Section #3: Momentum, Collisions, Heat

Question 1

What is the Equation to Center of Mass (C.O.M)

Answer 1

𝓧꜀ₒₘ = 1/M (Σmᵢ𝓧ᵢ)

Where:
M = Sum of masses
𝓧 = distance from origin

Question 2

What is the equation to Center of Mass of a structure with area ?

Answer 2

𝓧꜀ₒₘ = 1/Ａ • (𝓪ᵢ𝓧ᵢ)

Where :
Ａ = Sum of Areas
𝓧ᵢ = Centre of Mass for each area

Question 3

What is the equation to the Center of Mass of system of particles with velocity ?

Answer 3

𝓥꜀ₒₘ = 1/M (Σmᵢ𝓥ᵢ)

Where:
M = Sum of masses
𝓥ᵢ = Velocity of each particle

Question 4

What is the equation to Impulse & how is it derived ?

Answer 4

𝐉 = 𝘱₁ - 𝘱₀ = ∫ F(t) dt

This come from dividing the change of time (Δt) into our Momentum Function:

Δ𝘱 = mΔv → Δ𝘱/Δt = m(Δv/Δt)

Since (Δv/Δt) = Acceleration and ma = F , we not have Δ𝘱 = F/Δt

Convert deltas to differentials and take the intergral of both sides:

𝐉 = ∫ d𝘱 = ∫ F(t)dt

Question 5

Set up the equation for this problem:

A bullet is shot at Velocity: 𝓥₀ at Angle: θ₀. Once the bullet reaches maximum height, It explodes into two equally massed pieces. One piece is going at falls vertically and the other continues onward. How do we find how far the other piece travels ?

Answer 5

Start by finding the time of the bullet’s maximum vertical position (𝓥=0) using the velocity kinematics formula:

𝓥² = 𝓥₀² - 2(9.81m/s²)Δt
(Solving for t₁)

Then find when the bullet is expected to reach the ground using

Δy = 0 = 𝓥₀sinθ₀t - ½(9.81m/s²)t²
(Solving for t₂)

We can then use 𝘱₀ = 𝘱₁ relation to solve for the ending velocity of the second bullet:

m𝓥₀𝓍 = ½m₁𝓥₁𝓍 + ½m𝓥₂𝓍

With our 𝓥₀𝓍 and 𝓥₂𝓍 along with t₁ & t₂ we can now solve for our final distance:

d = 𝓥₀𝓍(t₁) + 𝓥₂𝓍(t₂)

Question 6

If a Bullet of mass = m₁ is shot into a Block of mass = m₂ causing the block to raise a height of h after sticking into the block.

How can initial velocity of our Bullet (𝓥₀) be found?

Answer 6

Start by acknowledging that energy is not lost within the system so we can use our Conservative energy function to find our final velocity (𝓥₁) :

½m𝓥₁² = maΔd. →
½(m₁ + m₂)𝓥₁² = (m₁ + m₂)gh
𝓥₁ = √(2gh)

Manipulate our Inelastic Collision function to solve for 𝓥₀:

𝓥₁ = (m₁𝓥₀) / (m₁ + m₂) →
𝓥₀ = (m₁ + m₂)𝓥₁ / (m₁) →

𝓥₀ = (m₁ + m₂) ⋅ √(2gh) / (m₁)

Question 7

Explain how to solve:

An unbalanced force on an object of m = 10kg increases from 0 - 50N in 4.0s causing the object initially at rest to move. What is the object’s speed at the 4.0s mark ?

Answer 7

Start by noting that 𝐉 = ∫ F(t) dt and the slope of F(t) follows formula: mt +b

if m = ΔF/Δt is the slope of F(t) = (50N/4s)t

Now take the integral of F(t) → ∫ (50N/4s)t dt = (12.5kg⋅m/s³)(t²/2) @ t = [0,4]

Set the Momentum Conservation Functions equal to each other:

m𝓥₁ - m𝓥₀ = ∫ F(t) dt → (10kg)𝓥₁ = (12.5kg⋅m/s³)(4²/2) → 𝓥₁ = (100kg⋅m/s) / (10kg)

𝓥₁ = 10m/s