Topic 1 - Momentum

𝑝 = 𝑚𝑣
●​ Momentum is the product of the mass and velocity of the object. (just product, not
dot product nor cross product)
●​ Linear momentum is a vector that points in the same direction as the object’s
velocity.
○​ Because momentum depends on the velocity of the object, and the velocity
depends on the choice of the reference frame, different observers will
measure different momenta for the same object.
𝑚
●​ The SI unit of linear momentum is [𝑘𝑔 ].
𝑠
●​ The total linear momentum of a system containing multiple objects is the vector
sum of the momenta (plural of momentum) of the individual objects:
Σ𝑝 = 𝑝1 + 𝑝2 + 𝑝3 +......
●​ Relationship between momentum and kinetic energy:
2
𝑚𝑣
𝐾𝐸 = 2
2
𝑚·𝑚𝑣
𝐾𝐸 = 𝑚·2
2 2
𝑚𝑣
𝐾𝐸 = 2𝑚
2
𝑝
𝐾𝐸 = 2𝑚
●​ Momentum in the above equation is ONLY magnitude. There is no direction.

Topic 2 - Impulse
●​ The calculus aspect of momentum (big boi style)
𝑑𝑝 𝑑
𝑑𝑡
= 𝑑𝑡
(𝑚𝑣)
𝑑𝑝 𝑑𝑣 𝑑𝑚
Use product rule
𝑑𝑡
= 𝑚 𝑑𝑡
+ 𝑑𝑡
𝑣
 𝑑𝑝 𝑑𝑣
Assume mass is constant ​ 𝑑𝑡
= 𝑚 𝑑𝑡
+ 0
𝑑𝑝
𝑑𝑡
= 𝑚𝑎
𝑑𝑝
𝑑𝑡
= Σ𝐹𝑒𝑥𝑡𝑒𝑟𝑛𝑎𝑙
●​ Because momentum and force are both vector quantities, we can express them in
their component forms:
𝑑 𝑝𝑥 𝑑 𝑝𝑦 𝑑 𝑝𝑧
Σ𝐹𝑒𝑥𝑡,𝑥 = 𝑑𝑡
​ ​ Σ𝐹𝑒𝑥𝑡,𝑦 = 𝑑𝑡
​​ Σ𝐹𝑒𝑥𝑡,𝑦 = 𝑑𝑡
●​ From the previous work, we see the derivative of momentum relative to time is
force. Therefore, the integral of force over a time interval is the change of
momentum.
𝑡𝑓 𝑡𝑓
𝑑𝑝
∫ 𝐹 𝑑𝑡 = ∫ 𝑑𝑡
𝑑𝑡
𝑡𝑖 𝑡𝑖
𝑡𝑓

∫ 𝐹 𝑑𝑡 = 𝑝𝑓 − 𝑝𝑖
𝑡𝑖
𝑡𝑓

∫ 𝐹 𝑑𝑡 = △𝑝
𝑡𝑖
●​ The change of momentum is defined as the impulse J:

𝐽 = △𝑝
 𝑡𝑓

𝐽 = ∫ 𝐹𝑒𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑑𝑡
𝑡𝑖
●​ For a constant external force:

𝐽 = 𝐹𝑒𝑥𝑡𝑒𝑟𝑛𝑎𝑙 ∆𝑡

Topic 3 - Conservation of Linear Momentum

●​ From the previous part, we know:

𝐽 = △𝑝
𝐽 = 𝑝𝑓 − 𝑝 𝑖
𝑝𝑖 + 𝐽 = 𝑝𝑓
(Recall energy conservation. Compare two conservation equations.)

●​ If there are multiple objects in the system, we should have multiple terms of
momentum.

(𝑝1,𝑖 + 𝑝2,𝑖 +...) + 𝐽 = (𝑝1,𝑓 + 𝑝2,𝑓 +...)

●​ If there are multiple external forces being exerted on the system, we should have
multiple terms of impulse.

(𝑝1,𝑖 + 𝑝2,𝑖 +...) + (𝐽1 𝑜𝑛 𝑠𝑦𝑠 + 𝐽2 𝑜𝑛 𝑠𝑦𝑠 +...) = (𝑝1,𝑓 + 𝑝2,𝑓 +...)

●​ When the sum of external forces exerted on the system is zero (aka impulse = 0),
the total linear momentum of a system is conserved.

𝑝𝑖 + 0 = 𝑝𝑓
𝑝𝑖 = 𝑝𝑓​ when Σ𝐹𝑒𝑥𝑡𝑒𝑟𝑛𝑎𝑙 = 0

。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。
How to approach a momentum-impulse question
 Step 1 - define the system, the initial and the final moment (and coordinate system).

Initial System Final

Step 2 - write the momentum-impulse equation

𝑝𝑖 + 𝐽 = 𝑝𝑓

Step 3 - consider the number of objects in the system and the number of external forces
exerted on the system.

(𝑝1,𝑖 + 𝑝2,𝑖 +...) + (𝐽1 𝑜𝑛 𝑠𝑦𝑠 + 𝐽2 𝑜𝑛 𝑠𝑦𝑠 +...) = (𝑝1,𝑓 + 𝑝2,𝑓 +...)

Step 4 - use the momentum-impulse bar chart

Topic 4 - Elastic and inelastic collision

Simulation

Elastic collision Inelastic collision

Elastic - bouncy - bounce off. Objects do not Inelastic collisions last more than an instant,
stick together at all. The collision happens in meaning objects will stick together for some
an instant. time during the collision.

In an elastic collision: In an inelastic collision:

●​ p is conserved; 𝑝𝑓 = 𝑝𝑖 ●​ p is conserved; 𝑝𝑓 = 𝑝𝑖
●​ KE is also conserved. 𝐾𝐸𝑓 = 𝐾𝐸𝑖 ●​ KE is NOT conserved. 𝐾𝐸𝑓 ≠ 𝐾𝐸𝑖

Total elastic collision is not realistic since Total inelastic collision means two objects
some kinetic energy (KE) has to be converted stick together and never split up even after
(in any event) to the internal energy (Uint). the collision.
Total inelastic collision is rather common.

●​ Most collisions in real life are between totally inelastic and totally elastic. They are
very difficult to analyze using simple equations.
●​ We focus on totally inelastic collisions and totally elastic collisions, which are
reasonable to deal with.

Elastic collision

𝑚1𝑣1,𝑖 + 𝑚2𝑣2,𝑖 + 𝐽 = 𝑚1𝑣1,𝑓 + 𝑚2𝑣2,𝑓

Final velocities are not necessarily the same.

Inelastic collision

𝑚1𝑣1,𝑖 + 𝑚2𝑣2,𝑖 + 𝐽 = (𝑚1 + 𝑚2)𝑣𝑓

Final velocities are the same.
Inelastic collision separation

(𝑚1 + 𝑚2)𝑣𝑖 + 𝐽 = 𝑚1𝑣1,𝑓 + 𝑚2𝑣2,𝑓

Initial velocities are the same.

Topic 5 - Collision in 2D
●​ Because momentum is a vector, it has x,y,(and z if necessary) components.
●​ For any inelastic collision (separation), we simply need to consider momentum in x
and y (and z if necessary) directions separately. Because the final (or initial)
velocities for the objects are the same, there is a limited amount of unknowns in
the equations, which means this type of questions
●​ For elastic collisions, things are more complicated as the final velocities are NOT
the same (most of the times). Usually they are difficult to solve if both 𝑣1,𝑓 and 𝑣2,𝑓
are unknown (both magnitude and directions). This will result in 4 different
unknowns in the same equation, effectively rendering the momentum conservation
equation useless.
○​ We will consider a specific situation where we can deal with less unknowns:
○​ Consider two identical objects colliding elastically on a flat surface. Right
before the collision, object 1 has an initial velocity of v1 while object 2 is
stationary. After the elastic collision, both objects went away in two
different directions. What is the angle between their final momentum?
Assume no friction.
 Pattern
When an object collides elastically with another stationary object that has equal mass,
the angle between their final velocities is 90 degrees.

Topic 6 - Ballistic Pendulum

A ballistic pendulum consists of a rigid beam connecting to a chamber that will catch the
projectile and move with it.
○​ A common assumption is the rigid beam has no mass.
○​ Kinetic energy is NOT conserved in a ballistic pendulum system.
。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。
* finish the Tarzan question on the testbook before doing this part.

Calculate what percent of kinetic energy is left after the collision.

●​ In order to calculate the percent of kinetic energy, we need to find the total
kinetic energy before and after the collision.
○​ Before the collision, only the projectile has kinetic energy;
○​ After the collision, both projectile and catcher have kinetic energy.

Moment 1 System Moment 2

Right before collision Projectile; catcher Right after collision
●​ We CANNOT use energy conservation for this part because some kinetic energy is
converted to the internal energy. The change of internal energy cannot be
 measured, nor is there an equation for Uint. But momentum is not affected by this
and is still conserved.

●​ This is an inelastic collision because the projectile and the catcher stick together
after the collision, therefore:
𝑚𝑃𝑣𝑃,1 + 𝑚𝐶𝑣𝐶,1 + 𝐽 = (𝑚𝑃 + 𝑚𝐶)𝑣2
○​ Initial momentum of the catcher is zero as it was not moving.
𝑡𝑓

○​ Impulse ≈ 0 because 𝐽 = ∫ 𝐹𝑒𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑑𝑡 and ∆𝑡 is very small. 𝐹𝑒𝑥𝑡𝑒𝑟𝑛𝑎𝑙 is the force

𝑡𝑖

the beam exerts on the catcher, which is NOT zero.

𝑚𝑃𝑣𝑃,1 + 0 + 0 = (𝑚𝑃 + 𝑚𝐶)𝑣2
●​ There are two unknowns in this equation: 𝑣𝑃,1 and 𝑣2. We cannot go further with just
this equation. So we will have to consider the next moment.

Moment 2 System Moment 3

Right after collision Projectile; catcher The catcher + projectile is at
the highest position
●​ For this part we would use energy conservation because there is no shape
change/temperature change/sound or light emitted.
𝐾𝑝+𝑐,2 + 𝑊 = 𝑈𝑔,𝑝+𝑐,3
𝑥𝑓

○​ Work = 0 because 𝑊 = ∫ ||𝐹(𝑥)|| 𝑑𝑥 (𝑐𝑜𝑠α) (dot product) and ⍺= 90°. The

𝑥𝑖

external force is the force beam on the catcher, which is NOT zero.
1 2
2
(𝑚𝑃 + 𝑚𝐶)𝑣2 + 0 = (𝑚𝑃 + 𝑚𝐶)𝑔𝑦3
1 2
2
𝑣2 = 𝑔𝑦3

𝑣2 = 2𝑔𝑦3
●​ Substitute v2 back to the momentum equation:
𝑚𝑃𝑣𝑃,1 = (𝑚𝑃 + 𝑚𝐶)𝑣2
𝑚𝑃+𝑚𝐶
𝑣𝑃,1 = 𝑚𝑃
2𝑔𝑦3
●​ Find total kinetic energy before and after collision:
 1 2 1 2
Σ𝐾𝑏𝑒𝑓𝑜𝑟𝑒 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛 = 𝑚𝑃𝑣𝑃,1 Σ𝐾𝑎𝑓𝑡𝑒𝑟 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛 = 2
(𝑚𝑃 + 𝑚𝐶)𝑣2
2

𝑚𝑃+𝑚𝐶 2 1
Σ𝐾𝑏𝑒𝑓𝑜𝑟𝑒 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛 =
1
𝑚𝑃( 2𝑔𝑦3) Σ𝐾𝑎𝑓𝑡𝑒𝑟 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛 = 2
(𝑚𝑃 + 𝑚𝐶)(2𝑔𝑦3)
2 𝑚𝑃

(𝑚𝑃+𝑚𝐶)
2 Σ𝐾𝑎𝑓𝑡𝑒𝑟 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛 = (𝑚𝑃 + 𝑚𝐶)(𝑔𝑦3)
Σ𝐾𝑏𝑒𝑓𝑜𝑟𝑒 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛 = 𝑚𝑃
(𝑔𝑦3)

Σ𝐾𝑎𝑓𝑡𝑒𝑟 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛 (𝑚𝑃+𝑚𝐶)(𝑔𝑦3)

% 𝐾𝐸𝑙𝑒𝑓𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 = Σ𝐾𝑏𝑒𝑓𝑜𝑟𝑒 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛
= (𝑚𝑃+𝑚𝐶)
2

𝑚𝑃
(𝑔𝑦3)

𝑚𝑃
% 𝐾𝐸𝑙𝑒𝑓𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 = 𝑚𝑃+𝑚𝐶