1. Chapter 7: Linear Momentum Linear momentum is: – the product of mass and velocity – Represented by the variable p – Equal to mv, where m is the mass of the object and v is its speed. (p = mv) – Conserved – A vector

2. More on the momentum vector Because velocity is a vector, momentum is a vector. The direction of the momentum depends on the direction of the velocity. The magnitude of the vector is p = mv.

3. The units of momentum Because p = mv, we would expect the units of momentum to be units of mass x units of velocity. Indeed, the units of momentum in SI units is kg*m/s, this unit has no special name. However, this tells you that you need kilograms and meters per second.

4. Example A 100,000kg truck is traveling east at a speed of 20m/s. Find the magnitude and direction of the momentum vector.

5. Solution Remember the magnitude of momentum is p = mv, so p = (100,000kg)(20m/s) = 2,000,000kgm/s. The direction of the momentum vector is the same direction of the velocity vector, so east in this example.

6. Changing momentum The only way to change the momentum of an object is to either change its mass or change its velocity. Remember that a change in velocity is called acceleration, which requires a force. So, changing momentum requires a force.

7. Momentum and Newton’s 2nd Let’s start with ΣF = ma – a = Δv/Δt So ΣF = m Δv/Δt – Δv = v – v 0 So ΣF = m(v – v 0 )/ Δt = – p = mv So ΣF = Δp/Δt

8. Example of momentum change Water leaves a hose at a rate of 1.5kg/s at a speed of 20m/s and is aimed at the side of a car. The water stops when it hits the car. What is the force exerted on the water by the car?

9. Solution Every second 1.5kg of water moves 20m/s. This means the water has a momentum of p = mv = (1.5kg)(20m/s) = 30kgm/s, which goes to 0 when it hits the car (because v = 0). F = Δp/Δt = (p final – p initial )/Δt = (0 – 30kgm/s )/1s = -30N

10. Question What happens if a car hits a semi head on?

11. Conservation of momentum Earlier I told you that momentum is conserved. What that means is “the total momentum before a collision equals the total momentum after” Momentum before = momentum after m 1 v 1 + m 2 v 2 = m 1 v 1 ’ + m 2 v 2 ’ The ‘ is read as “prime” and means after.

12. The law of conservation of momentum The law states this “The total momentum of an isolated system of bodies remains constant.” System = a set of objects that interact with each other Isolated system = the only forces present are those between objects in the system.

13. Example A 10,000kg railroad car traveling at a speed of 24.0 m/s strikes an identical car at rest. If the cars lock together as a result of the collision, what is their speed afterward?

14. Solution We start with p initial = p final p initial = m 1 v 1 + m 2 v 2 = (10,000kg)(24m/s) + (10,000kg)(0m/s) = 240,000kgm/s p final = (m 1 + m 2 )v ’ (it’s m 1 + m 2 because the cars linked up and became one object in the eyes of physics) (m 1 + m 2 )v ’ = 240,000 v ‘ = 240,000 / 20,000 = 12m/s

15. Example Calculate the recoil velocity of a 5.0kg rifle that shoots a 0.050kg bullet at a speed of 120m/s.

16. Solution Momentum is conserved, so start with p i = p f What is p initial ? Ask yourself the following: – What is the starting speed of the bullet before firing? – What is the starting speed of the rifle before firing? p initial = m B v B + m R v R = 0

17. Solution Cont. Now we need to set up p final and set it equal to 0. p final = m B v B ’ + m R v R ’ = (0.050kg)(120m/s) + (5.0kg)(v ’ R ) = 0 Solve for v ’ R v ’ R =

18. Interpretation of data Does our answer make sense? The rifle is moving 100 times slower than the bullet. How does that work?

19. Impulse During a collision the force on an object usually jumps from 0 to very high in a very short amount of time and then abruptly returns to 0. Let’s start with ΣF = Δp/Δt And solve it for Δp, Δp = FΔt = impulse

20. When do we care about impulse? Impulse is very helpful when we are working with large forces that occur in a very short amount of time. Examples: – A bat hitting a ball – Two particles colliding – Brief body contact

21. Example Calculate the impulse experienced when a 70kg person lands on firm ground after jumping from a height of 3.0m. Then estimate the average force exerted on the person’s feet by the ground, if the landing is stiff-legged and again with bent legs. With stiff legs the body moves 1cm during impact. With bent legs the body moves 50cm.

22. Solution We don’t know F so we can’t solve for impulse directly. But we know that impulse = Δp, and Δp = mv 2 – mv 1. This means we need to find v 1 (we know that v 2 will be 0) We can find v using conservation of energy: ΔKE = - ΔPE 1/2mv 2 – 0 = -mgΔy

23. Solution Algebra gives us Δp = 0 – (70kg)(7.7m/s) = -540Ns

24. Straight Legged In coming to a rest the body goes from 7.7m/s to 0 in a distance of 0.01m. The average speed during this period is (7.7m/s + 0)/2 = 3.8m/s = v Δt = d/v = 0.01m / 3.8m/s = 2.6E-3 s Impulse = FΔt = -540Ns so, F = -540Ns / 2.6E-3s = 2.1E5N

25. Bent Legs This is done just like the straight leg except d = 0.50m so Δt = 0.50m / 3.8m/s = 0.13s so, F = 540Ns/0.13s = 4.2E3N

26. Elastic Collisions An elastic collision is a collision in which kinetic energy is conserved. This means both kinetic energy and momentum are conserved. This is handy, because it gives us 2 equations we can solve simultaneously to find the two unknowns (the speed of each object after the collision)

27. The Math The two equations we need to solve are: v 1 – v 2 = v ’ 2 – v ’ 1 (derived from conservation of kinetic energy) and m 1 v 1 + m 2 v 2 = m 1 v ’ 1 + m 2 v ’ 2 (the conservation of momentum equation The strategy is to solve the first equation for either v ’ 2 or v ’ 1 plug that into the second equation.

28. Example A billiard ball of mass m moving with speed v, collides head-on with a second ball of equal mass at rest (v 2 = 0). What are the speeds of the two balls after the collision, assuming it is elastic?

29. Solution Conservation of momentum gives us: mv = mv ’ 1 + mv ’ 2, which we can divide by m to get: v = v ’ 1 + v ’ 2 (call this *) Now we use the first equation, v 1 – v 2 = v ’ 2 – v ’ 1, v = v ’ 2 – v ’ 1 (call this #) * - # gives us 0 = 2v ’ 1, so v ’ 1 = 0

30. Solution We can now substitute v ’ 1 = 0 into v = v ’ 2 – v ’ 1 and solve for v ’ 2 v ‘ 2 = v + v ’ 1 = v + 0 = v To summarize, Before collision: v 1 = v and v 2 = 0 After collision: v ’ 1 = 0 and v ’ 2 = v

31. Inelastic Collisions An inelastic collision is a collision in which kinetic energy is not conserved. If it is not conserved, then either KE f KE i In the former case, the energy of the objects is wasted as heat energy, sound energy, potential energy, or crushing the objects. In the later case, chemical or nuclear potential energy is released. (Think explosives)

32. Completely Inelastic Collisions When two objects completely stick together as a result of the collision, the collision is said to be completely inelastic. When this happens, the conservation of momentum becomes m 1 v 1 + m 2 v 2 = (m 1 + m 2 )v f

33. Ballistic Pendulum The ballistic pendulum is a device used to measure the speed of a projectile, such as a bullet. The projectile, of mass m, is fired into a block of mass M, which is suspended like a pendulum (M > m). As a result of the collision, the pendulum-projectile combination swings up to a maximum height h.

34. Solving the Ballistic Pendulum Let us determine the relationship between the initial speed of the projectile, v, and the height h. mv = (m + M)v ’ (i) KE 1 + PE 1 = KE 2 + PE 2 or ½(m + M)v ’2 + 0 = 0 + (m + M)gh (ii) so v ‘ =

35. Solution Continued Combining (i) and (ii) gives us