Chapter 7: Linear Momentum

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 – v0 • So ΣF = m(v – v0)/ 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 = (pfinal – pinitial)/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 • m1v1 + m2v2 = m1v1’ + m2v2’ • 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 pinitial = pfinal • pinitial = m1v1 + m2v2 = (10,000kg)(24m/s) +(10,000kg)(0m/s) = 240,000kgm/s • pfinal= (m1+ m2)v’ (it’s m1 + m2 because the cars linked up and became one object in the eyes of physics) • (m1 + m2)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 pi = pf • What is pinitial? • Ask yourself the following: • What is the starting speed of the bullet before firing? • What is the starting speed of the rifle before firing? • pinitial = mBvB + mRvR = 0

17. Solution Cont. • Now we need to set up pfinal and set it equal to 0. • pfinal = mBvB’ + mRvR’ = (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 = mv2 – mv1. • This means we need to find v1 (we know that v2 will be 0) • We can find v using conservation of energy:ΔKE = -ΔPE1/2mv2 – 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 exceptd = 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: • v1 – v2 = v’2 – v’1 (derived from conservation of kinetic energy) and • m1v1 + m2v2 = m1v’1 + m2v’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 (v2 = 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, v1– v2 = 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: v1 = v and v2 = 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 KEf < KEi orKEf> KEi • 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 m1v1 + m2v2 = (m1 + m2)vf

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) • KE1 + PE1 = KE2 + PE2 or • ½(m + M)v’2 + 0 = 0 + (m + M)gh (ii) so • v‘ =

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