This problem involves the conservation of momentum. Since the astronaut is initially motionless, the total momentum of the system (astronaut + canister) is zero before the throw. After the throw, we can use the conservation of momentum to calculate the astronaut's velocity.

Step 1: System Before the Throw

• Mass of astronaut: $m_a = 40 \, \text{kg}$
• Mass of one canister: $m_c = 5 \, \text{kg}$
• Initial velocity of astronaut: $v_a = 0 \, \text{m/s}$
• Initial velocity of canister: $v_c = 0 \, \text{m/s}$

The total initial momentum $p_{\text{initial}}$ of the system (astronaut + canister) is: $p_{\text{initial}} = (m_a + m_c) \times v_a = 0$ (Since both the astronaut and canister are initially motionless.)

Step 2: System After the Throw

• The astronaut throws the canister with a velocity of $v_{c'} = 3 \, \text{m/s}$ relative to her frame.
• After the throw, her velocity relative to the spaceship is $v_{a'}$ (which we need to find).

Now, using the law of conservation of momentum, we can set the total momentum before and after the throw equal to each other. The momentum before the throw was zero, so the total momentum after the throw must also be zero:

$m_a v_{a'} + m_c v_{c'} = 0$ Solving for $v_{a'}$ (the astronaut’s velocity after the throw): $v_{a'} = - \frac{m_c v_{c'}}{m_a}$ Substitute the given values: $v_{a'} = - \frac{5 \times 3}{40} = - \frac{15}{40} = -0.375 \, \text{m/s}$

Conclusion:

The astronaut’s velocity after throwing the canister is $-0.375 \, \text{m/s}$ (away from the canister, in the opposite direction of the throw).

Would you like a more detailed explanation of any step?

5 Related Questions:

1. What happens to the asteroid and canister after they collide and stick together?
2. How would the astronaut's velocity change if she threw the canister at a higher speed?
3. How does the mass of the astronaut affect her final velocity after the throw?
4. How is momentum conserved when the canister collides with the asteroid?
5. What would happen if the astronaut threw both canisters in opposite directions?

Tip:

Conservation of momentum is a powerful tool in space problems because there's no external force, making it easier to analyze isolated systems!