AP Physics, Third Quarter Notes

Unit P1-07-Impulse and Momentum Lesson Notes P1-07-1: Momentum and Impulse Name: Safiatu Contch State the definition of momentum. State the symbol and the units. Product of an objects mass + Velocity Momentum is a vector quantity. Explain what this means. Ly has direction P = mv Heuton Ingim Ns = unit S Lt or depends on it's v) Test Your Understanding: Momentum can be represented using a "momentum square diagram". Fill in the following square diagrams based on the description 6 NS A 2 kg toy car moves right at 3 m/s +6 +6 A 4 kg bowling ball moves left at 2 m/s 8 s (m / s) + 0 -1 +5 +4 +4 +3 +3 +2 +++ +2 +1 +1 0 0123456 m -2 -3 4 5 A 0.01 kg bullet travels upward at 400 m/s +6 + +++ +2 +1 0 + -1 4 Ans V (* 100m / s ) +4 +1+ +3 -2 -3 4 5 6 NS A 2000 kg car moves at 30 m/s to the west +6 +5 +4 V (* 10m / s ) +++ +1 0 3 6 + +++ +++ m ( 0.01-13 kg) -2 (* 1000 State the definition of impulse. State the symbol and units. Explain what impulse actually is. Impurse = change in momentum Delta*P =F Delta + How is impulse due to a constant force calculated? How is impulse due to a varying force calculated? Delta*p = F*Delta*t Multiply force by the time the force acts Variny force is represented by F vst grapa AP-area IF graph AP (m / s) (kg) (kg) - 7m N = Test Your Understanding: A toy car has mass 2 kg and initially moves forward with a speed of 3 m/s. Consider the following two cases. 1. The car crashes into a wall and stops as a result of the collision. 2. The car crashes into a wall and bounces backward with a speed of 3 m/s after the collision. In both cases, the car is only in contact with the wall for 2 seconds. Diagram these two situations and answer the questions that follow. Case 1 (Crash and stop): Car's Initial Momentum P=(2) (3)=6 + Impulse Applied to Car Car's Final Momentum + 171 Case 2 (Crash and Bounce): Car's Initial Momentum Impulse Applied to Car Car's Final Momentum 114 + 14 Egg-and-Sheet Example: If your teacher took an egg and threw it as hard as they could against a wall, the egg would shatter and make a mess everywhere. Instead, your teacher takes an egg and throws it as hard as they can against a bedsheet that is held vertically. State what you observe. Then explain why this happened using appropriate physical principles and vocabulary. Checklist 1 stated my observations 1 stated a les of physics that it always true connected the law of physics to the specific circumstances of the situation. I compared the two situations (stated what was the same in both cases) I contrasted the two situations ( different in both cases) stated what was I used physius vocabulary (momentum, sa force, velocity, speed, impulse, time) Car Crash Example: Cars in the 1950's were made out of solid steel so that they bounced off of each other when they collided. Cars today are built out of plastic materials that crumple when they collide. The speed limits during both time periods were about the same, and people in both time periods had about the same mass (on average). During which time period were injuries from accidents worse? Answer the question and explain your reasoning in a well-organized paragraph. Checklist: I answered the question directly 1 stated a law of physics that is always trae connected the law of physics to the specife circumstances of the station I compared the two situations (stated what was the same in both cers) I contrasted the two situations (stated what w different in both cases) I und physics vocabulary (momentum, ma force, velocity, speed, impulse, time) Example: A 0.03 kg falling ball reaches the floor moving with speed 5 m/s. When the ball is in contact with the floor, the net force acting on the ball increases linearly from zero to 8 N for 0.02 seconds, then decreases linearly back to zero for the next 0.04 seconds. The ball then leaves the floor with speed v. You are to determine the rebound speed v using the momentum square diagrams and F vs. / graph below. Ball's Initial Momentum + P=(. (3)(- 5) +6 Impulse Applied to Ball F(N) +12 5 +10 +8 +2 +1 0 2 +6 +4 (0.01 kg) + +2 -2 4 15 kg-m/s Ball's Final Momentum +6 +5 4 +3 +2 +1 0 (0.01 -2 -3 (0.01 kg) 01234567 6 -8 -10 -12 24 6 P 00 kg-m/s (a) Draw the square diagram for the initial momentum of the ball (remember the ball's direction). (b) Draw the graph of F vs. I described in the second sentence of the paragraph above. Then calculate the impulse applied to the ball. 24 Ng 1/7 * (8)(0.02) + 1/2 * (8)(0.04) = 0.2 + N_{7} (c) Draw the final momentum of the ball and calculate the rebound speed of the ball. mv / Delta * p = mvf (0.03)(- 4) + 0.24 = (0.05) * vf vf=^ 3m / r (d) What would net force applied to the ball was twice as strong and applied for half as much time? - (m / 2) (m / x) Delta*p = N_{s} p_{L} = Example: A 0.5 kg basketball is dropped from rest 1 m above the ground. The force that the ground exerts on the ball is shown below as a function of time when the ball bounces once off of the ground. m*v_{0} + f * ∞ * t = mv when the ball reaches the floor? 5(0) + 411 = (5) * v v =8.2 5 250 200 Upwant Force ( 50 D 100 150 0.02 0.04 Time (sec) 0.00 0.08 4.1125 150.00312 0.1 What impulse is applied to the ball? area under the F-1 graph 1/2 * (200)(0.64) = 4/5 How high does the ball go after bouncing? K = 0g 1/2 * m * v ^ 2 = mgy (0.5) * (3.5) ^ 2 = 10g y = 0.6 / m How fast does the ball move when the ball leaves the floor? Im -4.5m/s 오 triangle P + 4Ns m/5 m * v_{1} ^ 2 + Delta*rho = mvf (0.5)(- 4.5) + 4 = 0.5vf Vf = 3.5 * r/r International basketball regulations indicate that a basketball is properly inflated if the basketball bounces back up to between 53% and 64% of its drop height. Is this basketball properly inflated, under inflated, or over inflated? Explain your answer. 161/1 * 10 = 61x P1-07-2: Conservation of Momentum p = mv Delta*p = F*Delta*t = mv - m*v_{0} What does it mean for a system of objects to be "isolated" so that momentum can be conserved? Hint: It has t to do with "internal" and "external" forces. No unbalanced external force act for a significant amount of me A car of mass m initially moves to the right with speed v. If the car brakes to a stop, where does its momentum mv 80? Into the earth (friction) State the Law of Conservation of Momentum as an equation. Initial p=firal P m_{1}*v_{1}*d + m_{2}*v_{20} = m_{1}*v_{i} + m_{2}*v_{2} What would the Conservation of Momentum equation look like if we had three objects in our system? m_{1}*vi * 0 + m_{2}*v_{20} + m_{3}*v_{30} = m_{1}*v_{1} + m_{2}*v_{2} + m_{3}*v_{3} Fill in the table of "types of collisions". Type of Collision Word Description Relationship Between Initial and Final Kinetic Energy Relationship Between Initial and Final Momentum Explosion Gains in From external Usually as another energy type <= h i <<= hs p i = sum p F Elastic Total hi is the same as if Sigma*K_{i} = sum KF pi = Sigma*pf Inclastic Some of he is lost as is dissipated energy Sigma * |h|' > sum hf Sigma * p_{i} ^ x = Sigma*p_{f} Totally Obsects Stich togetter after the COMISION. Sigma*hi > Sigma*hf Test Your Understanding: A 2000 kg car and 4000 kg truck travel in opposite directions with speeds of 30 m/s. The car moves right and the truck moves left. If the two vehicles collide and lock together, how fast and in which direction will the wreckage be traveling? Make diagrams below to answer the question. p = mv Car's Initial Momentum (* 10m / s ) + +++ +5 6 Truck's Initial Momentum V (10 m/s) +6 ++ +5 System's Final Momentum V (* 10mh )\ +6 +4 +3 ++ + +++ +2 +2 +++ +1 +1 -1 ( -2 kg) -3 4 +5 +4 +3 +2 +1 = (* 1000 -1 -2 -3 -4 5 (* 1000 kg) 6 6 -6 2000 * 30 p_{e} = underline 60 000 kg-m/s Now of momentum the question (show an then plug-ins, then algebraic work, then the answer). m 2100... 10/20 + m + v_{10} =(m c+m+10 ) V v = (- 10 ^ m) / s forward with speed 6 m/s crashes into his 60 kg mother who is initially at rest and also on skates. The child pushes off of his mother so that the child now moves backward with speed 6 m/s. How fast and in what direction does the mother move after the interaction? Child's Initial Momentum + Mom's Initial Momentum Child's Final Momentum Mom's Final Momentum +6 (m / 0) v ci +5 +4 +3 +2 +1 C (10 ka) kg-m/s 16 +5 +4 +3 +2 +1 (+10 P kgm / s = +3 +2 -H + 0 (10 2 -3 -5 -6 p cc = underline -t0 kg-m/s P240 kg-m/s Now use a conservation of momentum equation to answer the question. How do you know, without making any calculations, that this collision is an "explosion"? p_{r} = - kgm / s 6000-10 p_{H} = - 1210050kgmVs 1000*-30 v mc (mk) m_{c} g ^ (mn) varphi(m / k) t ^ v = 0 -/ 1 m p_{a} = - m_{1} v ci What law of physics is Conservation of Momentum the same as? Newton's 3rd Law Derivation of Conservation of Momentum: For each line of the derivation, explain what mathematically was done. The first line is done for you as an example. - overline F tor * 1 = overline F ton 1 We start with Newton's Third Law, which states that the force that I exerts on 2 is equal and opposite to the force that 2 exerts on 1 - m_{2} vec a 2 =m 1 vec a 1 -m 2 vec v 2f - vec v 2i Delta*t =m 1 vec v if - vec v ii Delta t - m_{2} ( overline v 2f - overline v 2i )=m 1 ( overline v 1f - overline v 10 ) -m 2 overline v 2f +m 2 overline v 2i =m 1 overline v if -m 1 overline v 0i m_{1} vec v 0 +m 2 vec v 2i =m 1 vec v 1f +m 2 vec v 2f Test Your Understanding: Answer each yes/no question and then EXPLAIN YOUR REASONING. (a) Is it possible for a single object to be moving and yet have zero kinetic energy? 10 1 - 1/2 * m * n ^ 2 Soil V= then 70 (b) Is it possible for a system of objects to all be moving and yet No, it's impossible to add wpavposkv e ne t0=0 (c) Is it possible for a single object to be moving and yet have zero momentum? No 1 p= mosA v ne emptyset p ne0 (d) Is it possible for a system of objects to all be moving and yet the system has zero total momentum? Jes-moving in different Directions Example: A car traveling at 30 m/s forward strikes a barrier head-on and bounces backward moving at 10 m/s. The driver is 100 kg. If the car was in contact with the barrier for only 0.5 seconds, how much force was applied to the driver during the collision? (Show a solution using forces and motion, then show a solution using impulse and momentum.) Forces anst Motion Solurim V = V_{0} + at - 10 = 30 + a(0.5) - 3x - 30 F = m * d (100)(- 80) = 8 i00 (40 = 5)/2 /s2 Impulse and Monsentium Solution m*v_{0} + Delta*p = mv F = - 810lambda m*v_{0} + f' * Delta*t = mv 100(30) + r(5) = (100)(- 10) . 4 x0+F(.5=-1000