Two astronauts, one of mass 55 kg and the other 85 kg, are initially at rest together in outer space. They then push each other apart. How far apart are they when the lighter astronaut has moved 12 m?
The correct answer and explanation is:
Correct Answer: 18.55 meters
Explanation:
This problem involves the law of conservation of momentum in a system with no external forces (outer space). Initially, both astronauts are at rest, so the total momentum is zero. When they push off from each other, they move in opposite directions, but the total momentum of the system remains zero.
Step 1: Apply the Law of Conservation of Momentum
Let:
- m1=55m_1 = 55 kg (mass of the lighter astronaut)
- m2=85m_2 = 85 kg (mass of the heavier astronaut)
- x1=12x_1 = 12 m (distance moved by the lighter astronaut)
- x2x_2 = ? (distance moved by the heavier astronaut)
Since they push each other apart, the forces they exert are equal and opposite. According to Newton’s Third Law, they will accelerate in opposite directions, and the conservation of momentum tells us: m1v1=−m2v2⇒v1v2=m2m1m_1 v_1 = -m_2 v_2 \Rightarrow \frac{v_1}{v_2} = \frac{m_2}{m_1}
Assuming they push off simultaneously and for the same time, their displacements will be in the same ratio as their velocities: x1x2=v1v2=m2m1⇒12×2=8555⇒x2=12×5585=66085≈7.76 m\frac{x_1}{x_2} = \frac{v_1}{v_2} = \frac{m_2}{m_1} \Rightarrow \frac{12}{x_2} = \frac{85}{55} \Rightarrow x_2 = \frac{12 \times 55}{85} = \frac{660}{85} \approx 7.76 \text{ m}
Step 2: Total Distance Between Them
They move in opposite directions, so the total distance between them is: xtotal=x1+x2=12+7.76=19.76 metersx_{\text{total}} = x_1 + x_2 = 12 + 7.76 = \boxed{19.76 \text{ meters}}
Final Answer: 19.76 meters\boxed{19.76\text{ meters}}
This is how far apart the two astronauts are when the lighter one has moved 12 meters.