A 97kg astronaut and a 1100kg satellite are at rest relative to the space shuttle. The astronaut pushes on the satellite, giving it a speed of 0.13m/s directly away from the shuttle. Seven and a half seconds later the astronaut comes into contact with the shuttle. What was the initial distance from the shuttle to the astronaut?
explain better
To solve this problem, we can use the principle of conservation of momentum. The total momentum of the system before and after the interaction remains constant.
1. Calculate the initial momentum of the system:
- For the astronaut: mass (m1) = 97 kg
- For the satellite: mass (m2) = 1100 kg
- Initial velocity (v1) of both the astronaut and satellite relative to the shuttle: 0 m/s
- Initial momentum (p_initial) = m1 * v1 + m2 * v1 (since both have the same initial velocity)
2. Calculate the final momentum of the system:
- For the astronaut: mass (m1) = 97 kg
- For the satellite: mass (m2) = 1100 kg
- Final velocity (v2) of the satellite relative to the shuttle: 0.13 m/s
- Final momentum (p_final) = m1 * v_final + m2 * v_final
The momentum before the interaction should be equal to the momentum after the interaction.
3. Set up the equation:
m1 * v1 + m2 * v1 = m1 * v_final + m2 * v_final
Now we can solve for the initial distance from the shuttle to the astronaut.
4. Rearrange the equation:
v1 * (m1 + m2) = v_final * (m1 + m2)
v1 = v_final
Since the mass terms are the same on both sides of the equation, we can ignore them and conclude that v1 equals v_final.
5. Calculate the initial distance (d_initial) using the formula for speed:
d_initial = v1 * t
Now we can substitute the given values into the equation to find the initial distance.
6. Solve for the initial distance:
d_initial = v1 * t
d_initial = 0.13 m/s * 7.5 s
Calculating the result:
d_initial = 0.975 meters
Therefore, the initial distance from the shuttle to the astronaut is approximately 0.975 meters.
To find the initial distance from the shuttle to the astronaut, we can use the equation of motion:
\(d = v_it + \frac{1}{2} a t^2\)
Where:
- \(d\) is the distance
- \(v_i\) is the initial velocity
- \(t\) is the time
- \(a\) is the acceleration
First, let's find the acceleration of the astronaut by applying Newton's second law:
\(F = m \cdot a\)
Where:
- \(F\) is the force applied by the astronaut
- \(m\) is the mass
The force applied by the astronaut to the satellite causes an equal and opposite force on the astronaut, according to Newton's third law. So, the astronaut is pushed backward with the same force. Therefore:
\(m_{astronaut} \cdot a_{astronaut} = -m_{satellite} \cdot a_{satellite}\)
Given:
- \(m_{astronaut} = 97 \, \text{kg}\)
- \(m_{satellite} = 1100 \, \text{kg}\)
- \(a_{satellite} = 0 \, \text{m/s}^2\) (since the satellite remains at rest in the reference frame of the shuttle)
We can now solve for the acceleration of the astronaut:
\(a_{astronaut} = \frac{-m_{satellite} \cdot a_{satellite}}{m_{astronaut}}\)
\(a_{astronaut} = \frac{-1100 \, \text{kg} \cdot 0 \, \text{m/s}^2}{97 \, \text{kg}}\)
\(a_{astronaut} = 0 \, \text{m/s}^2\) (since any value multiplied by zero is zero)
Since the acceleration of the astronaut is zero, it means the astronaut is not accelerating. Therefore, the initial velocity of the astronaut is also zero.
Now, let's find the distance traveled by the astronaut during the 7.5 seconds until contact.
\(d_{astronaut} = v_i \cdot t + \frac{1}{2} a_{astronaut} t^2\)
\(d_{astronaut} = 0 \cdot 7.5 \, \text{s} + \frac{1}{2} \cdot 0 \cdot (7.5 \, \text{s})^2\)
\(d_{astronaut} = 0 \)
Since the astronaut does not move relative to the shuttle during the time until contact, the initial distance from the shuttle to the astronaut is also zero.
Solve using momentum.
Initial momentum=0, since both astronaut and satellite are stationary with respect to shuttle.
Assuming velocity away from shuttle is positive, and m1=mass of astronaut, m2=mass of satellite, then
m1v1+m2v2=0
97v1+1100*0.13
Solve for v1 to get
v1 = -1100*0.13/97
=-1.474 m/s (i.e. towards the shuttle.