, we can use the conversion factor \( 1 , \text{min} = 60 , \text{s} \):
\[ t \approx \frac{455.5 , \text{s}}{60} \approx 7.592 , \text{min} \]
Therefore, the time it will take for the astronaut to reach the shuttle is approximately 7.596 minutes (option 2).
1 answer