MTM Initial Challenge

Purpose: Find the relationship between the masses and velocities of the carts.
Each cart weighs 500g and each bar weighs 500g
Materials

• bar
• 2 carts
• calculator
• a clicker to separate the carts

Part A
1:1 40:40
1:2 54:26
1:3 60:20
1:4 65:15
2:3 48:32

Part B
1:1
.5:2.07
.33:3
.25:4.3
.66:1.5

We noticed that as the mass increases in one cart, the velocity on the other cart would move twice as slow, three times as slow, and then 4x as slow. The mass and velocity are directly, inversely proportional.

Rocket Lab

If you shot the rocket at a different angle where would it land?

Purpose

The purpose of this experiment was to accurately predict the distance a rocket would travel at any given angle. In order to make predictions, we used equations we learned in class. 

Materials

• rocket
• nails
• pressure cap
• angle blocks
• ball pump
• measurer
• goggles

horizontal equations

v=∆x/∆t
xf=vt+xi

vertical equations

a=∆v/∆t
∆x=1/2at^2+vit
vf=at+vi

a=-9.8m/s

v=∆x/∆t

v=29.9/4

v=7.475 m/s

cos(65)=7.475/x

vi=17.687 m/s

cos(45)=adj/17.687

vix=12.51 m/s

cos(55)=adj/17.687
cos(35)=adj/17.687
∆x=1/2at^2+vit
40.1=1/2(-9.8)t^2+17.687t
=-4.9t^2+17.687t-40.1
-b+/-√(b^2-4ac)/2a
-17.687+/-√(17.687^2-4(-4.9)(-40.1)/-9.8
-17.687+/-√-473.13
-17.687+/-21.75/-9.8
-39.4/-9.8
t=4.02s

Rocket landed @ 40.1 at 55 degree

Rocket landed @ 34.8m at 65 degree

Time in air for 55 degrees: 4.1s
percent error for length: 28%
percent error for time: 1%

Unit 5

This unit was a culmination of information we have learned from previous units, but this time we included horizontal vs. vertical directions. 
Horizontal equations
v=∆x/∆t
xf=vt+xi
models: CVPM, BFM
horizontal velocity=vx

Vertical equations
a=∆v/∆t
∆x=1/2at^2+vit
vf=at+vi
models: CAPM, UFPM
vertical velocity=vy

Meta Sheet

• If an object is moving downwards, the velocity and acceleration will be negative
• key words: upwards= positive and downwards=negative
• acceleration at the top of vertical= -10m/s^2 and velocity at top= 0m/s
• acceleration at the top of horizontal=0m/s^2 and velocity at top=constant
• for two dimensional maps, x-axis is always straight
• if problem gives you vi at an angle, you cannot use it for vertical & horizontal
• do NOT plug horizontal distances into vertical problem equations 
• remember in the quadratic equation: take square root and then solve +/- part

Example Problems 

A gum ball is thrown downward with an initial speed of 20m/s. Find the displacement of the gum ball after 4 seconds. Next, find the time it takes to reach a velocity of 50m/s and how long it would take to fall 100m.
Vertical
viy=-20m/s (because it's falling down)**
a=-10m/s^2

Displacement
∆x=1/2at^2+vit
∆x=1/2(-10)(4)^2+(-20)(4)
∆x=-80-80
∆x=160m

Time it takes to reach a velocity of -50m/s
vf=at+vi
-50=(-10)t+(-20)
-30=-10t
t=3s
**remember to change Vi, Vf, and a to negative if it's falling down or else you will get a completely different answer**

Time it takes to fall 100m **Negative 100m because it's falling**
∆x=1/2at^2+vit
-100=1/2(-10)t^2+(-20)t
0=-5t^2+(-20)t +100
-5=a -20=b 100=c
Quadratic equation time!