GENERAL PHYSICS 1 FINAL EXAM PRACTICE QUESTIONS HELP The mass of the Earth is 5.
ID: 1410676 • Letter: G
Question
GENERAL PHYSICS 1 FINAL EXAM PRACTICE QUESTIONS HELP
The mass of the Earth is 5.97e24 kg. The radius is 6.37e3 km The mass of the Moon is 7.35e22 kg. The radius is 1740km The average distance of the Earth and Moon is 3.84e5 km.
1. Use the potential energy (U) and kinetic energy (K) laws for gravity to derive the escape velocity from the Earth.
2. Use potential energy and kinetic energy to calculate how fast an object must go in order to reach a height of 1000.0 km above the Earth’s surface.
3. Use potential energy and kinetic energy to analyze the mechanical energy (E), potential energy (U) and kinetic energy (K) of an object with mass M attached to a spring with force constant k that moves with amplitude A. What are the energies when the spring is at maximum compression? What are the energies when the spring is unstretched (x=0)? What is the period (T) of motion? How fast is the object going at x=0 in terms of k, m and A?
Use potential energy and kinetic energy to analyze the ballistic pendulum, in which a block of mass M is hit with a bullet of mass m (they stick together). If you know the speed of the block after being hit, what is the kinetic, potential, and mechanical energies after being hit? What are they at the maximum height (h) of the pendulum? How is the speed of the block after being hit related to the speed of the bullet beforehand, assuming the block starts at rest?
Other problems.
What is the acceleration due to Earth’s gravity for the moon? What is the acceleration due to the Moon’s gravity for a 90-kg person on the Moon’s surface. In the class 1902 film A Trip to the Moon, the travelers return to the Earth by dropping off a cliff and then falling from the Earth to the Moon. Explain why that does or doesn’t work.
Explanation / Answer
1) Here, GMm/r = 1/2mv2
=> v = sqrt(2GM/r)
=> escape velocity = sqrt(2GM/r)
2) Here, 1/2 * m * v2 = GMm * (1/6.37 * 106 - 1/7.37 * 106)
=> v = sqrt(2 * 6.67 * 10-11 * 5.97 * 1024 * 2.13 * 10-8)
=> v = 4118.65 m/sec
=> fast an object must go = 4118.65 m/sec
3) Here, 1/2kx2 = 1/2mv2
=> period of motion = 2pi * sqrt(m/k)
=> velocity = (2pi/T) * sqrt(A2 - x2)
= sqrt(k/m) * sqrt(A2 - x2)
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