Transcript Document
Energy Revision Kinetic Ek = ½ mv2 Potential Ep = mgh Heat EH = cmT (changing temperature) Heat EH = ml (changing state) Electrical EE = Pt = IVt light Work done Ew = Fd energy sound nuclear E = Pt Energy is conserved, this means the total energy remains the same h At top of slope At bottom of slope Ek = 0 Ek = ½mv2 Ep = mgh Ep = 0 Assume no friction Total energy is conserved Ep at top = Ek at bottom mgh = ½mv2 gh = ½ v2 v = 2gh ie mass is not important With friction Total energy is conserved Ep at top = Ek at bottom + Ew mgh = ½mv2 + Fd Purple book Ex 2.9 Small mass, fast Or Large mass, slow Which would do the most damage? Momentum = mass x velocity momentum = mv Momentum is a vector quantity measured in kgm/s or kgms-1 Tutorial question 76 Collisions – trolleys stick together afterwards At start trolley 2 is stationary. Momentum before m1u1 kgm/s before m1 m2 v kg kg m/s 0.5 0.5 0.5 1 0.5 1.5 1 0.5 1 1 (m1 + m2) v Total momentum after kgm/s after Explosion – velocity before explosion is zero. Momentum m1 v1 before kg m/s m1v1 m2 kgm/s kg 0 0.5 0.5 0 0.5 1 0 0.5 1.5 0 1 0.5 0 1 1 v2 m 2 v2 Total m/s kgm/s momentum after The Principle of Conservation of momentum Total momentum remains the same provided there are no outside forces. Total momentum before = total momentum after Elastic collisions – where the total kinetic energy is conserved. Inelastic collisions – where the kinetic energy is not conserved eg. Some of the kinetic energy is changed into heat and sound energy A B 1. Trolley A with a mass of 2kg and a velocity of 1m/s, collides head on with trolley B, mass 2kg moving at 2m/s. Velcro causes them to stick together. (a) What is their speed after the collision? (b) Is it an elastic or inelastic collision? Tutorial question 77 to79 2. Two rubber balls collide head on as shown. The red ball rebounds at 1m/s. (a)What is the velocity of the green ball? (b) Is it an elastic or inelastic collision? A B mass 4kg mass 2kg velocity 2m/s velocity 3m/s Tutorial questions 80 to 84 3. A field gun of mass 1000kg fires a shell of mass 5kg with a velocity of 100m/s. Calculate the recoil velocity of the gun. Tutorial questions 85 to 88 SAQs up to 71 Impulse – on one object 1. Impulse = average force x time = Ft Impulse is a vector quantity measured in newton seconds (Ns) 2. Impulse = change in momentum of the one object = mv - mu Impulse is also measured in kgm/s or kgms-1 3. Impulse = area under the force time graph force time This means that Impluse = Ft = mv – mu = area under force time graph Things to beware of F is average force, not the maximum. Direction and sign of velocities ie rebound Impact time is often very short and can be given in milliseconds (ms) Mass is often given in grams 1. A car with mass 600 kg and velocity of 40 m/s skids and crashes into a wall. The car comes to rest 50 ms after hitting the wall. Calculate the average force on the car during the collision. F/N 2. During a game of hockey a stationary ball of mass 150 g is struck by a player. The graph shows how the force on the ball varies with time. (a) Calculate the impulse on the ball. (b) Calculate the speed which it leaves the stick. 1200 3 6 t/ms (c) A softer ball is hit and leaves the stick with the same velocity. Sketch its force time graph. Aim: To find the average force exerted by a cue on a snooker ball. Measurements d, diameter of ball = m t1, contact time = s t2, time to go through light gate = m, mass of ball = kg Calculations Velocity of ball after collision, v = d/t2 = Average force, F = (mv – mu)/t1 = Tutorial questions 88 to 99 SAQs up to 79 s Wearing no seatbelt The person continues to move forward at a constant speed. Newton’s first law Until they collide with the dashboard etc, stopping them suddenly. F = (mv – mu)/t so short time means large average force Wearing a seat belt The person is brought to a stop at the same time as the car is stopping. The stopping time is increased as the car crumples and the seatbelt has some give. The force is also on the parts of the body where it will do the least harm F = (mv-mu)/t the same change in momentum but a longer time means a smaller average force. Tutorial questions 88 to 99 SAQs up to 79 Points to note 1. The rate the fuel is ejected at is 30 kg/s. This means in a time of 1 second there is a mass of 30 kg 2. change in momentum of the rocket = - change in momentum of fuel Newton’s Third Law Tutorial questions 88 to 99 SAQs up to 79