MOMENTUM!

Momentum Defined

p = m v• p = momentum vector• m = mass• v = velocity vector

Momentum Facts

• Linear Momentum• p = m v• Momentum is a vector quantity!• Velocity and momentum vectors point in

the same direction.• SI unit for momentum: kg · m /s (no

special name).

Momentum Facts

• Momentum is a conserved quantity (this will be proven later).

• A net force is required to change a body’s momentum.

• Momentum is directly proportional to both mass and speed.

• Something big and slow could have the same momentum as something small and fast.

Momentum Examples

10 kg 3 m /s

10 kg30 kg · m /s

5 g

9 km /s

p = 45 kg · m /s at 26º N of E

26º

Equivalent Momenta

Bus: m = 9000 kg; v = 16 m /s p = 1.44 ·105 kg · m /s

Train: m = 3.6 ·104 kg; v = 4 m /s p = 1.44 ·105 kg · m /s

Car: m = 1800 kg; v = 80 m /s p = 1.44 ·105 kg · m /s

continued on next slide

Equivalent Momenta (cont.)• The train, bus, and car all have different masses

and speeds, but their momenta are the same in magnitude. The massive train has a slow speed; the low-mass car has a great speed; and the bus has moderate mass and speed. Note: We can only say that the magnitudes of their momenta are equal since they’re aren’t moving in the same direction.

• The difficulty in bringing each vehicle to rest--in terms of a combination of the force and time required--would be the same, since they each have the same momentum.

Newton’s 2nd Law in terms of momentum

p = mv, F = Δp/ΔtPractice

1) Calculate the momentum of a 110kg rugby player running the 100m in 15s.

2) Calculate the force that the rugby player in the previous example imparts on your body if, during a tackle you halve his speed in the time of 0.2s

Answers

1) v = 100/15= 6.67, p = 110 x 6.67 = 733kgm/s2) F = 0.5 x 733/0.2 = 1830N

Impulse• When two object collide, their velocities

change and since momentum is a function of an object’s velocity, the momentum changes. This change in momentum is called impulse(I).

• Same unit as momentum, kgm/s• Can be defined as a force times a time.• Units can also be Ns

Impulse• I = Δp• I = p2 – p1

• I = mv – mu• I = mΔv

F = maF = Δp/ΔtF = I/ Δt since I = ΔpI = FΔt

Impulse Practice• A fielder catches a 0.4kg cricket ball travellin at

a velocity of 20m/s and brings it to a stop. Calculate a)The impulse(change in momentum) of the cricket

ballb)The force on the fielder’s hand if he stops the ball

with a rigid arm, where the time for the ball to come to rest is 0.02s

c)The force on the fielder’s hand if he brings the ball to rest more gradually, say in a time of 0.5s

Answers

a) I = mΔv, 8kgm/sb) I = FΔt, 400Nc) I = FΔt, 16N

Conservation of Momentum in 1-DWhenever two objects collide (or when they exert forces on each other

without colliding, such as gravity) momentum of the system (both objects together) is conserved. This mean the total momentum of the objects is the same before and after the collision.

before: p = m1 v1 - m2 v2

after: p = - m1 va + m2 vb

m1 m2

v1 v2

(Choosing right as the +

direction, m2 has - momentum.)

m1 m2

va vb

m1 v1 - m2 v2 = - m1 va + m2 vb

Directions after a collisionOn the last slide the boxes were drawn going in the opposite direction

after colliding. This isn’t always the case. For example, when a bat hits a ball, the ball changes direction, but the bat doesn’t. It doesn’t really matter, though, which way we draw the velocity vectors in “after” picture. If we solved the conservation of momentum equation (red box) for vb and got a negative answer, it would mean that m2 was still moving to the left after the collision. As long as we interpret our answers correctly, it matters not how the velocity vectors are drawn.

m1 v1 - m2 v2 = - m1 va + m2 vb

m1 m2

v1 v2

m1 m2

va vb

Sample Problem 1

1. A rifle fires a bullet into a giant slab of butter on a frictionless surface. The bullet penetrates the butter, but while passing through it, the bullet pushes the butter to the left, and the butter pushes the bullet just as hard to the right, slowing the bullet down. If the butter skids off at 4 cm/s after the bullet passes through it, what is the final speed of the bullet?(The mass of the rifle matters not.)

7 kg

v = 0700 m/s

35 g

7 kgv = ?

35 g4 cm/s

continued on next slide

Sample Problem 1 (cont.)

7 kg

v = 0700 m/s

35 g

7 kgv = ?

35 g4 cm/s

p before = 7 (0) + (0.035) (700)

= 24.5 kg · m /s

Let’s choose left to be the + direction & use conservation of momentum, converting all units to meters and kilograms.

p after = 7 (0.04) + 0.035 v

= 0.28 + 0.035 v

p before = p after 24.5 = 0.28 + 0.035 v v = 692 m/s

v came out positive. This means we chose the correct direction of the bullet in the “after” picture.

Sample Problem 2

2) Same as the last problem except this time it’s a block of wood rather than butter, and the bullet does not pass all the way through it. How fast do they move together after impact?

7 kg

v = 0700 m/s

35 g

v7. 035 kg

(0.035) (700) = 7.035 v v = 3.48 m/s

Note: Once again we’re assuming a frictionless surface, otherwise there would be a frictional force on the wood in addition to that of the bullet, and the “system” would have to include the table as well.

Proof of Conservation of MomentumThe proof is based on Newton’s 3rd Law. Whenever two objects collide (or exert forces on each other from a distance), the forces involved are an action-reaction pair, equal in strength, opposite in direction. This means the net force on the system (the two objects together) is zero, since these forces cancel out.

M mFF

force on m due to Mforce on M due to m

For each object, F = (mass) (a) = (mass) (v / t ) = (mass v) / t = p / t. Since the force applied and the contact time is the same for each mass, they each undergo the same change in momentum, but in opposite directions. The result is that even though the momenta of the individual objects changes, p for the system is zero. The momentum that one mass gains, the other loses. Hence, the momentum of the system before equals the momentum of the system after.

Conservation of Momentum applies only in the absence of external forces!

In the first two sample problems, we dealt with a frictionless surface. We couldn’t simply conserve momentum if friction had been present because, as the proof on the last slide shows, there would be another force (friction) in addition to the contact forces. Friction wouldn’t cancel out, and it would be a net force on the system.The only way to conserve momentum with an external force like friction is to make it internal by including the tabletop, floor, or the entire Earth as part of the system. For example, if a rubber ball hits a brick wall, p for the ball is not conserved, neither is p for the ball-wall system, since the wall is connected to the ground and subject to force by it. However, p for the ball-Earth system is conserved!

Sample Problem 3

EarthM

apple

An apple is originally at rest and then dropped. After falling a short time, it’s moving pretty fast, say at a speed V. Obviously, momentum is not conserved for the apple, since it didn’t have any at first. How can this be?

m

F

F

answer: Gravity is an external force on the apple, so momentum for it alone is not conserved. To make gravity “internal,” we must define a system that includes the other object responsible for the gravitational force--Earth. The net force on the apple-Earth system is zero, and momentum is conserved for it. During the fall the Earth attains a very small speed v. So, by conservation of momentum:

V

v

m V = M v

Sample Problem 4

before

after

3 kg 15 kg10 m/s 6 m/s

3 kg 15 kg4.5 m/s v

A crate of raspberry donut filling collides with a tub of lime Kool Aid on a frictionless surface. Which way on how fast does the Kool Aid rebound? answer: Let’s draw v to the right in the after picture. 3 (10) - 6 (15) = -3 (4.5) + 15 v v = -3.1 m/sSince v came out negative, we guessed wrong in drawing v to the right, but that’s OK as long as we interpret our answer correctly. After the collision the lime Kool Aid is moving 3.1 m/s to the left.

Conservation of Momentum in 2-D

m1m2

v1

v2

m1 m2

vavb

12

a b

To handle a collision in 2-D, we conserve momentum in each dimension separately. Choosing down & right as positive:

before: px = m1 v1 cos1 - m2 v2 cos2

py = m1 v1 sin1 + m2 v2 sin2

after: px = -m1 va cosa + m2 vb cos b

py = m1 va sina + m2 vb sin b

Conservation of momentum equations:

m1 v1 cos1 - m2 v2 cos2 = -m1 va cosa + m2 vb cos b

m1 v1 sin1 + m2 v2 sin 2 = m1 va sina + m2 vb sin b

Conserving Momentum w/ Vectors

m1m2

p1

p 2

m1 m2

p a p b

12

a b

BEFORE

AFTER

This diagram shows momentum vectors, which are parallel to their respective velocity vectors. Note p1 + p 2

= p a + p b and p before = p after as conservation of momentum demands.

p1

p 2

p before

p a

p b

p after

Exploding Bomb

A c m e

before

after

A

c

m

eA bomb, which was originally at rest, explodes and shrapnel flies every which way, each piece with a different mass and speed. The momentum vectors are shown in the after picture.

continued on next slide

Exploding Bomb (cont.)

Since the momentum of the bomb was zero before the explosion, it must be zero after it as well. Each piece does have momentum, but the total momentum of the exploded bomb must be zero afterwards. This means that it must be possible to place the momentum vectors tip to tail and form a closed polygon, which means the vector sum is zero.

If the original momentum of the bomb were not zero, these vectors would add up to the original momentum vector.

2-D Sample Problem

0.3 kg

34 m/s40

152 gA mean, old dart strikes an innocent mango that was just passing by minding its own business. Which way and how fast do they move off together?

before

after

5 m/s

452 g

v

152 (34) sin 40 = 452 v sin

152 (34) cos 40 - 300 (5) = 452 v cos

Working in grams and taking left & down as + :

Dividing equations : 1.35097 = tan

= 53.4908Substituting into either of the first two equations : v = 9.14 m/s

4132.9736

Alternate Solution

5168

40

1500

p

Shown are momentum vectors (in g m/s). The black vector is the total momentum before the collision. Because of conservation of momentum, it is also the total momentum after the collisions. We can use trig to find its magnitude and direction.

Law of Sines :

40

sin 1500

Law of Cosines : p2 = 5168 2 + 1500

2 - 2 5168 1500 cos 40 p = 4132.9736 g m/s

Dividing by total mass : v = (4132.9736 g m/s) / (452 g) = 9.14 m/s

= sin 40 = 13.4908

Angle w/ resp. to horiz. = 40 + 13. 4908 53.49

Comments on Alternate Method• Note that the alternate method gave us the exact

same solution.• This method can only be used when two objects

collide and stick, or when one object breaks into two. Otherwise, we’d be dealing with a polygon with more sides than a triangle.

• In using the Law of Sines (last step), the angle involved (ß) is the angle inside the triangle. A little geometry gives us the angle with respect to the horizontal.

Angular Momentum

r

v

m

Angular momentum depends on linear momentum and the distance from a particular point. It is a vector quantity with symbol L. If r and v are then the magnitude of angular momentum w/ resp. to point Q is given by L = r p = m v r. In this case L points out of the page. If the mass were moving in the opposite direction, L would point into the page.

The SI unit for angular momentum is the kg m2 / s. (It has no special name.) Angular momentum is a conserved quantity. A torque is needed to change L, just a force is needed to change p. Anything spinning has angular has angular momentum. The more it has, the harder it is to stop it from spinning.

Q

Angular Momentum: General Definition

r

v

m

If r and v are not then the angle between these two vectors must be taken into account. The general definition of angular momentum is given by a vector cross product: L = r p

This formula works regardless of the angle. As you know from our study of cross products, the magnitude of the angular momentum of m relative to point Q is: L = r p sin = m v r. In this case, by the right-hand rule, L points out of the page. If the mass were moving in the opposite direction, L would point into the page.

Q

Moment of InertiaAny moving body has inertia. (It wants to keep moving at constant v.) The more inertia a body has, the harder it is to change its linear motion. Rotating bodies possess a rotational inertial called the moment of inertial, I. The more rotational inertia a body has, the harder it is change its rotation. For a single point-like mass w/ respect to a given point Q, I = m r

2.

I = m r 2

mr

For a system, I = the sum of each mass times its respective distance from the point of interest.

r2

m2

r1

m1

I = mi ri

2

= m1 r12 + m2 r2

2

Q

Q

Moment of Inertia ExampleTwo merry-go-rounds have the same mass and are spinning with the same angular velocity. One is solid wood (a disc), and the other is a metal ring. Which has a bigger moment of inertia relative to its center of mass?

mm

r r

answer: I is independent of the angular speed. Since their masses and radii are the same, the ring has a greater moment of inertia. This is because more of its mass is farther from the axis of rotation. Since I is bigger for the ring, it would more difficult to increase or decrease its angular speed.

Angular AccelerationAs you know, acceleration is when an object speeds up, slows down, or changes directions. Angular acceleration occurs when a spinning object spins faster or slower. Its symbol is , and it’s defined as:

= / tNote how this is very similar to a = v / t for linear acceleration.Ex: If a wind turbine spinning at 21 rpm speeds up to 30 rpm over 10 s due to a gust of wind, its average angular acceleration is 9 rpm / 10 s. This means every second it’s spinning 9 revolutions per minute faster than the second before. Let’s convert the units:

9 rpm10 s = 9 rev / min

10 s = 9 rev

min 10 s

= 9 (2 rad) (60 s) 10 s

= 0.094 rad / s2

Since a radian is really dimensionless (a length divided by a length), the SI unit for angular acceleration is the “per second squared” (s-2).

Torque & Angular Acceleration

Newton’s 2nd Law, as you know, is Fnet = m a

The 2nd Law has a rotational analog: net = I

A force is required for a body to undergo acceleration. A “turning force” (a torque) is required for a body to undergo

angular acceleration.

The bigger a body’s mass, the more force is required to accelerate it. Similarly, the bigger a body’s rotational inertia,

the more torque is required to accelerate it angularly.

Both m and I are measures of a body’s inertia (resistance to change in motion).

Linear Momentum & Angular Momentum

If a net force acts on an object, it must accelerate, which means its momentum must change. Similarly, if a net torque acts on a body, it undergoes angular acceleration, which means its angular momentum changes. Recall, angular momentum’s magnitude is given by

L = m v r

rvm

So, if a net torque is applied, angular velocity must change, which changes angular momentum.

proof: net = r Fnet = r m a

= r m v / t = L / t So net torque is the rate of change of angular momentum, just as net force is the rate of change of linear momentum. continued on next slide

(if v and r are perpendicular)

Linear & Angular Momentum (cont.)

Here is yet another pair of similar equations, one linear, one rotational. From the formula v = r , we get

L = m v r = m r (r ) = m r 2 = I

This is very much like p = m v, and this is one reason I is defined the way it is.

In terms of magnitudes, linear momentum is inertia times speed, and angular momentum is rotational inertia times

angular speed.

L = I p = m v

Spinning Ice SkaterWhy does a spinning ice skater speed up when she pulls her arms in?

Suppose Mr. Stickman is sitting on a stool that swivels holding a pair of dumbbells. His axis of rotation is vertical. With the weights far from that axis, his moment of inertia is large. When he pulls his arms in as

he’s spinning, the weights are closer to the axis, so his moment of inertia gets much smaller. Since L = I and L is conserved, the product of I and is a constant. So, when he pulls his arms in, I goes down, goes up, and he starts spinning much faster.

I = L = I

Comparison: Linear & Angular Momentum

Linear Momentum, p• Tendency for a mass to

continue moving in a straight line.

• Parallel to v.• A conserved, vector quantity. • Magnitude is inertia (mass)

times speed.• Net force required to change it.• The greater the mass, the

greater the force needed to change momentum.

Angular Momentum, L• Tendency for a mass to continue

rotating.• Perpendicular to both v and r.• A conserved, vector quantity.• Magnitude is rotational inertia

times angular speed.• Net torque required to change it.• The greater the moment of

inertia, the greater the torque needed to change angular momentum.