Wagner OnaPhysics Year 12

Gun powder rockets

Definitions:

Kinetic energy is the energy of motion or the energy that a body possesses by virtue of being in motion.1 The equation for kinetic energy is (KE = 0.5 • m • v2) where m = mass of object and v = speed of object

Gravitational energy is the potential energy held by and object because of its high position compared to a low position, it is energy associated with gravity/ gravitational force2. The equation for this is PEgrav = mass • g • height

The law of Conservation of energy states that the total energy of an isolated system does not changed, it is conserved over time.3 The basic formula for conservation of energy is Energy spent in one act = Energy gained in the related act4

Efficiency generally talking is to how far we are getting the particular outcome for the given input with as much less waste as possible. It is defined as the ratio of useful work done to heat or energy absorb by a system. 5 There are two equations to this,

Efficiency formula in terms of Work is given by

Efficiency formula in terms of Energy is given by

Exercises: second page. Use the conservation of energy and the equations below to answer the following questions:

1 http://faculty.wwu.edu/vawter/PhysicsNet/Topics/Work/WorkEngergyTheorem.html

2 http://simple.wikipedia.org/wiki/Gravitational_energy

3 http://en.wikipedia.org/wiki/Conservation_of_energy

4 http://physics.tutorvista.com/energy/conservation-of-energy.html

5 http://formulas.tutorvista.com/physics/efficiency-formula.html

Wagner OnaPhysics Year 12

1. Assuming that the rocket uses all of its gunpowder in one initial explosion calculate the initial vertical velocity of the rocket for it to reach an altitude of 400m (use our equations of motion to answer this).V: 0S: 400T: ?U: ?A: 9.81V2=U2+2as Rearrange to find U √v2−2as=U

√02−2×−9.81×400=88.6m/ sAnswer: 88.6m /s

2. What mass of fuel will the rocket need to reach an altitude of 400m (you will need to make some estimates here of the mass of gunpowder needed and use trial and error to work this out)Kinetic energy: 1/2xmxv2added masses on the rocket 200+150g which was 350g then divided by 1000 to get kg and used Ek equation 1/2x0.35x88.582=1373.122J

*The rocket is pushed upwards with a gunpowder rocket of efficiency 25%, energy density of 3.0 MJkg-1, this means that when 1kg of gunpowder explodes it releases 3.0 million Joules of energy.

Therefore equation used: Efficiency X density X Unknown Grams X 1 million= gunpowder needed. Hence tried with 1 kg = (1/4x3x1)x(1x106)=750000 (answer way off 1373 joules)Tried with 0.0020 kg = (1/4x0.0020x1) x (1x106) = 1500 (answer close)Tried with 0.00181 kg = (1/4x0.0181x1) x (1x106) = 1357.5 (closes value)Then tried with 0.00183 kg = (1/4x0.0183x1) x (1x106) = 1372.5 (closest value)Therefore it needs a mass of 0.00183 kg or 1.83 grams of fuel to reach an altitude of 400m.

3. How much kinetic energy will the rocket need at ground level?Ek=1/2mv2

1/2x0.35x88.582=1373.122J4. How much gravitational potential energy will the rocket gain?

Egp=mg∆h0.35x9.81x400= 1373.4JThe conservation of energy concept is correct here, the energy does not disappear it stays isolated and conserved over time. This shows chemical energy converted into kinetic energy, as seen on question 3 and 4.