Momentum & Impulse AP/Honors Physics 1

Mr. Velazquez

Linear Momentum

• An object’s linear momentum is a simple quantity defined as its mass times its velocity vector.

• As the product of a scalar and a vector, momentum is a vector quantity; this means we must specify a direction.

• We can think of an object’s momentum as its “stopping power” or “drive.” The higher momentum is, the greater effect the object will have on anything it collides with.

𝒑 = 𝒎𝒗

The SI units of momentum are 𝐤𝐠 ⋅ 𝐦 𝐬 . There is no special name for this unit.

Linear Momentum

Which has the greater momentum, a 150-gram baseball moving at 52 m/s, or a 6.50-kg bowling ball travelling at 1.4 m/s? Assume they move in the same direction.

𝑝 = 𝑚𝑣 𝑝 = 0.15 kg 52 m s

𝒑 = 𝟕. 𝟖 𝐤𝐠𝐦 𝐬

𝑝 = 𝑚𝑣 𝑝 = 6.5 kg 1.4 m s

𝒑 = 𝟗. 𝟏 𝐤𝐠𝐦 𝐬

𝑣 𝑣

Linear Momentum

“Everyone has a plan, until they get punched in the face.”

–Mike Tyson

Based on what we know about momentum, what could a fighter do to reduce the impact of a punch?

Change in Linear Momentum

The change in an object’s momentum is defined as:

This may seem obvious, but a little tinkering with Newton’s force equation reveals an interesting connection:

∆𝒑 = 𝒎∆𝒗

𝐹 = 𝑚𝑎

𝐹 = 𝑚∆𝑣

∆𝑡

𝑭∆𝒕 = 𝒎∆𝒗

Change in momentum

Force multiplied by the change in time; called the impulse

Impulse

• The impulse experienced by an object is equal to the change in momentum experienced by an object. • As a result of Newton’s Second Law, impulse is also

equivalent to the force acting on an object multiplied by the time interval.

• In many ways, momentum and impulse are related to each other in the same manner that work and energy are related. The former describes a change in the latter.

𝑰 = ∆𝒑 = 𝒎∆𝒗 = 𝑭∆𝒕

Change in Linear Momentum Calculate the impulse experienced by a 50-kg gymnast who lands on firm ground after falling (from rest) from a height of 3.0 m.

We can use a kinematic equation to find the speed of the gymnast just before she hits the ground:

𝑣2 = 𝑣02 + 2𝑎𝑠

𝑣 = 2(9.81 ms2 )(3.0 m)

𝒗 = 𝟕. 𝟕 𝐦 𝐬

Change in Linear Momentum Calculate the impulse experienced by a 50-kg gymnast who lands on firm ground after falling (from rest) from a height of 3.0 m.

Now we know the speed the gymnast will have when she first hits the ground (v=7.7 m/s). In an instant, this will be reduced to zero. Her impulse (change in momentum) will therefore be:

∆𝑝 = 𝑚∆𝑣

∆𝑝 = 50 kg 7.7 m s − 0 m s

∆𝒑 = 𝟑𝟖𝟎 𝐤𝐠𝐦 𝐬

Change in Linear Momentum

A 5.5-kg cannonball is fired from a cannon via an explosion reaction inside, which applies a 56,000-N force (assumed constant) to the cannonball for 1/120th of a second. Calculate the final velocity of the cannonball, assuming it starts from rest.

Change in momentum is equal to the impulse. Impulse is equal to force multiplied by the time interval.

∆𝑝 = 𝑚∆𝑣 = 𝐹∆𝑡 𝑚∆𝑣 = 𝐹∆𝑡

∆𝑣 =𝐹∆𝑡

𝑚

∆𝑣 =56,000 N 1

120 s

(5.5 kg)= 𝟖𝟓 𝐦 𝐬

Conservation of Momentum • Recall that the change in momentum (impulse) is equal to

the force acting on an object multiplied by the time interval.

• Logically, if the net force acting on an object (or objects) does not change over a period of time, then the total momentum will not change.

• This means if there are no external forces acting on a system of objects, the total momentum will be conserved. This is the Law of Conservation of Momentum.

• This physical law is especially important when understanding collisions.

∆𝑝 = 𝐹∆𝑡

𝒑𝟏 = 𝒑𝟐 𝒑𝑨𝟏 + 𝒑𝑩𝟏 = 𝒑𝑨𝟐 + 𝒑𝑩𝟐 CoM for Two Objects CoM for One Object

Conservation of Momentum: One Dimension • Consider two billiard balls,

A and B, colliding with each other.

• We assume the net external forces acting on the balls are zero; they will only be in contact with each other.

• The momentum of each ball will change, but the total momentum of the system will not.

A B

A B

A B

𝐹𝐵𝐴 𝐹𝐴𝐵

𝑣𝐴 𝑣𝐵

𝑣𝐴 𝑣𝐵

BEFORE COLLISION

DURING COLLISION

AFTER COLLISION

𝑝𝐴1 + 𝑝𝐵1 = 𝑝𝐴2 + 𝑝𝐵2

𝑚𝐴𝑣𝐴1 +𝑚𝐵𝑣𝐵1 = 𝑚𝐴𝑣𝐴2 +𝑚𝐵𝑣𝐵2

Conservation of Momentum: One Dimension Two billiard balls, A and B, of masses 100 g and 120 g respectively, head toward each other on a frictionless table, as shown. Before colliding, ball A has a velocity of 3.6 m/s to the right and ball B has a velocity of 1.4 m/s to the left. After the collision, the velocity of ball B has is 2.5 m/s to the right. What is the velocity of ball A after the collision?

A B 𝑣𝐴 𝑣𝐵 BEFORE COLLISION

A B 𝑣𝐴 𝑣𝐵

AFTER COLLISION

𝑝𝐴1 + 𝑝𝐵1 = 𝑝𝐴2 + 𝑝𝐵2

𝑚𝐴𝑣𝐴1 +𝑚𝐵𝑣𝐵1 = 𝑚𝐴𝑣𝐴2 +𝑚𝐵𝑣𝐵2

0.1 kg 3.6m s + (0.12 kg)(−1.4m s ) = (0.1 kg)𝒗𝑨𝟐 + (0.12 kg)(2.5 m s )

+

0.36 kgm s + −0.17 kgm s = 0.1 kg 𝒗𝑨𝟐 + (0.30 kgm s )

𝒗𝑨𝒇 = −𝟏. 𝟏 𝐦 𝐬

Conservation of Momentum: One Dimension A 2,000-kg train car traveling at 24.0 m/s strikes another train car of identical mass, at rest. If the cars lock together as a result of the collision, what is their speed afterward?

A B 𝑣𝐴

A B 𝑣𝐴+𝐵

𝑚𝐴𝑣𝐴 +𝑚𝐵𝑣𝐵 = 𝑚𝐴+𝐵𝑣𝐴+𝐵

𝑣𝐴+𝐵 =𝑚𝐴𝑣𝐴𝑚𝐴+𝐵

𝑣𝐴+𝐵 =(2,000 kg)(24.0 m s )

(2,000 kg + 2,000 kg)= 𝟏𝟐. 𝟎 𝐦 𝐬

Notice this is exactly half of car A’s initial velocity.

Elastic vs. Inelastic Collisions • Recall that during a collision

where no external forces are acting on a system, total momentum is conserved. What about kinetic energy?

• For billiard balls, we saw that momentum was conserved.

• If the objects are hard enough, and no heat or other energy is produced by the collision, total kinetic energy will be conserved as well.

• These collisions, in which total kinetic energy is conserved, are called elastic collisions.

A B 𝑣𝐴1 𝑣𝐵1

BEFORE COLLISION (ELASTIC)

A B

DURING COLLISION

A B 𝑣𝐴2 𝑣𝐵2

AFTER COLLISION (ELASTIC)

𝐾1 = 𝐾2 𝐾𝐴1 + 𝐾𝐵1 = 𝐾𝐴2 + 𝐾𝐵2

𝟏

𝟐𝒎𝑨𝒗𝑨𝟏

𝟐 +𝟏

𝟐𝒎𝑩𝒗𝑩𝟏

𝟐 =𝟏

𝟐𝒎𝑨𝒗𝑨𝟐

𝟐 +𝟏

𝟐𝒎𝑩𝒗𝑩𝟐

𝟐

Elastic Collisions in One Dimension

A B 𝑣𝐴1 𝑣𝐵1

BEFORE COLLISION (ELASTIC)

A B 𝑣𝐴2 𝑣𝐵2

AFTER COLLISION (ELASTIC)

𝐾1 = 𝐾2

𝐾𝐴1 + 𝐾𝐵1 = 𝐾𝐴2 + 𝐾𝐵2

1

2𝑚𝐴𝑣𝐴1

2 +1

2𝑚𝐵𝑣𝐵1

2 =1

2𝑚𝐴𝑣𝐴2

2 +1

2𝑚𝐵𝑣𝐵2

2

𝑚𝐴𝑣𝐴12 −𝑚𝐴𝑣𝐴2

2 = 𝑚𝐵𝑣𝐵22 −𝑚𝐵𝑣𝐵1

2

𝑚𝐴 𝑣𝐴12 − 𝑣𝐴2

2 = 𝑚𝐵 𝑣𝐵22 − 𝑣𝐵1

2

𝑚𝐴 𝑣𝐴1 − 𝑣𝐴2 𝑣𝐴1 + 𝑣𝐴2 = 𝑚𝐵(𝑣𝐵2 − 𝑣𝐵1)(𝑣𝐵2 + 𝑣𝐵1)

𝑣𝐴1 + 𝑣𝐴2 = 𝑣𝐵2 + 𝑣𝐵1

𝒗𝑨𝟏 − 𝒗𝑩𝟏 = 𝒗𝑩𝟐 − 𝒗𝑨𝟐

𝑝1 = 𝑝2

𝑝𝐴1 + 𝑝𝐵1 = 𝑝𝐴2 + 𝑝𝐵2

𝑝𝐴1 − 𝑝𝐴2 = 𝑝𝐵2 − 𝑝𝐵1

𝒎𝑨 𝒗𝑨𝟏 − 𝒗𝑨𝟐 = 𝒎𝑩(𝒗𝑩𝟐 − 𝒗𝑩𝟏)

divide the equations

Head-on Elastic Collision

Elastic Collisions in One Dimension A billiard ball A of mass 𝑚 moving with speed 𝑣 collides with another billiard ball B of equal mass which is at rest (𝑣𝐵 = 0). If this is an elastic collision, what are the speeds of both balls just after the collision?

A B 𝑣

BEFORE COLLISION (ELASTIC)

A B 𝑣𝐴 𝑣𝐵

AFTER COLLISION (ELASTIC)

Conservation of momentum gives us: 𝑝1 = 𝑝2

𝑚𝑣 = 𝑚𝑣𝐴 +𝑚𝑣𝐵 𝑣 = 𝑣𝐴 + 𝑣𝐵

Conservation of Kinetic Energy (for head-on elastic collisions) gives us:

𝑣𝐴1 − 𝑣𝐵1 = 𝑣𝐵2 − 𝑣𝐴2 𝑣 = 𝑣𝐵 − 𝑣𝐴 i ii

Subtracting equation (ii) from equation (i) leaves us with: 0 = 2𝑣𝐴 → 𝒗𝑨 = 𝟎

Now, substituting into equation (ii): 𝑣 = 𝑣𝐵 − 𝑣𝐴 → 𝒗𝑩 = 𝒗

A 𝑣

Elastic Collisions in One Dimension A marble of mass 15.0 g moving at 265 m/s collides elastically with a shot-put of mass 4.50 kg which is at rest. Find the velocity of each object after the collision.

A B 𝑣𝐴1

BEFORE COLLISION (ELASTIC)

𝑣𝐴2 𝑣𝐵2

AFTER COLLISION (ELASTIC)

Conservation of momentum gives us: 𝑝1 = 𝑝2

𝑚𝐴𝑣𝐴1 + 0 = 𝑚𝐴𝑣𝐴2 +𝑚𝐵𝑣𝐵2 𝑚𝐴𝑣𝐴1 = 𝑚𝐴𝑣𝐴2 +𝑚𝐵𝑣𝐵2

Conservation of Kinetic Energy (for head-on elastic collisions) gives us:

𝑣𝐴1 − 𝑣𝐵1 = 𝑣𝐵2 − 𝑣𝐴2 𝑣𝐴2 = 𝑣𝐵2 − 𝑣𝐴1 i ii

Substituting equation (ii) into equation (i): 𝑚𝐴𝑣𝐴1 = 𝑚𝐴 𝑣𝐵2 − 𝑣𝐴1 +𝑚𝐵𝑣𝐵2

A B

2𝑚𝐴𝑣𝐴1 = 𝑣𝐵2 𝑚𝐴 +𝑚𝐵 → 𝑣𝐵2 =2𝑚𝐴𝑣𝐴1𝑚𝐴 +𝑚𝐵

=2(0.015 kg)(265 m s )

0.015 kg + 4.50 kg= 𝟏. 𝟕𝟔 𝐦/𝐬

Elastic Collisions in One Dimension A marble of mass 15.0 g moving at 265 m/s collides elastically with a shot-put of mass 4.50 kg which is at rest. Find the velocity of each object after the collision.

A B 𝑣𝐴1

BEFORE COLLISION (ELASTIC)

𝑣𝐴2 𝑣𝐵2

AFTER COLLISION (ELASTIC)

Now we have the post-collision velocity for the heavier shot-put (𝒗𝑩𝟐 = 1.76 m/s). We can use this with equation (ii) to find the same for the marble:

𝑣𝐴2 = 𝑣𝐵2 − 𝑣𝐴1 𝑣𝐴2 = 1.76 𝑚 𝑠 − 265 𝑚 𝑠

𝒗𝑨𝟐 = −𝟐𝟔𝟑 𝒎 𝒔

A B

The negative confirms that the marble will now be moving in the opposite direction, and its speed has slowed, since some momentum has been transferred to the shot-put.

Exit Ticket: Momentum & Collision

Billiard ball A of mass 0.060 kg heads toward a second Billiard ball B of mass 0.090 kg. Before they collide, ball A has a velocity of 2.50 m/s to the right, and ball B has a velocity 1.15 m/s to the right. After they collide elastically, what is the speed and direction of each ball after the collision?

A B 𝑣𝐴1 𝑣𝐵1

BEFORE COLLISION (ELASTIC)

A B 𝑣𝐴2 𝑣𝐵2

AFTER COLLISION (ELASTIC)

Homework: DUE 12/11

Pg. 267-268, #4-32 (Multiples of 4)

Elastic Collisions in One Dimension

A B 𝑣𝐴1 𝑣𝐵1

BEFORE COLLISION (ELASTIC)

A B 𝑣𝐴2 𝑣𝐵2

AFTER COLLISION (ELASTIC)

Conservation of Energy Equation

𝒗𝑨𝟏 − 𝒗𝑩𝟏 = 𝒗𝑩𝟐 − 𝒗𝑨𝟐

Conservation of Momentum Equation

𝑝1 = 𝑝2

𝒎𝑨𝒗𝑨𝟏 +𝒎𝑩𝒗𝑩𝟏 = 𝒎𝑨𝒗𝑨𝟐 +𝒎𝑩𝒗𝑩𝟐

or

𝒎𝑨 𝒗𝑨𝟏 − 𝒗𝑨𝟐 = 𝒎𝑩(𝒗𝑩𝟐 − 𝒗𝑩𝟏)

Head-on Elastic Collision

Combine both these equations (using substitution, division, addition, etc.) to

find missing velocities.

𝒑 = 𝒎𝒗

Momentum

∆𝒑 = 𝒎∆𝒗 = 𝑭∆𝒕

Impulse

Elastic Collisions in One Dimension

A B 𝑣1

BEFORE COLLISION (ELASTIC)

A B 𝑣𝐴2 𝑣𝐵2

AFTER COLLISION (ELASTIC)

Final Velocities of A and B

𝒗𝑨𝟐 =𝒎𝑨 −𝒎𝑩

𝒎𝑨 +𝒎𝑩𝒗𝟏

𝒗𝑩𝟐 =𝟐𝒎𝑨

𝒎𝑨 +𝒎𝑩𝒗𝟏

Head-on Elastic Collision (if B is initially at rest)

WARNING: These formulas only work under the specified conditions. (i.e. if B starts at rest, and A starts with initial

velocity 𝒗𝟏)

Elastic vs. Inelastic Collisions • Nearly all collisions are inelastic to some degree.

• Kinetic energy will either be less or greater after the collision.

• When two objects stick together after a collision, the collision is said to be completely inelastic.

• Initial kinetic energy will be transformed into other types of energy consistent with conservation of momentum.

A B 𝑣𝐴

A B 𝑣𝐴+𝐵

Inelastic Collisions A 2,000-kg train car traveling at 24.0 m/s strikes another train car of identical mass, at rest. If the cars lock together as a result of the collision, what is their speed afterward, and how much initial kinetic energy has been transferred to other forms?

A B 𝑣𝐴

A B 𝑣𝐴+𝐵

Recall from our previous answer that the final velocity was 𝒗𝑨+𝑩 = 𝟏𝟐. 𝟎 𝐦 𝐬

𝐾1 =1

2𝑚𝐴𝑣𝐴

2 =1

22,000 kg 24.0 m s

2 = 𝟓𝟕𝟔 𝐤𝐉

𝐾2 =1

2𝑚𝐴+𝐵𝑣𝐴+𝐵

2 =1

24,000 kg 12.0 m s

2 = 𝟐𝟖𝟖 𝐤𝐉

So exactly half of the kinetic energy has been converted to alternate forms (probably thermal or sound energy)

Inelastic Collisions: Ballistic Pendulum To measure the speed of a bullet, we first measure the mass 𝑚 of the projectile, then fire it into a large block of mass 𝑀 suspended as a pendulum. As a result of the collision, the bullet and pendulum swing upward and the maximum height ℎ of the pendulum can be measured. Determine the relationship between the projectile speed 𝒗 and the vertical height 𝒉.

M 𝑣

𝑚

𝑣𝑀 = 0 𝑀 +𝑚

ℎ 𝑣𝑀+𝑚

Inelastic Collisions: Ballistic Pendulum

M 𝑣

𝑚

𝑣𝑀 = 0

𝑀 +𝑚

ℎ 𝑣𝑀+𝑚

Conservation of Momentum (Before and After Collision)

𝑝1 = 𝑝2 𝑚𝑣 = 𝑀 +𝑚 𝑣𝑀+𝑚

𝑣𝑀+𝑚 =𝑚𝑣

𝑀 +𝑚

Conservation of Energy (After Collision)

𝐾1 + 𝑈1 = 𝐾2 + 𝑈2 1

2𝑀 +𝑚 𝑣𝑀+𝑚

2 + 0 = 0 + 𝑀 +𝑚 𝑔ℎ

1

2𝑀 +𝑚

𝑚𝑣

𝑀 +𝑚

2

= 𝑀 +𝑚 𝑔ℎ

𝑚2𝑣2

𝑀 +𝑚 2= 2𝑔ℎ

𝑣2 =𝑀 +𝑚

𝑚

2

2𝑔ℎ

𝒗 =𝑴+𝒎

𝒎𝟐𝒈𝒉

Exit Ticket, Part 1 (DO THIS NOW) A cart of mass 225 kg moving at 15.5 m/s crashes into a second cart of mass 315 kg which is at rest. If the carts stick together after the collision, what is the shared velocity of the carts after the crash, and how much kinetic energy has been “lost” ?

A B 𝑣𝐴

A B 𝑣𝐴+𝐵

Formulas:

𝑝1 = 𝑝2

𝐾1 −𝑊𝑙𝑜𝑠𝑡 = 𝐾2

Collisions in Two Dimensions The most obvious example of this is billiards (pool). Here’s Donald Duck and a billiards expert: https://www.youtube.com/watch?v=cdVRvYnqj6w A less obvious (and more ridiculous-looking) example is from the strange world of Olympic curling: https://www.youtube.com/watch?v=miB7HzUvmM0

Collisions in Two Dimensions For simplicity (and because it’s unlikely you’ll see alternate examples in tests) we will focus on the example of an elastic non-head-on collision where one object moving with velocity 𝑣0 (called the “projectile”) strikes a second object at rest (called the “target”).

A

B

B

A

+y

+x

𝑣0

𝑣𝐴

𝑣𝐵

𝜃𝐴

𝜃𝐵

Split the momentum into horizontal and vertical components. Each is conserved.

Horizontal (𝒙) Momentum:

𝑝𝑥1 = 𝑝𝑥2 𝒎𝑨𝒗𝟎 = 𝒎𝑨𝒗𝑨 𝐜𝐨𝐬𝜽𝑨 +𝒎𝑩𝒗𝑩 𝐜𝐨𝐬𝜽𝑩

Vertical (𝒚) Momentum:

𝑝𝑦1 = 𝑝𝑦2

0 = 𝑚𝐴𝑣𝐴 sin 𝜃𝐴 −𝑚𝐵𝑣𝐵 sin 𝜃𝐵 𝒎𝑨𝒗𝑨 𝐬𝐢𝐧 𝜽𝑨 = 𝒎𝑩𝒗𝑩 𝐬𝐢𝐧 𝜽𝑩

Collisions in Two Dimensions If the collision is elastic, kinetic energy will also be conserved, so we can apply the energy equation, and we now have three equations, which we can use to solve for three unknowns:

A

B

B

A

+y

+x

𝑣0

𝑣𝐴

𝑣𝐵

𝜃𝐴

𝜃𝐵

Conservation of Kinetic Energy

(Elastic Collision):

1

2𝑚𝐴𝑣0

2 =1

2𝑚𝐴𝑣𝐴

2 +1

2𝑚𝐵𝑣𝐵

2

𝒎𝑨𝒗𝟎𝟐 = 𝒎𝑨𝒗𝑨

𝟐 +𝒎𝑩𝒗𝑩𝟐

Collisions in Two Dimensions

A

B

B

A

+y

+x

𝑣0

𝑣𝐴

𝑣𝐵

𝜃𝐴

𝜃𝐵

Horizontal (𝒙) Momentum:

𝒎𝑨𝒗𝟎 = 𝒎𝑨𝒗𝑨 𝐜𝐨𝐬𝜽𝑨 +𝒎𝑩𝒗𝑩 𝐜𝐨𝐬𝜽𝑩

Vertical (𝒚) Momentum:

𝒎𝑨𝒗𝑨 𝐬𝐢𝐧 𝜽𝑨 = 𝒎𝑩𝒗𝑩 𝐬𝐢𝐧 𝜽𝑩

Kinetic Energy (Elastic Collision):

𝒎𝑨𝒗𝟎𝟐 = 𝒎𝑨𝒗𝑨

𝟐 +𝒎𝑩𝒗𝑩𝟐

We can use these three equations together to solve for three separate variables, assuming the conditions are met (elastic, non-head-on, 2nd object at rest) and the other variables are given.

Collisions in Two Dimensions

A

B

B

A

+y

+x

𝑣0

𝑣𝐴

𝑣𝐵

𝜃𝐴

𝜃𝐵

Suppose object A is a curling stone of mass 18.0 kg, and it has an initial velocity 𝑣0 = 5.15 m/s as it heads toward another curling stone B of mass 20.0 kg which is at rest. After the collision, stone A’s velocity is angled 32.5° above it’s initial direction. Find the velocities of stones A and B after the collision, including the direction of stone B.

Given: 𝑚𝐴 = 18.0 kg 𝑚𝐵 = 20.0 kg 𝑣0 = 5.15 m s 𝜃𝐴 = 32.5°

Find: 𝑣𝐴 = ? 𝑣𝐵 = ? 𝜃𝐵 = ?

Exit Ticket, Part 2

e

p

p

e

+y

+x

𝑣0

𝑣𝑒

𝑣𝑝

𝜃𝑒

𝜃𝑝

An electron (𝒎𝒆 = 𝟗. 𝟏𝟏 × 𝟏𝟎−𝟑𝟏 𝐤𝐠) is fired at a speed of 𝒗𝟎 = 𝟐. 𝟗𝟏 × 𝟏𝟎𝟖 𝐦 𝐬 and collides

elastically with a proton (𝒎𝒑 = 𝟏. 𝟔𝟕 × 𝟏𝟎−𝟐𝟕 𝐤𝐠)

which is at rest. The angle of the electron’s trajectory after the collision is later measured to be 𝜽𝒆 = 𝟕𝟐. 𝟓° past its original direction. Find the velocities of the electron and proton after the collision as well as the direction of the proton’s velocity.

Given: 𝑚𝑒 = 9.11 × 10−31 kg 𝑚𝑝 = 1.67 × 10−27 kg

𝑣0 = 2.91 × 108 m s 𝜃𝑒 = 72.5°

Find: 𝒗𝒆 = ? 𝒗𝒑 = ?

𝜽𝒑 = ? Horizontal (𝑥) Momentum:

𝑚𝐴𝑣0 = 𝑚𝐴𝑣𝐴 cos 𝜃𝐴 +𝑚𝐵𝑣𝐵 cos 𝜃𝐵

Vertical (𝑦) Momentum: 𝑚𝐴𝑣𝐴 sin 𝜃𝐴 = 𝑚𝐵𝑣𝐵 sin 𝜃𝐵

Kinetic Energy (Elastic Collision):

𝑚𝐴𝑣02 = 𝑚𝐴𝑣𝐴

2 +𝑚𝐵𝑣𝐵2

Don’t Forget: HW Problem Set

Momentum Pg. 267-268, #4-32 (multiples of 4)