1. Momentum and Impulse So far we’ve studied the properties of a single object; i.e. its motion and energy How do we analyze the motion of two or more objects that interact with each other??

2. Momentum It’s the product of mass times velocity p = mv Units: (kg) x (m/s) = kg·m/s Momentum is a technical term for something we already know! – If a train and a car are going the same speed, it’s harder to stop the train. The greater mass of the train gives it more momentum than the car. – A bullet fired from a gun has more penetrating power than one thrown by hand. Even though they have the same mass, the first bullet has more momentum due to its higher velocity.

3. Give it a try!! Which vehicle has more momentum, a 1500kg truck moving at 0.3m/s or a 105kg go-cart moving at 5 m/s? – Truck p = mv  (1500kg) x (0.3m/s) = 450 kg·m/s – Go  cart p = mv  (105kg) x (5m/s) = 525 kg·m/s

4. Interaction When two objects interact they exert forces on each other. Newton’s Third Law states that these forces must be equal and opposite.

5. Change in Momentum For one of these objects, Newton’s First Law gives F = ma  a = F/m If we use the average acceleration for this case a avg =  v/  t then  v/  t = F avg /m  m  v = F avg  t Looking at the left side of this equation m  v = m(v f – v i ) = mv f – mv i = p f – p i =  p

6. Impulse-Momentum Theorem Therefore, the average force times the time interval over which it acts is equal to the change in momentum.  p = F avg  t – We call F avg  t the impulse that acts on an object. Units: (force) x (time) = N·s Are N·s equal to kg·m/s? Do dimensional analysis to find out.

7. Impulse, Day-to-Day Like momentum, you already understand how this idea works, now you have a scientific name for it! If a constant net force acts on an object (say a box is pulled along a slippery floor); the longer you apply the force, the greater will be the change in the object’s speed. Similarly, if you apply a force to an object for a specific amount of time (say a push on a swing), the greater the force, the greater will be the change in the object’s speed.

8. Give it a try!! A golf club strikes a 46g golf ball for 0.5ms, the ball leaves the face of the club at 70 m/s. Find the average force that the club exerts on the ball during the impact. (the ball is initially at rest)

9. Solution m  v = F avg  t  F avg = m  v/  t F avg = (.046kg)(70m/s – 0m/s)/(5x10 -4 s) F avg = 6440 N = 1448 lbs What then is the average force that the ball exerts on the club? Newton’s Third Law says -6440 N

10. Conservation of Momentum Whenever two objects interact, it has been found that the sum of their momentum is the same before and after the interaction. 2,f p tot,i = p tot,f  m 1 v 1,i + m 2 v 2,i = m 1 v 1,f + m 2 v 2,f This is called the Law of Conservation of Momentum – only true if there are no net external forces acting on the objects during their interaction.

11. Types of Interaction (Explosions) One object splitting into two or more parts Two objects being separated by some force Rockets!!!

12. Types of Interaction (Collisions) whether it’s a collision in the day-to-day sense (car accidents) or not (catching a baseball)

13. Explosive Interactions (One-Dimensional) Imagine an object of mass m at rest exploding into two parts of mass m 1 and m 2. Conservation of momentum tells us that the initial momentum of the system must equal the final momentum of the system. p i = p f or mv = m 1 v 1,f + m 2 v 2,f but, since mv = 0 then m 1 v 1,f + m 2 v 2,f = 0 – therefore m 1 v 1,f = -m 2 v 2,f The minus sign tells us that the two parts must be moving in the opposite direction (makes sense right!!)

14. Give it a try!! Neil is a 150kg astronaut floating at rest in outer space. He decides he wants his picture taken with the Earth behind him, so he throws his camera to Buzz, another astronaut floating near-by. a) Is Neil still at rest after throwing the camera? No, conservation of momentum says that he must move in the opposite direction in order to have equal and opposite momentum from the camera.

15. Give it a try!! (continued) b) If the camera has a mass of 0.80kg and it moves away with a velocity of 12m/s to the left, what is Neil’s velocity after he throws it? (m 1 = 0.80kg, v 1,f = 12m/s, m 2 = 150kg, v 2,f = ?) p i = p f = 0, so m 1 v 1,f + m 2 v 2,f = 0 or m 1 v 1,f = -m 2 v 2,f which gives us -(m 1 v 1,f )/m 2 = v 2,f so, v 2,f = -0.064m/s c) How far will Neil be from the spot where he threw the camera after 1 hour? d = v 2,f t given t = 3600s, then d = 230m

16. Collision Types Elastic Collisions – Momentum is conserved – Kinetic energy is conserved too! No permanent deformation, no sound, no friction Inelastic Collisions – Momentum is conserved – Kinetic energy is not conserved Possible permanent deformation, sound, or friction between objects Work done by non-conservative forces

17. Give it a try!! Two balls are rolling along a table with negligible friction. One ball, with a mass of 0.250kg, has a velocity of 0.200m/s eastward. The other ball, with a mass of 0.100kg, has a velocity of 0.100m/s eastward. The first ball hits the second from directly behind. If the final velocity of the first ball is 0.143m/s eastward, what is the final velocity of the second ball? Is this an elastic collision?

18. Solution (part 1) Using conservation of momentum m 1 v 1,i + m 2 v 2,i = m 1 v 1,f + m 2 v 2,f solving for v 2,f v 2,f = (m 1 v 1,i + m 2 v 2,i - m 1 v 1,f )/m 2 v 2,f = 0.243m/s eastward (Remember, velocity is a vector quantity!!)

19. Solution (part 2) Elastic collisions mean kinetic energy is also conserved KE i = KE f – or ½m 1 (v 1,i ) 2 + ½m 2 (v 2.i ) 2 = ½m 1 (v 1,f ) 2 + ½m 2 (v 2,f ) 2 – Before the collision KE i = 0.00550J – Afterward KE f = 0.00551J Yes! It is an elastic collision.

20. Inelastic collisions Two possible outcomes The objects bounce apart afterwards The objects stick together afterwards (perfectly inelastic)

21. Perfectly Inelastic Collisions What is conserved? How do you think the final velocities of the objects after the collision will be related to each other? Remember, velocity is a vector. You have to take direction into account!!

22. Give it a try!! A 5kg lump of clay traveling at 10m/s to the left strikes a 6kg lump of clay moving at 12m/s to the right. Find the final velocity of the resulting object if they stick together. How much kinetic energy is lost in the collision?

23. Solution (part 1) Given: m 1 = 5kg m 2 = 6kg v 1,i = -10m/s v 2,i = 12 m/s m 1 v 1,i + m 2 v 2,i = (m 1 + m 2 )v f (m 1 v 1,i + m 2 v 2,i )/ (m 1 + m 2 ) = v f v f = +2m/s or 2m/s to the right

24. Solution (part 2) Kinetic energy is not conserved so let’s find  KE.  KE = KE f – KE i KE i = ½m 1 (v 1,i ) 2 + ½m 2 (v 2,i ) 2 = 682J KE f = ½(m 1 + m 2 )(v f ) 2 = 22J  KE = 22J – 682J = -660J The energy did not disappear! It was converted to heat energy in the clay and probably into sound energy.

25. Collisions (Two Dimensions) Momentum is a vector like velocity. Use conservation of momentum along each axis separately when solving problems. – Pick an orientation for your coordinate system that simplifies the problem

26. Give it a try!! A 60kg man is sliding east at 0.5m/s on a frozen pond man (assume it’s frictionless). A friend throws him a 5kg bag of salt to help him stop. If the bag’s velocity is 5m/s to the north, what is the man’s velocity bag after catching the bag?

27. Solution Given: 1: m 1 = 60kg, v 1,i,x = 0.5m/s, v 1,i,y = 0 2: m 2 = 5 kg, v 2,i,x = 0, v 2,i,y = 5m/s Use conservation of momentum for each component X-axis – m 1 v 1,i,x + m 2 v 2,i,x = (m 1 + m 2 )v f,x – (m 1 v 1,i,x + 0)/ (m 1 + m 2 ) = v f,x – v f,x = 0.46m/s Y-axis – m 1 v 1,i,y + m 2 v 2,i,y = (m 1 + m 2 )v f,y – (0 + m 2 v 2,i,y )/ (m 1 + m 2 ) = v f,y – v f,y = 0.38m/s

28. Solution Find the magnitude and direction of the velocity Magnitude – Use the Pythagorean Theorem – v = 0.60m/s Direction –  = tan -1 (v y /v x ) –  = 40 0 So, v = 0.60m/s @ 40 0 N of E

29. Impulse vs Conservation Of Momentum Impulse: only correct if applied to one of the objects in an interaction –  p = F avg  t Conservation of momentum: only correct if applied to all the objects in an interaction – p tot,i = p tot,f 2,f m 1 v 1,i + m 2 v 2,i = m 1 v 1,f + m 2 v 2,f

30. continued putting the object 2 terms on the left, and object 1 terms on the right: 2,f m 2 v 2,i - m 2 v 2,f = m 1 v 1,f - m 1 v 1,i p 2,i – p 2,f = p 1,f – p 1,i -  p 2 =  p 1 -F avg,2  t 2 = F avg,1  t 1 - therefore, from Newton’s third law, the magnitude of the impulse on each object in an interaction is the same

31. The Ballistic Pendulum One projectile, one hanging object Perfectly Inelastic Collision Collision: – conservation of momentum!!! Pendulum Swing: – conservation of energy!!! Different initial and final points for each part. before after

32. Give it a try!! A 100g bullet is fired into a 1.35kg block of wood. If the block rises 25cm, how fast was the bullet going when it hit the block? 25cm

33. Solution There are two parts – The perfectly inelastic collision – The upswing of the block/bullet Knowns: m 1 =.100kg, m 2 = 1.35kg,  y =.25cm – Need velocities to analyze the collision so let’s examine the up swing (i.e. conservation of energy) ME i = ME f (be careful choosing initial and final points) – what forms of energy are present?? K i = U f or ½mv i 2 = mgh f so v i 2 = 2gh f v i = 2.21m/s

34. Solution Now we know the speed of the block/bullet after the collision. So v f = 2.21m/s, v 2 = 0, v 1 = ?, m = m 1 + m 2 = 1.45kg we get m 1 v 1 + m 2 v 2 = mv f v 1 = mv f /m 1 or v 1 = 32m/s