This is a conservation of momentum problem. Since the astronaut and the toolbox are in space (where no external forces act), the total momentum before and after the push must remain constant. This principle can be mathematically expressed as:

$\text{Total momentum before} = \text{Total momentum after}$

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Givens:

• Mass of the astronaut, $m_A = 80 \, \text{kg}$
• Mass of the toolbox, $m_T = 15 \, \text{kg}$
• Initial velocity of both the astronaut and the toolbox: $u_A = u_T = 0 \, \text{m/s}$ (they are initially at rest)
• Final velocity of the toolbox after being pushed: $v_T = 1.6 \, \text{m/s}$
• Final velocity of the astronaut (to be found): $v_A \, ?$

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Solution:

Since there are no external forces, the total momentum before and after the push remains the same.

1. Initial total momentum: $p_{\text{initial}} = (m_A \cdot u_A) + (m_T \cdot u_T) = (80 \cdot 0) + (15 \cdot 0) = 0 \, \text{kg} \cdot \text{m/s}$

2. Final total momentum: $p_{\text{final}} = (m_A \cdot v_A) + (m_T \cdot v_T)$

   We need to find $v_A$. Using conservation of momentum: $0 = (80 \cdot v_A) + (15 \cdot 1.6)$

3. Solve for $v_A$:

   $0 = 80 \cdot v_A + 24$

   $80 \cdot v_A = -24$

   $v_A = \frac{-24}{80} = -0.3 \, \text{m/s}$

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Final Answer:

The final velocity of the astronaut is:

$v_A = -0.3 \, \text{m/s}$

The negative sign indicates that the astronaut moves in the opposite direction to the toolbox (to the left, assuming the toolbox moves to the right).

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Do you want more details or have any questions?
Here are 5 related questions you might explore:

1. What is the principle of conservation of linear momentum?
2. How would the final velocities change if the toolbox had more mass?
3. What role does Newton's third law play in this scenario?
4. How does conservation of momentum work in two-dimensional collisions?
5. What happens to the center of mass of the astronaut-toolbox system after the push?

Tip: In space, even a small force can generate significant motion because there is no resistance to slow objects down.