What do you think has more oomph a

• What do you think has more oomph – a roller-skate or a Mack truck? • What is Newton’s First Law? • Law of Inertia • Says objects resists changes to their state of motion – an object at rest wants to stay at rest, and an object in motion wants to stay in motion.

• Momentum is all about the object in motion, wanting to stay in motion. • Momentum is a quantitative (numerical) way to describe how much it wants to stay in motion.

• What does its motion depend on? • Its mass. (More mass = more momentum. ) • And its velocity. (More velocity = more momentum. • Momentum is the product between mass and velocity. • What’s the symbol for momentum? • Its ρ. Looks like a lower case p, but its actually the greek letter “row. ”

New Formula!! • ρ = mv • Momentum = mass x velocity • This is a way of associating a number with how much it wants to stay in motion. • Another way to describe it would be to ask how hard it would be to stop its motion. • For example…

ρ = mv Momentum • Is it a vector or scalar? • Vector! • Being a vector quantity, it has a direction, and the direction is very important when doing momentum calculations.

ρ = mv Momentum • What are the units for momentum? • Mass (kg) • Velocity (m/s) • ρ = kg • m/s

Simple Momentum • This car is moving at 20 m/s from left to right. It has a mass of 1000 kg. • Its momentum = 1000 kg × 20 m/s = 20, 000 kg m/s from left to right. • So - if the question asks you to calculate the momentum, it’s simply p=mv.

What if you aren’t given a direction? • Such as…What is the value of the momentum of a 10 kg ball rolling down a bowling alley at a speed of 5 m/s? • p = 10 kg x 5 m/s = 50 kg m/s • In this example above, we only looked at the value of the momentum. This is why we used the word speed.

Try these yourself: • 1. What’s the momentum of a 109 kg football player running at a top speed of 8 m/s? • 872 kgm/s • 2. What’s the momentum of a 9. 8 g bullet traveling at 730 m/s? • 7. 154 kgm/s

Let’s look at change in momentum: • A car with a mass of 1500 kg accelerates from 9 m/s to 16 m/s. What’s its change in momentum? • Change in means final minus initial • Δp = pf – po • Δp = mvf – mvo • Δp = 1500 kg(16 m/s) – 1500 kg(9 m/s) • Δp = 10, 500 kgm/s

Lets look at another “change in momentum”. • Ex. Ball bouncing off a wall. • It leaves the wall at the same speed as before. • Must establish a direction, so let’s call to the left negative, and to the right positive.

What is the change in momentum? • If the speed is the same, is the change in momentum zero?

Let’s take a close look…

Final Initial • • • m=4 kg v= 3 m/s Δp = pf – po Δp = mvf–(-mvo) Δp = mvf + mvo vf=vo (given) Δp = +2 mv Δp=2(4 kg)(3 m/s) Δp = 24 kgm/s

Rebounds: • In all problems involving rebounds, the change in momentum ends up being the initial and final added together! • This is because you have to stop in from moving in one direction and start it moving in the opposite direction.

How can you change an object’s momentum? • • Δp = Δmv You either change the_____ velocity In which case Δp = mΔv or you change the ______ mass. In which case Δp = (Δm)v

Case where the change in momentum means the velocity changes: • • • Δp = mΔv Combine with F = ma, (Recall that F is the net force acting) F = m(Δv/ Δt) Rearranging F(Δt) = mΔv Therefore, F(Δt) = Δp

Impulse Defined: • • F(Δt) = Δp F(Δt) = is called an impulse Impulse = a change in momentum So an external force acting on a system results in a change in momentum (velocity) of the system.

Impulse Example • A car is involved in a collision in which it is brought to a standstill from a speed of 24 m/s. The driver of mass 85 kg is brought to rest by his seat belt in a time of 400 ms. • a) Calculate the average force exerted on the driver by his seat belt.

Solution • Calculate the average force exerted on the driver by his seat belt. • • First we need to work out the change in momentum (impulse) • Δp = mvf - mvo = 0 - 85 kg x 24 m/s = 2040 Ns • • Now we can work out the force • F = Δp/ Δt = 2040 Ns ÷ 0. 400 s = 5100 N •

• If we compare this force to his weight and hence work out the “g-force, ” • • The weight of the driver = mg = 85 kg x 9. 8 = 834 N • The g-force = 5100 N ÷ 834 N = 6. 1 g • • What is the likelihood of serious injury? • • This will reduce the likelihood of serious injury as the body withstand accelerations of up to 8 g. Aerobatic pilots regularly pull 5 to 6 g in their manoeuvers.

Example: Students tossing a softball. • What do they do when they catch the ball?

Example student jumping off desk: • • m = 65 kg If he jumps from 1 m, velocity on impact is V = √ 2 g(1 m) = 4. 5 m/s What provides the force that stops the student? • The floor.

Landing… • The student can land either by bending his knees, or stiff-legged. • Is the change in momentum the same in both cases? • Yes, Δp = mvf – mvo • If the landing is stiff-legged, Δt = 0. 001 s • If the landing is with bent knees, Δt=0. 5 s

Let’s calculate the force on the student from the floor in the two different landing situations: • In both cases, Δp = m(vf – vo) • Δp=65 kg (0 – (-4. 5 m/s)) = 292. 5 kgm/s • F(Δt) = Δp • F = Δp/Δt • 1) Stiff-legged: F = 292. 5/0. 001 s = 292, 500 N! • 2) Bent knee: F = 292. 5/0. 5 s = 585 N

Try this yourself • Determine the average force on a tennis ball hit by a racket. The ball has a mass of 0. 14 kg and an initial speed of 30 m/s to the left. It rebounds from the racket with a speed of 40 m/s in the opposite direction and is in contact with the racket for 0. 002 s.

Solution • Don’t forget to establish a direction. Let’s call to the right positive. • F(Δt) = Δp • F = Δp/Δt Impulse = change in momentum • F = m(Vf – Vo)/t • F = (0. 14 kg)(40 m/s-(-30 m/s)) / (0. 002 s) • F = 4900 N to the right

Assignment • Handout Problems 1 – 15 pages 1 -3 • Sect 7. 1 #1 -7 page 7

Conservation of Momentum • An important principle: • The total momentum of a system remains constant provided that no external forces act on the system. • This means the total momentum before = total momentum after. • po = p f

But. . • What’s a system and how do you figure out if an outside force is acting? • Good question!

Let’s define a “system: ” • Carts on Track • Skateboards • A system is whatever you choose. Draw a circle around it. If there are 2 objects, your system can be both objects, or it can be just one of them. Then determine if an external force acts on the system.

Another Question: • If momentum is conserved, does it mean that each object in the system retains it’s same momentum? • No, it means the total momentum of the system stays constant, but how it is shared can change.

If the net external force is zero, then you conserve momentum. • How? • F(Δt) = Δp • • If F = 0, then Δp = ? Zero This means no change in momentum! This means the total momentum before = total momentum after. • po = p f

Collisions • In collision, you let the system be all the objects involved. • Momentum is always conserved in collisions.

So if the net external force is zero, • momentum before = momentum after. • This includes the momentum of all the objects in the system before and after. • If there are only 2 objects, it looks like: • m 1 vo 1 + m 2 vo 2 = m 1 vf 1 + m 2 vf 2.

Newton’s Cradle

What about conservation of energy? • Total energy before = total energy after (energy is conserved). • If KE is conserved, then we also can say that • ½m 1 vo 12 + ½m 2 vo 22 = ½m 1 vf 12 + ½m 2 vf 22. • And the collision is considered “elastic”.

Henry and Isabelle Ice Skate • Henry (m=80 kg) is going 5 m/s and runs into Isabelle (m=60 kg), who was at rest. • If they hook arms and move off together, how fast will they be going? • Momentum Before = Momentum After

• m 1 vo 1 + m 2 vo 2 = m 1 vf 1 + m 2 vf 2. • (80 kg)(5 m/s) +0 = (80 kg)vf + (60 kg) vf • Or since vf is the same for both since they lock arms, we can write it: • (80 kg)(5 m/s) +0 = (80 kg + 60 kg)vf • vf = 2. 86 m/s

The diagram shows two cars at a fairground, before and after bumping into each other. One car and driver has a total mass of 500 kg, while the other car and driver has a total mass of 400 kg.

• • (a) What is (i) the total kinetic energy before the collision; (ii) the total kinetic energy after the collision. (iii) the total loss in kinetic energy.

• • (i) The total kinetic energy before the collision; Kinetic Energy = 1/2 mv 2 Kinetic Energy of blue car = 1/2 x 500 kg x (5 m/s)2 = 6250 J Kinetic Energy of yellow car = 1/2 x 400 kg x (2 m/s)2 =800 J Total energy = 6250 J + 800 J = 7050 J

• • (ii) the total kinetic energy after the collision. Kinetic Energy of blue car = 1/2 x 500 kg x (3 m/s)2 = 2250 J Kinetic Energy of yellow car = 1/2 x 400 kg x (4. 5 m/s)2 = 4050 J Total energy = 2250 J + 4050 J = 6300 J

• (iii) the total loss in kinetic energy. • Total loss = 7050 J - 6300 J = 750 J • • (b) Is this an elastic collision? • It is inelastic, because there has been energy lost. •

A second collision is shown below: • What is the speed of the 500 kg car after the collision?

Solution • Momentum before = momentum after • m 1 vo 1 + m 2 vo 2 = m 1 vf 1 + m 2 vf 2. • • Momentum before = (500 kg x 5 m/s) + (400 kg x 2 m/s) = 2500 kg m/s + 800 kg m/s = 3300 kg m/s

• Momentum after • = (500 kg x v 1 f m/s) + (400 kg x 4 m/s) • = 500 v 1 f kg m/s + 1600 kg m/s

• • • Therefore: 3300 kg m/s=500 v 1 f kg m/s + 1600 kg m/s 500 v 1 f = 3300 kg m/s - 1600 kg m/s 500 v 1 f = 1700 kg m/s v 1 f = 1700 ÷ 500 = 3. 4 m/s In this example, we have not had to worry about signs as the motion has been in the same direction.

Ballistics Example:

• A bullet of mass 45 g is travelling horizontally at 400 m/s when it strikes a wooden block of mass 16 kg suspended on a string so that it can swing freely. The bullet is embedded in the block.

• Calculate: • a) The velocity at which the block begins to swing; • b) The height to which the block rises above its initial position; • c) How much of the bullet’s kinetic energy is converted to internal energy. •

• a) The velocity at which the block begins to swing: • • Momentum before = momentum after • m 1 vo 1 + m 2 vo 2 = m 1 vf 1 + m 2 vf 2. • Momentum before = (0. 045 kg x 400 m/s)+(0) • Momentum before = 18 kg m/s • • Momentum after = total mass of bullet and wood x speed - they move together • Momentum after=(16. 0 kg + 0. 045 kg) x vf m/s • Momentum after = 16. 045 vf kg m/s • Therefore: • 16. 045 vf kg m/s = 18 kg m/s • vf = 18 kg m/s ÷ 16. 045 • vf = 1. 12 m/s

• b) The height to which the block rises above its initial position; • We use the conservation of energy principle: • Kinetic energy = potential energy • 1/2 mv 2 = mgΔh • m's cancel out. • Rearranging: • Δh = v 2/2 g • = (1. 122)2 ÷ (2 x 10 m/s 2) • = 0. 063 m •

• c) How much of the bullet’s kinetic energy is converted to internal energy. • • We need to know the kinetic energy of the bullet: • KE = 1/2 mv 2 = 1/2 x 0. 045 kg x (400 m/s)2 = 3600 J • • Now we need to know the kinetic energy of the block and bullet: • KE = 1/2 x 16. 045 kg x (1. 12 m/s)2 = 10. 1 J • • Kinetic energy lost = 3600 J 10. 1 J = 3590 J • This energy is not destroyed, but converted into internal energy – heat and sound and maybe deformation.

Elastic vs. Inelastic Collisions • If ΔKE is considered zero, it’s “elastic”. • If ΔKE is not zero, the collision is “inelastic. ” • If some kinetic energy is lost, converted into heat or light, then the collision is inelastic.

Think about two objects travelling in the same direction. The table below shows the properties of the objects:

• There are two important principles here: • 1. Conservation of momentum: • Total momentum before = total momentum after • m 1 vo 1 + m 2 vo 2 = m 1 vf 1 + m 2 vf 2.

• • and 2. Conservation of Energy Total energy before = total energy after • ½m 1 vo 12 + ½m 2 vo 22 = ½m 1 vf 12 + ½m 2 vf 22+ E • The term E is the energy that is lost in the collision. In a perfectly elastic collision E = 0.

Elastic Collision – Linear Finding Relative Velocities

Derive the following: • Using conservation of momentum and conservation of energy, show that for a general case of an elastic collision in which ball 1 (with velocity vo 1) hits ball 2 which is at rest (v 2 = 0), the relationship between the initial and final speeds of the balls is as follows: • v 1 f(m 1+m 2) = v 1 o(m 1 -m 2) • v 2 f = v 1 o ( 2 m 1/(m 1+m 2)

Center of Mass

• Link - Honda • http: //www. youtube. com/watch? v=T 9 leh. Hx v-C 8 • Mikewu 610 accord

Assignment • Handout #16 – 21 • Linear momentum and its conservation, and impulse #1 -5

16. Full Back & Defensive Tackle • A 95 kg full back running at 8. 2 m/s collides in mid air with a 128 kg defensive tackle moving in the opposite direction. • • • A) what was full back’s momentum before the collision? B) what was the change in the full back’s momentum? C) What was the change in the tackle’s momentum? D) what was the tackle’s original momentum? E) how fast was the tackle moving originally? • 779, -779, 6. 09 m/s

17. Marbles • • • Glass ball A: m. A = 0. 005 kg v. AO = 0. 20 m/s v. AF = 0. 08 m/s Glass ball B: m. B = 0. 010 kg v. BO = 0. 10 m/s • • • What is ball A’s original momentum? What is ball A’s Δp? What is ball B’s Δp? What is the momemtum of B after the collision? What is ball B’s speed after the collision?

18. Van/Car Collision • A 2575 kg van runs into the back of an 825 kg car at rest. They move off together at 8. 5 m/s. Assuming no friction with the ground, find the initial speed of the van.

19. Bullet/Block • A 15 g bullet is shot into a 5085 g wooden block standing on a frictionless surface. The block with the bullet in it, acquires a velocity of 1. 0 m/s. Calculate the velocity of the bullet before striking the block.

20. Rollerbladers • Standing face to face, they push off each other. Rollerblader A has a mass of 90 kg, B has a mass of 60 kg. Find the ratio of their velocities after their hands lose contact. Which has the greater speed?

21. Impluse • What impulse is needed to stop a 70 kg basketball player traveling at 6 m/s?

Collisions in 2 D, Explosions

Collision – page 7 • M 1 = 0. 150 kg, v. O 1 = 0. 90 m/s, @50˚ as shown in picture • M 2 = 0. 26 kg, v. O 2 = 0. 54 m/s along +x • Vf 2 = 0. 70 m/s @ 35 ˚ below the +x axis • A) calculate Vf 1 and the angle for M 1 • B) calculate Vf if collision is completely inelastic.

• Need to find V 1 FX and V 1 FY and use Pythagorean theorem to get V 1

First – break into x and y components • v 10 X = (. 9)sin 50 = 0. 6894 m/s • v 10 Y = (. 9)cos 50 = -0. 5785 • v 20 X = 0. 54 m/s • V 2 f. X = (. 7)cos 35 = 0. 5734 m/s • V 2 f. Y = -(. 7)sin 35 = -0. 4 m/s

POX = PFX • M 1 V 10 x + M 2 V 20 x = M 1 V 1 FX + M 2 V 2 FX • Want to find V 1 FX • {(. 15 kg)(. 6894 m/s) + (. 26 kg)(. 54 m/s)(. 26)(. 5734)}/(. 15 kg) = V 1 FX • V 1 FX = 0. 63 m/s

POY = PFY • M 1 V 10 Y + M 2 V 20 Y = M 1 V 1 FY + M 2 V 2 FY • Want to find V 1 Fy • -(. 15 kg)(. 5785 m/s) + (. 26 kg)(. 4 m/s)/(. 15 kg) = V 1 FY • V 1 Fy = 0. 12 m/s (0. 117 m/s has to round)

Vf 1 • Vf 1 = Square root of V 1 Fx 2 + V 1 Fy 2 • Vf 1 = 0. 64 m/s • ϴ = tan-1(. 12/. 63) = 11 degrees above +x

If they stick together: • • • M 1 V 10 x + M 2 V 20 x = (M 1+ M 2)VFX VFx = 0. 5946 m/s M 1 V 10 y + M 2 V 20 Y = (M 1+ M 2)VFY VFy = -0. 21165 Vf = square root of Vfx 2 + Vfy 2 = 0. 6311 ϴ = tan-1(. 21165/. 5946) = 19. 6 degrees below +x axis

Hand drawn problem sheet • • • 1. 6. 67 m/s 2. 3. 48 m/s @ 73. 3 degrees 3. 10 m/s 4. 5 m/s @ 36. 9 dgrees S of E 5. 11. 4 m/s @ 21 degrees S of E 6. 4. 9 m/s @ 28 degrees S of E

Review • 1. Agree • 2. Disagree – If F=0 then the Δp = 0. • 3. Disagree – larger. Objects stop then start in the other direction. • 4. Agree • 5. Agree p=mv • 6. Agree FΔt = mΔv

• 7. Disagree – it means the total momentum stays constant, but how it is shared can change. • 8. Disagree – if there is an initial v in one of the objects before, then they can not both be at rest after. • 9. Agree

• 10. Disagree – vf would be half as much. • 11. Disagree – it would have to be zero since the initial momentum equals zero. • 12. Disagree – only external forces can change momentum of a system.

AP Review • Explain how the laws of conservation restrict the manner in which systems can change due to interactions.

• How would you design an experiment that exhibits the conservation of linear momentum along with analysis of error?

• Given a situation where conservation of momentum appears to the broken, theorize a solution (perhaps involving interactions that were not apparent at first) that supports the validity.

• Describe what might happen a system if an external agent exerts a force on it.

• How would you graphically interpret data from a momentum experiment?

• Show narratively or pictorially, how the structure of a system determines its properties and how external agents can cause the system to change.