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1/14/2015 MOMENTUM S Momentum S Question: How is the momentum of a body changed? Answer: By the application of a force F! What is the magnitude of the momentum of a 28-g sparrow flying with a speed of 8.4 m/s? S What is the momentum possessed by a car with a mass of 2350 kg if it is traveling at 125 km/h? S The velocity must be expressed in units of m/s, so we do the conversion: = 34.72 m/s NEWTONS SECOND LAW WAS WRITTEN IN MOMENTUM Momentum S Momentum p is defined as the product of mass and velocity, mv. Units: kg m/s p = mv m = 1000 kg How much momentum does an 800 Kg car have that travels at 4.5 m/s (~10 mi/hr)? S Momentum p = (1000 kg)(16 m/s) p = 16,000 kg m/s v = 16 m/s 1 1/14/2015 How fast would a 200 gram bullet have to travel to have the same amount of momentum as the car in previous question? S S You can forget about all forces that are internal Or isolated in a system. If a 1Kg projectile travelling at 20 m/s strikes a .5 Kg target originally at rest and they stick together, with what velocity do they travel off together? S Vector Nature of Momentum S Consider the change in momentum of a ball that is dropped onto a rigid plate: + vf vo A 2-kg ball strikes the plate with a speed of 20 m/s and rebounds with a speed of 15 m/s. What is the change in momentum? p = m(vf –vo)= (2 kg) [(15 m/s) - (-20 m/s)] p = 2kg (35 kg m/s) Does the angel of reflection effect the momentum? Momentum S p = 70 kg m/s No Momentum Before = Momentum After S • We don’t know why bodies maintain a relationship between mass and velocity, • We do know that momentum is conserved in the interactions between bodies • This will lead us to the law of the conservation of momentum (like energy) 0 = Procket - Pgas BUT AS MASS DECREASED VELOCITY INCREASES 2 1/14/2015 Momentum and Force S IMPULSE S • If a particle changes its momentum then a force must have acted on it • If a force acts on a particle then momentum must change. • If all forces sum to zero then (all internal forces-not external) then COM would hold. The impulse is the change in momentum. Momentum: p = mv The most general statement of Newton’s 2nd S COURSE TARGETS S Total force acting on a body = momentum / time. It is more general than ∑F = ma because it allows for the mass m to change with time also(2nd law keeps mass constant) Note: if m is constant, it becomes: http://www.youtube.com/watch?v=yUpiV2I_IRI&safe=active S S Goodnight GoodnightSucka Sucka Goodnight Sucka Goodnight GoodnightSucka Sucka Goodnight Sucka Impulse = Change in “mv” 3 1/14/2015 S S F=dp/dt • Offer to drop hammer or beaker brush on hands • Offer to hit hands with brush or wood handle side of beaker • Jump up ask them if they would want to land legs locked versus bend legs (more force less time – momentum change is same. Impulse Calculus Connection Force is the time derivative of momentum We also discussed power, the time derivative of energy: P = dE / dt. S Newton’s Second Law S Impulse Changes Velocity S How are momentum and impulse defined? Momentum = mass x velocity or P = mv. Impulse refers to a change in momentum. What is the difference in momentum of a Mack truck and a roller skate and why? The Mack truck has more mass and greater momentum. A tennis ball of mass 0.045 kg is hit at a speed of 45 m/s The racket was in contact with the ball for 0.0035s. Find (a) the impulse imparted to the tennis ball, and (b) the average force exerted on the ball by the racket. S Consider a mallet hitting a ball: F 4 1/14/2015 IMPULSE Impulse from a Varying Force S S Normally, a force acting for a short interval is not constant. It may be large initially and then play off to zero as shown in the graph. In the absence of calculus, we use the average force Favg. F As force increases on a mass, so Impulse: a force F does that masses acceleration. increases for a smaller time interval t. time, t The unit for impulse is the Newton-second (N s) S Impulse can often be shown as AUC on an F vs. t curve: ) Momentum is constant, so increasing the time requires the force to be less (nice on knees The face of a golf club exerts an average force of 4000 N for 0.002 s. What is the impulse imparted S to the ball? Impulse: F P = (4000 N)(0.002 s) t If we multiply the average force by the Average force constant force that would give the same result as the actual force Impulse and Momentum S Two flexible balls collide. The ball B exerts an average force of 1200 N on ball A. How long were the balls in contact if the impulse is 5 N s? S Impulse = Change in momentum Impulse: A force F acting on a ball for a time t increases its momentum mv. Momentum is constant, so increasing the time requires the force to be less (nice on knees) A B t = 0.00420 s The impulse is negative; the force on ball A is to the left. Unless told otherwise, treat forces as average forces. 5 1/14/2015 A 50-g golf ball leaves the face of the club at 20 m/s. If the club is in contact for 0.002 s, what average force acted on the ball? S What impulse is required to stop a 0.25 kg baseball traveling at 52 m/s? (b)If the ball is in the fielder’s mitt for 0.10 seconds as it is being stopped, what average force was exerted on the ball? S F (0.002 s) = (0.05 kg)(20 m/s) Average Force: F = 500 N A 0.145-kg baseball pitched at 39.0 m/s is hit on a horizontal line drive straight back toward the pitcher at 52 m/s. If the contact time between bat and ball is .003 s. calculate the average force between the ball and bat during contact. S A 500-g baseball moves to the left at 20 m/s striking a bat. The bat is in contact with the ball for 0.002 s, and it leaves in the opposite direction at 40 m/s. What was average force on ball? S + m = 0.5 kg F - 20 m/s + 40 m/s t Towards the pitcher My wife’s 1500-kg car at -15.0 m/s hits a garage door that does not move (oh I wish) but the cars final velocity is +2.6m/s. If the contact time between bumper and door is 0.15 s. calculate the a) impulse and b) the average force. Continued . . . S S + m = 0.5 kg + F - 20 m/s 40 m/s t F(0.002 s) = (0.5 kg)[40 m/s -(-20 m/s)] F(0.002 s) = (0.5 kg)[60m/s] F(0.002 s) = 30 kg m/s F = 15,000 N 6 1/14/2015 S You (mass = 55 kg) run into boulder at 8.0 m/s. You stop in 0.010 s. What force acts on you? A constant friction force of 25 N acts on a 65kg skier for 20 s. What is the skier’s change in velocity? S If the skier moves to the right, then the speed will decrease, because the friction force is to the left. F = -44,000N Now you run into a haystack, it takes 3.5 s to stop. What force acts? The skier loses 7.7 m/s of speed. S S 1. How much momentum does a 20 gram spitball traveling towards a teacher at 3 m/s have? 2. If a 100 Kg linebacker running at 5 m/s hits a 60 Kg quarterback and slows down to 1 m/s, how fast does the quarterback fly back? 3. If the same 100 Kg linebacker going 5 m/s hits a 60 Kg quarterback and “wraps him up” and they travel backwards together, at what velocity do they go back? 4. So why do football coaches tell their defense to “wrap up” when they hit? Water splashes back! 5. If Mr. C (65 Kg of solid muscle) stands on a skateboard and throws an 8 Kg bowling ball at 3 m/s, how fast will he go in the opposite direction? S S force as the time rate of change of momentum: F = dP / dt. If we assume for the moment that the mass of an object is not changing (not necessarily true: consider a rocket!), we see that F = m dV / dt = m a, It is important to keep in mind the two alternate "definitions" of force in terms of derivatives. As a time derivative of momentum, force is necessary to change an object's momentum. Alternatively, when an object's momentum is changed by contact with another object, it exerts a force on that object (remember this the next time you get hit with a ball!). 7 1/14/2015 S http://www.youtube.com/watch?v=XjwO9InuFJk CONSERVATION OF MOMENTUMS Law of Conservation of Momentum Total (vector) momentum before collision = total (vector) momentum after collision. For 2 colliding objects, (zero external force) the total momentum is conserved (constant) throughout the collision. The total (vector) momentum before the collision = the total (vector) momentum after the collision. CONSERVATION OF MOMENTUM S CONSERVATION OF MOMENTUMS HOLDS for ALL collisions! But only If all forces sum to zero then: S What is the resultant momentum of S this system? 5kg at 1 m/s 5kg at 1 m/s 8 1/14/2015 What is the resultant momentum of S this system? 5kg at 1 m/s Donald Crider-Phelps pushes off a boat to cross the English channel (Uh yeah but he goes the wrong way.) He weighs 109 kg and the boat weighs 1025kg. Both are initially at a velocity of 15m/s. If the DC-P dives against the boats motion at 5 m/s what was the resultant velocity of the boat? (uh, no fat jokes here people) S 2 kg at 3 m/s My grandfather was thrown from a jeep. He weighed 120kg. The jeep weighed 3000 kg. The jeep was initially moving at 80km/hr and a changed to 79km/hr when he flew out. What was his speed? S A bow hunter walks across a frozen lake (frictionless). He sights his game. He weighs 60 kg and the arrow weighs 500g. Both are initially at rest. The arrow is fired at 50.0 m/s what was the resultant velocity of the Archer? Luckily for his game the hunter hasn’t taken physics S 80km/hr = 22.2m/s 79.88km/hr = 21.94m/s G is Grandad, J is the Jeep Note that the masses do not have to be converted to kg, since all masses are in the same units, and a ratio of masses is what is significant. Michael Phelps pushes off a boat to cross the English channel (everybody needs a little head start you know.) He weighs 85 kg and the boat weighs 1025kg. Both are initially at rest. If the Michael moves at 5 m/s what was the velocity of the boat? S An Elephant Gun (for display purposes only)weighs 10.kg and is at rest. A bullet weighs (for target practice only) 50 grams and exits the barrel at 500MPH S 9 1/14/2015 My new 3800-kg long bed truck coasts along with a constant speed of 8.6 m/s. Snow begins to fall vertically and fills the bed at 3.5 kg/min. Ignoring friction with the tracks, what is the speed of the car after 90.0 min? S S N An atomic nucleus initially moving at 420 m/s emits an alpha particle in the direction of its velocity, and the remaining nucleus slows to 350 m/s If the alpha particle has a mass of 4.0 amu and the original nucleus has a mass of 222 amu, what speed does the alpha particle have when it is emitted? masses do not have to be converted to kg, since all masses are in the same units, and a ratio of masses is what is significant. S S S S A is alpha particle, N is the nucleus If a 1 Kg projectile traveling at 20 m/s strikes a 20 Kg target and the projectile moves back at 10 m/s, how fast does the target move? 10 1/14/2015 S S S S S S 11 1/14/2015 S • Advantage of bending knees when landing! Impulse: p = -540 N s m =70 kg, h =3.0 m, F = ? c) Knees bent: v = 7.7 m/s to v = 0 in d = 50 cm (0.5m) v = (½ ) (7.7 +0) = 3.8 m/s Time t = d/v = 0.13 s F = |p/t| = 4.2 103 N (Net force upward on person) From free body diagram, F = Fgrd - mg 4.9 103 N S Leg bone does not break!!! Example 7-6 S • Advantage of bending knees when landing! a) m =70 kg, h =3.0 m Just before he Impulse: p = ? hits the ground Ft= p = m(0-v) First, find v (just before hitting): KE + PE = 0 Just after he m(v2 -0) + mg(0 - h) = 0 hits the ground v = 7.7 m/s Impulse: p = -540 N s Opposite the person’s momentum • Advantage of bending knees when landing! Impulse: p = -540 N s m =70 kg, h =3.0 m, F = ? b) Stiff legged: v = 7.7 m/s to v = 0 in d = 1 cm (0.01m)! v = (½ ) (7.7 +0) = 3.8 m/s Time t = d/v = 2.6 10-3 s F = |p/t| = 2.1 105 N (Net force upward on person) From free body diagram, F = Fgrd - mg 2.1 105 N Enough to fracture leg bone!!! S Problem 17 • Impulse: p = change in momentum wall. • Momentum || to wall does not change. Impulse will be wall. Take + direction toward wall, vi = v sinθ Impulse = p = mv v = - v sinθ = m[(- v sinθ) - (v sinθ)] f = -2mv sinθ = - 2.1 N s. Impulse on wall is in opposite direction: 2.1 N s. ENERGY AND COLLISIONS S S 12 1/14/2015 COURSE TARGETS S COLLISION GUIDELINES S • EVEN IF KE IS NOT CONSERVED, THE TOTAL ENERGY HAS TO BE CONSERVED • THE MOMENTUM MUST BE CONSERVED • AN EXPLOSION IS CONSIDERED ELASTIC • KE CAN BE CHANGED: i.e. GOES INTO INTERNAL FRICTION OF CHANGING THE MATTER Elastic – Inelastic Collisions S Completely elastic Inelastic Totally inelastic All KE& p conserved Increasing amounts of KE “lost” in collision Standard p conservation problems S • only 1 unknown in before-after situation • often done algebraically • But some we don’t know two variables, usually the primed • look for cancellation for mass variable if one is m and the other 3m p conserved, large amounts of KE lost objects stick together COLLISION GUIDELINES S • Prime is given to indicate the momentum after the collision • Kinetic energy can be destroyed/changed, but momentum cannot be (in absence of external forces.) only 1 object with combined total mass DEMO S • TENNIS BALL, PLAY DOUGH, BASEBALL DROP • DROP AT SAME TIME AND • Nature has no choice, it has to give the 2X larger mass ½ the velocity. And the ½ mass 2x the velocity. • Collisions are more complex if we don’t know one of the resultant velocities. 13 1/14/2015 INELASTIC S Inelastic Collision S 1.0-kg cart and a 2-kg dropped brick S S collision between a 8000-kg engine and a 2000-kg flatcar • KE IS NOT CONSERVED FOR INELASTIC - MOMENTUM IS ALWAYS CONSERVED FOR COLLISIONS. • Completely Inelastic Collisions = Inelastic collisions in which the 2 objects collide & stick together. collision between a 3.0-kg loaded cart and a 2-kg dropped brick S Inelastic 1kg at 20 m/s S 2 kg at 10 m/s What is the total KE before and after the collision 14 1/14/2015 S S 2 ball of clay collide in a perfectly inelastic collision. One has a mass of 4.0 kg, the second has a mass of 3.5 kg. The first one has a velocity of 3.4 m/s, the second has a velocity of – 4.8 m/s. Velocity after the collision? A 9300-kg trolley traveling at 15.0 m/s strikes a locomotive at rest. The two stick together and move off with a speed of 6.0 m/s. What is the mass of the locomotive ? Chin-Chin the crazy panda envies his brother’s toilet seat Go-Kart. In a fit of rage he and his race Go-Kart weight 195-kg moving at 2.5 m/s to the east plow Inelastically into the unattended toilet seat Go-Kart(101-kg). What is new velocity of Chin-Chin? Talk about not my brothers keeper? S S A clown car of mass 1000 kg travels east at 30 m/s, and collides with a 3000 kg demon truck traveling west at 20 m/s. (a) If the collision is completely inelastic, how fast are the car and truck going, and in what direction, after the collision? (b) What was the kinetic energy is lost in the collision? S energy is a scalar rather than a vector S 15 1/14/2015 An SUV of mass 1800 kg travels east at 15 m/s, and collides with a 200 kg mini traveling west. The wreck moves 7 m/s east after collision. Police want to know how fast the mini was moving. (a) If the collision is completely inelastic, how fast is the mini before the collision? (b) What kinetic energy is lost in the collision? energy is a scalar rather than a vector S S S Jump right into this and exaplain superposition and use the previous problem for this. Then jump into Hands on lab the very next day. DO MOMENTUM HAND ON S LAB HERE Have lab partners solve the 4 problems while the others roll carts Put answers on the board over the period 2D INELASTIC COLLISIONS S 2D INELASTIC SCENARIO #1 S Perpendicular Impact 16 1/14/2015 X: 2D INELASTIC SCENARIO #1 S energy is a scalar rather than a vector S y: V = (Vx,Vy) An eagle (4.3 kg) flying due east at 1.8 m/s hits a USAF Thunderbird (5600kg) flying at 10.2m/s due north. Determine the resultant vector (v’ and ) S As Mr. Crider is sinking the 8 ball to win the Stadium pool tourney Ms. Bridge (the coug she is) rolls a sticky wad of gum (137g) at 1.2 m/s and strikes it perpendicular inelastically The cue ball is (.25kg) moving West at 2.9 m/s. In what direction, and with what speed, are they moving after the collision? Solve for energy lost. S While timing stoplights, Mr. Crider car gets T-boned (not again!) His car weights 1500kg and is rolling north at 2 m/s. The physics student (also timing lights) that plows into him has a vehicle weight of 1200kg and was moving east 15 m/s. What is the resultant vector (angle and magnitude)? S S 17 1/14/2015 S S DO MOMENTUM HAND ON S LAB HERE Have lab partners solve the 4 problems while the others roll carts Put answers on the board over the period S http://www.youtube.com/watch?v=0vbDam6EcOA S 1. An eagle (4.3 kg) flying due east at 1.8 m/s hits a USAF Thunderbird (5600kg) flying at 10.2m/s due north. Determine the resultant vector (v’ and q) 2. As Mr. Crider is sinking the 8 to win the SHS pool tourney. The 8 ball (.25kg) is moving WEST at 2.9 m/s. Ms. Bridge (the coug she is) rolls a coug wad of gum (137g) (NORTH) at 1.2 m/s and strikes it perpendicular & inelastically In what direction, and with what speed, are they moving after the collision? What is KE energy loss? 3. A 1000 kg car traveling at 30 m/s, 30° south of east, collides with a 3000 kg truck heading northeast at 20 m/s. The collision is completely inelastic. How fast, and in what direction, and how much energy is lost? LIMITS TO COM??? S If the projectile is >> massive than the target, the velocity of the target particle after the collision will be about twice that of the projectile velocity and the projectile velocity will be essentially unchanged. 4. A 105 kg bike and rider at 12 m/s, 23° west of north, collides with a 60 kg runner heading 18° east of north at 1 m/s. The collision is completely inelastic. How fast, and in what direction, and how much energy lost? 18 1/14/2015 2D INELASTIC SCENARIO #2 S S Acute Impact S A 1000 kg car traveling at 30 m/s, 30° south of east, collides with a 3000 kg truck heading northeast at 20 m/s. The collision is completely inelastic. How fast, and in what direction, and how S S energy is a scalar rather than a vector S much energy is lost? m1 v1 v2 m2 19 1/14/2015 A 105 kg bike and rider at 12 m/s, 23° west of north, collides with a 60 kg runner heading 18° east of north at 1 m/s. The collision is completely inelastic. energy is a scalar rather than a vector S S How fast, and in what direction, and how much energy lost? v2 m2 v1 m1 Note that the cos is here is used to find y S Explosions S S Center of Mass S Note that the sin is here is used to find x • A bomb explodes into 3 identical pieces. Which of the following configurations of velocities is possible? (a) 1 (b) 2 v v m m v V m m (1) (c) both v v m m (2) 20 1/14/2015 Center of Mass S S S S • No external forces, so P must be conserved. • Initially: P = 0 • In explosion (1) there is nothing to balance the upward momentum of the top piece so Pfinal 0. v mv mv m mv v ELASTIC COLLISIONS v m m Elastic collision & kinetic energy is conserved (1) S Center of Mass Total Kinetic energy (KE) is conserved!! S • No external forces, so P must be conserved. • All the momenta cancel out. • Pfinal = 0. mv v mv mv m v v m m (2) 21 1/14/2015 Momentum Vs. Energy S In a bizarre carnival activity you are required to catch a tennis ball. This you easily do. Next, a solid metal (iron) ball of the same diameter is to be thrown to you. You are given the following choices: same kinetic energy, same velocity, same momentum. Which of them would be the best choice to give you an easy catch? Let: Tennis ball mass = 0.250 kg, v = 10.0 m/s. Mass of iron ball = 10 kg Same KE: KE same: • S The two balls of the same mass will exchange velocities. This will always be true for 1-D elastic collisions of objects of equal mass. Same KE: Same momentum: Same P: Momentum Vs. Energy S Car A has a mass of 2000 kg and is traveling at 10 m/s north. Car B has a mass of 1000 kg and is traveling at 20 m/s north. Both cars have the same momentum pA = (2000 kg)(10 m/s) = 20,000 kg-m/s, direction = north pB = (1000 kg)(20 m/s) = 20,000 kg-m/s, direction = north However, they have different kinetic energies KEA = ½ (2000 kg)(10 m/s)2 = 100,000 J KEB = ½ (1000 kg)(20 m/s)2 = 200,000 J COEFF OF RESTITUTION S S SCENARIO #1 • 2 balls striking with different masses and velocities Restitution equation derivation • solving for both final velocities requires a system of 2 equations • ALWAYS set up momentum conservation equation • 2nd equation is K conservation equation • restitution equation yields a “relative velocity” relationship by taking a ratio of K conservation equation to P conservation equation • we will use them both to predict results of collisions From KE conservation (1) S Get like m’s next to each other • See Notes from COM Dividing both sides by ½ and factoring m1 and m2 : Rearranging Eq. (2) From the difference of two squares Eq. (3) Dividing equation (3) by equation (4) yields: 22 1/14/2015 Dogs have no patience in pool. Bowser hits his ball(m1=170g at 3.1 m/s east) as Shep hits his. (m2=192g at 3.4 m/s west). Find final velocities if they strike head on elastically. S S The Ball return at Chalet Bowl sends two balls at once. As the first (m1=10kg at .40 m/s) strikes the end of the return and bounces back it strikes the second ball (m2=9kg at .60 m/s). Find final velocities after they strike head on elastically. S S A stadium basketball player throws a junior basketball (m1=1.5kg at 2.30 m/s) at the same time another player throws a 2nd ball back (m2=1.85kg at 1.30 m/s). Find final velocities after they strike head on elastically. S 1d Elastic collisions: Rather than directly use momentum For Elastic Collisions ONLY S We use conservation + KE conservation Momentum conservation: (1) along with conservation of energy: (2) (3) • (3) is equivalent to momentum conservation + KE conservation, since (3) was derived from these conservation laws! The Rubber Ball Brothers decide for an elastic chest bump. Lil’ bro (m1=2.50kg at 4.75 m/s east) and Big bro (m2=2.85kg at 8.60 m/s west). Find their final velocities. 23 1/14/2015 A stadium volleyball player serves a ball (m1=2.5kg at 3.10 m/s) at the same time another player smashes a 2nd ball back (m2=2.5kg at 4.70 m/s). Find final velocities after they strike head on elastically. S S Which way will white ball go? S What if they are the same size? S What would happen if the car and truck were both made out of rubber and the collision was elastic, with no loss of kinetic energy. In this case the calculations are a lot more complicated, because we have to combine the energy conservation equation with the momentum conservation equation: momentum before = momentum after! S S m1v1 + m2v2 =m1(v1) + m2 (v2) the vector sum is constant! Rule # 1: Momentum is always conserved in a collision! Rule # 2: KE is conserved for elastic collisions only. 24 1/14/2015 What if they aren’t the same size? S SCENARIO #2 S • Ball 1 Striking a Ball 2 at rest B Substitute this relationship into the momentum conservation equation for the collision. Different Impacts S • Bowling Ball (m1)>>>>>(m2) hits Ping Pong Ball (m2) A lawn bowling ball (m1) of mass 0.440 kg moving east (+x) direction) with a speed of 3.30 m/s collides head-on with a 0.220-kg billiard ball (m1) at rest. If the collision is perfectly elastic, what will be the speed and direction of each ball after the collision? S • Ping Pong Ball (m1)<<<<<< (m2) hits Bowling Ball (m2) • Bowling Ball (m1) = hits Bowling Ball (m2) 1d Elastic collisions: Rather than directly use momentum For Elastic Collisions ONLY S We use conservation + KE conservation Momentum conservation: (1) along with conservation of energy: a 0.2 kg billiard ball (m2) at rest is struck by a ping pong ball (m1) of mass .0027 kg moving west with a speed of 2.0 m/s. If the collision is perfectly elastic, what will be the speed and direction of each ball after the collision? S (2) (3) • (3) is equivalent to momentum conservation + KE conservation, since (3) was derived from these conservation laws! 25 1/14/2015 A 0.2 kg billiard ball (m1) moving east with a speed of 2.0 m/s strikes a ping pong ball (m1) of mass .0027 kg at rest is collides head-on with. If the collision is perfectly elastic, what will be the speed and direction of each ball after the collision? S SCENARIO #3 S • 2 ball A Striking a second ball at rest B and one of the resulting velocities is given What is the loss in energy after the collision? A 0.2 kg billiard ball (m1) moving east with a speed of 2.0 m/s strikes head-on a 0.2 kg billiard ball (m1) at rest. If the collision is perfectly elastic, what will be the speed and direction of each ball after the collision? S A bowling ball of mass 5.9 kg that is moving with a speed of 11.3 m/s west collides head-on and elastically with another ball initially at rest. Afterward the incoming bowling ball bounces backward with a speed of 6.2 m/s Calculate (a) the velocity of the target ball after the collision, and (b) the mass of the target ball. S S S • This case implies that a train moving at 60 miles/hr which hits a small rock on the track could send that rock forward at 120 miles/hr if the collision were head-on and perfectly elastic. It also implies that if the speed of the head of your golf club is 110 miles/hr, the limiting speed on the golf ball off the tee is 220 miles/hr in the ideal case. That is, the limiting speed of the ball depends strictly on the clubhead speed, and not on how much muscle power you put into the swing. 26 1/14/2015 SCENARIO #2 HEAD-ON S Comment on Energy Conservation S • We have seen that the total kinetic energy of a system undergoing an inelastic collision is not conserved. Have to solve completely elastic problem with 2 equations 2 unknowns – Energy is lost: • Heat (bomb) Bending of metal (crashing cars) • Kinetic energy is not conserved since work is done during the collision! • In some cases we will say that Kinetic Energy is conserved to solve a problem. 1 object at rest is easy, no object at rest a bit harder Elastic collisions of same S masses rebound at 90deg from each other BALLISTIC PENDULUM S There is no difference in the fact that the block and bullet are moving vertically instead of horizontally. The collision is still totally inelastic and conserves momentum, and the energy is still conserved as PE in the rising of the block and embedded bullet after the collision. BALLISTIC PENDULUM S S Momentum Conservation Energy Conservation (after impact) 27 1/14/2015 A bullet is fired (m1=.02kg at 375 m/s) into a ballistic pendulum (m2=1.35kg). How high will it travel? If the rope is 5m, what is the S A bullet is fired (m1=.045kg) into a ballistic pendulum (m2=3.25kg). How fast is it moving if it climbs .9m? S A 10.5 g bullet is fired into a 8.50 kg wooden block that is hanging straight down, suspended by a 1.50 m length of light line. The bullet stays in the block. The block swings outward, so that the line it hangs from makes an angle of 7.00 to the vertical. (a) What is the velocity of the bullet before it strikes the block? (b) What is the loss of energy in the collision? S STUDENT MISCONCEPTIONS S Momentum is not a vector. Conservation of momentum applies only to collisions. Momentum is the same as force. Moving masses in the absence of gravity do not have momentum. The center of mass of an object must be inside the object. Center of mass is always the same as the center of gravity. Momentum is not conserved in collisions with "immovable" objects Momentum and kinetic energy are the same. A bullet is fired vertically into a 1.40-kg block of wood at rest directly above it. If the bullet has a mass of 29.0 g and a speed of 510 m/s how high will the block rise after the bullet becomes embedded in it? S FORMULAS ON AP S 28 1/14/2015 EXAM TIME S 29