PHYS_2014_Homework_5_soln

PHYS_2014_Homework_5_soln - PHYS2014 Benton Fall 2010 OSU...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
PHYS2014 Fall 2010 Benton OSU Physics Dept. 1 Physics 2014: General Physics I Solutions to Homework Assignment 5 1. Astronomers predict that on Friday 13 April 2029 the asteroid 99942 Apophis is going to pass within 30,000 km of the Earth. Given the fact that, on average, the Moon is 385,000 km distant from the Earth, this asteroid is going to pass pretty darn close. 99942 Apophis has a mass of 2.7 × 10 10 kg and is expected to be traveling at a speed of 12.6 km/s relative to the Earth at the time of closest approach. What if the astronomers goofed in their calculations and the asteroid is actually going to hit the Earth? With what kinetic energy would 99942 Apophis impact the Earth in terms of megatons of TNT. A detonation of 1 megaton of TNT releases 4.2 × 10 15 J of energy. The largest nuclear weapon ever detonated (the Tsar Bomba test by the former Soviet Union) was about 50 megatons, while the largest nuclear weapon the U.S. ever detonated (the Castle/Bravo test) was 15 megatons. Solution: ( ) 2 2 10 18 18 15 1 1 m 2.7 10 kg 12,600 4.3 10 J 2 2 s 4.3 10 J 1021 Mton J 4.2 10 Mton K mv = = × = × × = × This is equal to 68 of the largest hydrogen bombs the US ever detonated…all at once and all in the same place! Let’s hope the astronomers don’t get it wrong. 2. A 5.0 kg mortar shell is fired at an angle of 35 ° with respect to the horizontal and has an initial speed of 120 m/s. Neglecting air resistance, what is the potential energy of the shell with respect to the ground at the top of its trajectory? Solution: Using Conservation of Energy, we can assume that all the kinetic energy of the mortar shell at the beginning of its trajectory is converted into gravitational potential energy. However, it will only be that fraction of the kinetic energy that goes into countering gravity and takes the shell to height, h . This will be proportional to the sin of the angle. ( ) ( ) 2 2 1 sin 2 1 5.0kg 120m s sin35 11,844J 2 i gf i gf gf K U m v mgh U U θ = = = = = r We check our answer by finding the maximum height reached by the mortar shell using kinematics and plugging that into our equation for gravitational potential energy.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
PHYS2014 Fall 2010 Benton OSU Physics Dept. 2 2 f v ( ) ( ) ( ) ( ) 2 2 2 2 2 sin 2 120m s sin35 sin 242m 2 2 9.8m s 5.0kg 9.8m s 735m 11,844J i i gf v gh v h g U mgh θ = + = − = − = ⋅ − = = = r 3. You are watching a National Geographic Special on television. One segment of the program is about archer fish which inhabit streams in southeast Asia. This fish actually spits drops of water at insects to knock them into the water so it can eat them. The commentator states that the archer fish keeps its mouth at the surface of the pond and squirts a jet of water from its mouth at a speed of 4 m/s. You watch an archer fish shoot a juicy mosquito off a leaf and into the stream. You estimate that the leaf was about 3/4 of a meter above the stream. You wonder at what minimum angle from the horizontal must the fish shoot a water droplet in order to hit the mosquito. Since you have time during the commercial, you quickly calculate this angle using conservation of energy.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/10/2012 for the course PHYS 2014 taught by Professor Nandi during the Fall '08 term at Oklahoma State.

Page1 / 10

PHYS_2014_Homework_5_soln - PHYS2014 Benton Fall 2010 OSU...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online