An astronaut in her space suit has a total mass of m1 = 76.9 kg, including suit and oxygen tank. Her tether line loses its attachment to her spacecraft while she's on a spacewalk. Initially at rest with respect to her spacecraft, she throws her oxygen tank of mass m2 = 12.0-kg away from her spacecraft with a speed v = 7.70 m/s to propel herself back toward it.
(a) Determine the maximum distance she can be from the craft and still return within 2.00 min (the amount of time the air in her helmet remains breathable).
(b) Explain in terms of Newton's laws of motion why this strategy works.
To solve the problem, we can use the principle of conservation of momentum. The momentum before the oxygen tank is thrown is equal to the momentum after the oxygen tank is thrown. The momentum is given by the equation:
(m1 + m2) * v1 = m1 * v2
Where:
m1 = 76.9 kg (mass of the astronaut)
m2 = 12.0 kg (mass of the oxygen tank)
v = 7.70 m/s (speed of the oxygen tank)
v1 = velocity of the astronaut with respect to the spacecraft before throwing the oxygen tank
v2 = velocity of the astronaut with respect to the spacecraft after throwing the oxygen tank
(a) To determine the maximum distance she can be from the craft and still return within 2.00 min, we can use the equation for displacement:
v2 = u + at
Where:
u = initial velocity of the astronaut with respect to the spacecraft
a = acceleration of the astronaut
t = time
Rearrange the equation to solve for displacement:
s = u * t + 0.5 * a * t^2
The initial velocity can be determined by rearranging the momentum equation:
v1 = (m1 + m2) * v2 / m1
Substituting the given values:
v1 = (76.9 + 12.0) * v2 / 76.9
Now, we can substitute this value of v1 into the equation for displacement:
s = (76.9 + 12.0) * v2 * t / 76.9 + 0.5 * a * t^2
We're given that the time is 2.00 minutes, which we need to convert to seconds:
t = 2.00 * 60 = 120 seconds
We also know that the acceleration is due to the force applied by throwing the oxygen tank:
F = m2 * v / t
a = F / m1
Substituting in the given values:
F = (12.0 kg) * (7.70 m/s) / (120 s) ≈ 0.6167 N
a = (0.6167 N) / (76.9 kg) ≈ 0.008 N/kg
Substituting the values of v2 and a into the equation for displacement:
s = (76.9 + 12.0) * v2 * t / 76.9 + 0.5 * a * t^2
Now we can plug in the values and solve for the maximum distance:
s = (76.9 + 12.0) * (7.70 m/s) * (120 s) / 76.9 + 0.5 * (0.008 N/kg) * (120 s)^2
s ≈ 144.44 meters
Therefore, the maximum distance she can be from the craft and still return within 2.00 min is approximately 144.44 meters.
(b) This strategy works based on Newton's laws of motion. According to the law of conservation of momentum, the total momentum before and after an event remains constant, unless acted upon by an external force. When the astronaut throws the oxygen tank, she propels herself in the opposite direction with an equal magnitude of momentum.
This is in accordance with Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. The force exerted by the astronaut's throw causes the oxygen tank to move in one direction, while the astronaut moves in the opposite direction. This allows the astronaut to generate an equal and opposite momentum, propelling her toward the spacecraft.
To determine the maximum distance the astronaut can be from the craft and still return within 2.00 minutes, we need to consider the conservation of momentum.
(a) Initially, the total momentum of the astronaut and the oxygen tank is zero since they were at rest. When she throws the oxygen tank away, the momentum of the tank -m2*v is in the opposite direction to the astronaut's momentum. According to the law of conservation of momentum, the total momentum remains constant.
Given:
Mass of the astronaut, m1 = 76.9 kg
Mass of the oxygen tank, m2 = 12.0 kg
Speed at which the oxygen tank is thrown, v = 7.70 m/s
Let the final velocity of the astronaut be V.
The change in momentum of the astronaut is given by:
Δp = m1*V - (-m2*v)
= m1*V + m2*v
Since the total momentum remains zero,
0 = m1*V + m2*v
Solving for V, we get:
V = - (m2*v) / m1
Now, we can calculate the maximum distance the astronaut can be from the craft.
We know that distance = speed * time.
Given:
Time, t = 2.00 minutes = 120 seconds
Distance = V * t = - (m2*v) / m1 * t
Substituting the given values, we can calculate the maximum distance.
(b) According to Newton's third law of motion, for every action, there is an equal and opposite reaction. When the astronaut throws the oxygen tank away, she experiences a reaction force in the opposite direction. This reaction force propels her forward due to Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
By throwing the oxygen tank away with a certain speed, the astronaut experiences an equal but opposite force that propels her forward. This allows her to conserve momentum and use the reaction force to propel herself back toward the spacecraft.
The law of conservation of linear momentum
0=mv-MV
V=mv/M= 12•7.7/76.9 =1.2 m/s
S=Vt=1.2•2•60 = 144.2 m