April 6th, 2011

Key: x^2 = x squared; Vo = V nott (aka initial velocity); * = answer

Well, we have a quiz tomorrow and we reviewed for that. I just realized that I'm posting a little late for you to look at this and learn more from it. Woops. Not that I'm going to be much help anyways. Anywho, let's get to it.

We basically went over a worksheet that looks strikingly similar to this one that I found on google images on the first try:
a. So we've gone over this twenty million times, but in case your memory is a bit foggy, the marbles motion is parabolic because the ball's horizontal velocity, due to negligible friction and inertia, will be constant. However, after it leaves the table, since the only vertical force acting on it is gravity, it will accelerate towards the floor at 9.8m/s/s.

b.
While it is on the rail, we know that the ball is moving horizontally at a constant velocity of 10 m/s. This means no unbalanced forces, so the force diagram will have two arrows: an up arrow for the normal force the rail is exerting on the ball, and a down arrow for gravity. These two arrows are equal in length because I'm pretty sure the ball isn't accelerating upwards or downwards anytime soon.
When it leaves the table, since we are ignoring air resistance and inertia is not a force, there will only be one down arrow on your force diagram for gravity. (This computer doesn't have paint or anything similar to it, so no pretty drawings today :( ) It's horizontal movement is still constant because an outside force hasn't prevented the ball's inertia to keep it moving forward. As for vertical, it is accelerating downward.

c. Once the ball leaves the table, calculate how long it will take for the ball to hit the floor.
Okay, so here we use the equation: change in position = 1/2at^2+Vot
We are only talking about vertical movement. We know that the table is 1.5m tall, it's acceleration is 9.8 (gravity), and that it's initial velocity is 0, because the ball wasn't moving vertically at all before it left the table. So we plug this information in.

• 1.5m = 1/2(9.8m/s/s)t^2 + 0t (Then divide by 4.9 to get t^2 alone.)
• t^2 = 3.06 (Find the square root of both sides)
  
• t = .55s*

d. In the time you have calculated in part c, how far will the ball travel horizontally before hitting the floor?
Alright, so now we're talking about horizontal movement, but now we have t, which we found in the last problem. We also know that initial velocity is 10m/s, because that was the velocity left the table, and will always be the horizontal velocity until an outside force acts on it. Also, since velocity is constant, we know that acceleration is 0. So plug that in to our equation again:

• changeX=1/2(0)(3.06) + 10*.55s
  
• X = 5.5m*

e. Suppose the table was doubled in height to 3.0m.
i. Determine the horizontal range of the marble as it falls to the floor.
ii. What effect does doubling the height have on the range of the marble?
iii. What other factors affect the range of the sphere?

i. Okay, this is basically combining c and d, except we have a new vertical height. We need to find how long it takes for the marble to hit the floor so that we can plug that time into another equation to see where, in regards to horizontal direction, the marble will be.
So, once again, thinking about vertical movement, we plug in our facts into the equation. (See c for more explanation if necessary.)

• 3m = 1/2(9.8)t^2 = 0t
• t^2 = .612
• t = .782s

So take that time and plug it into our horizontal direction equation

• changeX = 1/2(0)t^2 + 10*.782
• x = 7.82m*

ii. As we can see, doubling the height increased the horizontal range.
iii. Other factors that could affect the range of the sphere:

• Horizontal velocity, because, given the same amount of time it takes to hit the floor, the marble would move faster to go farther or slower to come up short.
• Increased friction or air resistance, since it would slow the ball down and change its range
• *** One thing that would NOT change it's horizontal range would be the weight or mass of the ball, as long as the velocity was still the same because, as we've seen and tested ourselves, if you drop two objects varying in weight at the same height, they will always land at the same time, meaning none of of the variables have changed, producing the same horizontal range.

So yep. That's all. I would do more, like go to the back of the worksheet, but I'm lazy sick.

-Norman


SCIENCE!!!!

Angeline's Review, April 5

Hey guys! So today we continued discussing horizontally launched projectiles. Remember that if a projectile is horizontally launched at the same time an object is dropped, these two objects will hit the ground at the same time. This is because no matter what, gravity is still acting on the objects with an acceleration of 9.8 m/s/s. Even though the projectile is also moving forward, gravity is still acting on it. I'll post the answers to the problems we did in class later. I thought that this was at least a start :). Sorry the print above is really small. If you can't read it, the answer is .63 m.

Wednesday, March 23, 2011

Today, we reviewed. That's it.

Am I done? Great. I've got other things to do.

We went over the, ahem...

Unbalanced force Particle Model:
Newton's Second Law Review Problems

...worksheet.
Using a new, groundbreaking technique where the students put the answers on the blackboard, we answered problems 3, 4, and 5 in class:

3.

So, our baseball player can throw at 100 miles per hour (mi/hr) by extending his arm a distance of 1.5 meters (m). If his ball weighs 0.15 kilograms (kg), what is the unbalanced Force he must exert on the ball?

We will use this equation to find unbalanced Forces:



I hope you enjoy my childish scrawl.

M meaning our mass (which in this case is 0.15 kg) and A meaning our acceleration.

But, we need to find our A, which we measure in m/s/s.

However, you'll notice that we have an issue with measurements. We'll use measurements in m and seconds (s), but the question starts out with a measurement in mi/hr.

Homey don't play that.


None of you will really get this reference.

Which brings me to why I left Chemistry class: Dimensional Analysis.
This horrid mathematical device was the general tool in converting the terrible mi/hr to our darling meters per second (m/s).


Just in case you've forgotten how to do this, I'll try to explain this in a way that won't make me have another panic attack.
So, we set up 100 mi/hr in the upper left. According to our sheet, 1 mi is about 1600 m. So, we set the ratio (1600 m/1 mi) up next to it. Mi goes on the bottom because it's sort of like multiplying fractions, and the old measurements are supposed to cancel. We need to get rid of that measurement to convert our given.
So, we can successfully convert our mi to m by multiplying 100 and 1600. But, now we need s.
Because hr is below on the left, it has be be above on the right, again so it can cancel out. below it goes 60 minutes (min). The ratio is (1 hr/60 min). But, we want s, so we add another ratio (1 min/60 s), so we can convert our measurement to seconds.

I might have made this more confusing.



Multiply 100 by 1600. Then divide it by 60 twice. You about 44.4 m/s.

We'll use that to find the acceleration with this equation:


Velocity (V) squared equals Initial Velocity (Vo) squared plus 2 times A times change in position (ΔX).
Our V is 44.4 m/s. Our Vo is assumed to be zero. Our ΔX is 1.5 m minus zero m, which is 1.5 m.


So, with our acceleration, now we go back to our first equation and plug in our numbers:


98.6 Newtons (N) are exerted on the ball to throw it at 100 mi/hr.

4.

After staring from rest, a 0.50 kg ball falls 30 m. It has contact with the ground for 2 milliseconds (.002 s) before bouncing back up and reaching 20 m. We need to find the force exerted on the ball as it hit the ground.

Because we have a mass again, we'll use our Funbal = MA equation.
We need to find the acceleration as it hit the ground, which would be between our two velocities.


Oh my God, NO!!

Yes, two velocities. There is when it is falling initially and after it bounces on the ground. Luckily, we can use the same equation from before:



Or, at least I hope it is.

So, I put 9.8 in for A.

"But then, we have acceleration! Why do we need to figure out what it is if we have it?"

Well, very bold student, we have acceleration, while the ball is in the air, while the acceleration due to the force of gravity acts upon it, not when it hits the ground.

So, we get 24.25 m/s from the first one and 19.8 m/s from the second. We take these two and stick them into another equation:


Good ol' change in Velocity (ΔX) over change in Time (ΔT).
If we consider the bounce upwards to be "moving forward," then the ball dropping would be considered "moving backwards." That's why the velocity is negative now.
Doing this equation leads us to the ungodly acceleration of 22,025 m/s/s.

Plugging our numbers back into our Funbal equation:


According to maths, our answer is 11,012.5 N.

5.

Another sled problem. That's all I should have to say.

There's a pair of kids in a sled. One of them is having too much fun. With the sled, their mass is 100 kg. A man with no muscle mass is pulling the sled by way of a rope at a 25° angle. The tension on the rope is 120 N. The friction force is 15 N. We have to find the acceleration.

We can use the same formula we have been using, but this time we have to change it to solve for A instead:


We have a mass (100 kg), so we just need our unbalanced Force. We make a Force diagram:


X is our force forward, which we need to find for the unbalanced force.
The reason the normal force doesn't equal the force of gravity is because of that 50 N over on the right. The rope is pulling the sled to the right and up, to the ground isn't pushing up against the sled as much as it would is it were at rest.

To find X, we use Trigonometry:


We subtract 15 N from 108.76 N and get 93.76 N, our unbalanced force.
We plug it into our adapted Funbal equation:


The acceleration is .94 m/s/s, the child was having way too much fun, and this entry is finished.

IM FINALLY BLOGGING!

okay so sorry its taken so long. Anyways..

3/16

Mr. Dickie put multiple practice problems on the board that I payed very close attention to just for you guys (and my grade:/)

#1 the guy up there^ pulling the car (picture it with wheels). mass of car= 1.5kg; person pulling with a force of 10N. find acceleration.

a= Fun/mass SO a=10N/1.5kg and you get 6.67 m/s^2 as the acceleration

#2 okedoke so the picture with the finger pushing the block is next. needed the acceleration. Given: finger pushing with 10N; mass of block=1.5kg. oh and theres M (mew thing)=0.15

first we need to find friction. so we do Force of Friction= (Mlittlek)(Normal Force)

=(0.15)(15N) to get frictional force of 2.25 N

Next we need the acceleration so we take the Force of the pushing (10N) and subtract the force of friction (2.25) to get 7.75N. Then divide that by the mass (1.5kg) and we get 5.17m/s^2 as the acceleration.

#3 so there is a pulley system. the car on top is 1 kg and the weight pulling it is 0.2kg. to find the acceleration we take the Fun/mass.

a=2N/1.2kg equalssssss 1.67 m/s^2

#4 ELEVATOR PROBLEMMM

So Mr. Dickie, with a mass of 78kg is in an elevator that accelerates at 3m/s^2 when it starts moving up. **this time we are given the acceleration and want the Fun!

Fun= (a)(m)

Fun=(3m/s^2)(78kg) = 234N

You may think you are done but wait! you have to add 234 and 780 to get your grand total of 1014N as your normal force !

#5 Mr Dickie is in an elevator moving up at 2.7m/s^2. the elevator slows down and a rate of 3m/s^2. oh and he still has the same mass of 78kg.

So to do this problem we dont even need the 2.7 m/s/s part. ignore that. We just find the force unbalanced again which is still 234N and this time instead of adding that to 780 we SUBTRACT it from 780 to get 546 as the normal force

3/21

So we basically just worked on review sheet for our TEST ON THURSDAYY:O

that is all.

Jon Sheppard's Class Review for Tuesday March 22, 3011

Okay, so today be basically did the first 2 problems on the "Newton's Second Law Review Problems.

#1. An 80 kg water skier is being pulled by a boat with a force of 220 N causing the skier to accelerate at 1.8 meters per second per second. Find the drag force on the skier.

so the normal force and the gravitational forces are each 800 N(up and down arrows), and the force of tension of the rope on the skier is 220 N. To find the drag force, use the funbal=(a)(m)-1.8 X 80 = 144=funbal. So the drag force is 220-144=76 N.

#2. A 2000 kg car is slowed down uniformly from 20m/s to 5m/s in 4 seconds. Determine the average unbalanced force on the car during this time, and how far the car traveled while slowing down.

Well, to find the Funbal, you first have to find the acceleration, which is V=at+vo-5=a(4)+20=-3.75m/s/s. Then funbal=(a)(m)-(3.75)(2000)=-7000 N. Then the distance covered is Chang in X=1/2(at)^2 - -3.75 X 4= 15^2= 225m.

Thank you for your time and cooperation, i really appreciate it ;)

Love,

yours truly, Jon Sheppard

Accelerometer Project

Paulina and I didn't realize that we had to put this on the blog, hence why it's about 43789462839 days late. Our sincerest apologies.

Tuesday, March 15

Today in physics we reviewed unbalanced forces. We started with this problem:Using this information, we had to find the acceleration. First, we made a force diagram for both the weight and the cart. The force diagram for the cart looked like this:

and the force diagram for the weight looked like this:

To find the acceleration we used the formula acceleration=Funbalanced/mass. The unbalanced force is 1 N, and the mass is 0.6 kg, so the equation would look like: acceleration=1 N/0.6 kg. When you solve the equation, the acceleration is 1.67 m/s/s.