A 90 kg astronaut Travis is stranded in space at a point 12 m from his spaceship. In order to get back to his ship, Travis throw
1 answer:
Answer:
Explanation:
This is a recoil problem, which is just another application of the Law of Momentum Conservation. The equation for us is:
which, in words, is
The momentum of the astronaut plus the momentum of the piece of equipment before the equipment is thrown has to be equal to the momentum of all that same stuff after the equipment is thrown. Filling in:
Obviously, on the left side of the equation, nothing is moving so the whole left side equals 0. Doing the math on the right and paying specific attention to the sig fig's here (notice, I added a 0 after the 4 in the velocity value so our sig fig's are 2 instead of just 1. 1 is useless in most applications).
0 = 90.0v - 2.0 and
2.0 = 90.0v so
v = .022 m/s This is the rate at which he is moving TOWARDS the ship (negative was moving away from the ship, as indicated by the - in the problem). Now we can use the d = rt equation to find out how long this process will take him if he wants to reach his ship before he dies.
12 = .022t and
t = 550 seconds, which is the same thing as 9.2 minutes
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1. 0.42 Hz
The frequency of a simple harmonic motion for a spring is given by:
where
k = 7 N/m is the spring constant
m = 1 kg is the mass attached to the spring
Substituting these numbers into the formula, we find
2. 2.38 s
The period of the harmonic motion is equal to the reciprocal of the frequency:
where f = 0.42 Hz is the frequency. Substituting into the formula, we find
3. 0.4 m
The amplitude in a simple harmonic motion corresponds to the maximum displacement of the mass-spring system. In this case, the mass is initially displaced by 0.4 m: this means that during its oscillation later, the displacement cannot be larger than this value (otherwise energy conservation would be violated). Therefore, this represents the maximum displacement of the mass-spring system, so it corresponds to the amplitude.
4. 0.19 m
We can solve this part of the problem by using the law of conservation of energy. In fact:
- When the mass is released from equilibrium position, the compression/stretching of the spring is zero: , so the elastic potential energy is zero, and all the mechanical energy of the system is just equal to the kinetic energy of the mass:
where m = 1 kg and v = 0.5 m/s is the initial velocity of the mass
- When the spring reaches the maximum compression/stretching (x=A=amplitude), the velocity of the system is zero, so the kinetic energy is zero, and all the mechanical energy is just elastic potential energy:
Since the total energy must be conserved, we have:
5. Amplitude of the motion: 0.44 m
We can use again the law of conservation of energy.
- is the initial mechanical energy of the system, with
being the initial displacement of the mass and
being the initial velocity
is the mechanical energy of the system when x=A (maximum displacement)
Equalizing the two expressions, we can solve to find A, the amplitude:
6. Maximum velocity: 1.17 m/s
We can use again the law of conservation of energy.
- is the initial mechanical energy of the system, with
being the initial displacement of the mass and
being the initial velocity
is the mechanical energy of the system when x=0, which is when the system has maximum velocity,
Equalizing the two expressions, we can solve to find , the maximum velocity: