Chapter+2

= =

= = toc =Section 1=

Calculating Velocities for Different frames of Reference
11/22/2010 Trigonometry: SOH CAH TOA SinTheta = opposite/hypotenuse Theta = Sin^1(opposite/hypotenuse)

CosTheta = Adjacent/hypotenuse Theta = Cos^1(adjacent/hypotenuse)

TanTheta = opposite/adjacent Theta = Tan^1(opposite/adjacent)

http://www.sumopaint.com/home/

1a.

Physics To Go: Velocity
11/23/2010 1a. The ball will roll forever, for as long as there is no force acting upon it 1b. Newton's first law of motion states that every object is in a state of rest unless it is acted upon by an external force 2. The ball will reach 20 cm 3. No because there is gravity and friction everywhere, unless the environment is in space, a zero-g vacuum environment. 4. The puck will accelerate until it hits its intended target, then will slow down dramatically 5. The person sees the ball traveling at 7 m/s 6. The velocity of the javelin relative to the ground is 14.5 m/s 7a. The relative velocity of the cart moving forward is 8m/s 7b. The velocity is 3.2m/s 7c. The cart is going 5.6 m/s relative to the ground moving forward, but it is 2.4m/s moving to each side going perpendicular 8. The speed of the arrow as it leaves the horse is 103m/s 9a. 21.21 feet 9b. 43.857 feet 9c. 57.955 feet 9d. 172.105 feet 10a. Newton's first law in sports can be seen in hockey, baseball, and football. The object is at rest in baseball when the ball is in the pitcher's glove, and the object is in motion when the pitcher throws the ball. In hockey, the puck is not in motion when it is first dropped and stays on the ice for a couple of milliseconds, the puck is in motion, however, the rest of the game. In football, if you can say a player is an object, the player does not move at the beginning of the game until the quarter back calls the ball, and the player(object) is in motion after the qb calls it. 10b.

= = =Section 2=

Investigate: Measuring Motion
1a. The dots are evenly spaced 1b. The dots are further apart and evenly spaced. 1c. The dots are closer together and evenly spaced. 1d. The dots will start closer together and get further and further from each other, becoming unevenly spaced. 7a. The paper was moving through the machine at a constant speed, so the dots must be evenly spaced because every dot has been imprinted at certain time intervals. 7b. Yes, the speed was constant because there is about the same amount of space between each pair of dots.

8a. 8b. The student's speed was accelerating 8c. Yes, the trend is exponential since the space between each pair keeps getting larger and larger

9a. The average distance in strip length is about 1 cm, due to the more dots being closer together, then the dots further apart. 9b. The change in lengths are vaguely constant, so the student did somewhat move with a constant acceleration

10a. 10b. The pattern on the graph for decreasing speed is the opposite of the pattern of the graph for increasing speed. The dots are closer together at the beginning of a graph increasing speed, but they are closer together at the end of a graph decreasing speed. 10c. The acceleration is once, again, -1cm.

11a. The graph of the person increasing speed is +1cm, the graph of the person decreasing speed is -1cm.

Physics Plus: Large Accelerations at Slow Velocity
12/2/2010 1. (-0.5 - 0.5)/0.01 = -100 m/s^2. Negative means that the direction is being changed. 100m/s^2 means that the force is equal to 10 g's(9.8m/s^2). The fastest roller coaster is probably around 4g's. Fighter pilots experience 10g's of force. 2. (-0.5 - 0.5)/0.20 = -5m/s^2. About half a g. 3a. -5m/s^2 3b. -5m/s^2 3d. -5m/s^2

Physics Talk: Acceleration
12/2/2010 Acceleration is a change in the velocity of an object over time. Average speed is the distance traveled divided by the time it took to travel that distance. Instantaneous speed is the speed measred during an instant. The speed as the time interval approaches, but does not become zero.

Checking Up: Acceleration
12/2/2010 1) a Constant Speed they are evenly spaced out. b. Positive Acceleration the distance between the tickers gradually increase c. Negative Acceleration Distance between tickers decrease gradually 2) 400m/50s = 8m/s 3. The speed at a certain instant is instantaneous speed, and average speed is the speed of the entire distance covered. 4. A = (100-0)/10 A = 10m/s

Physics to Go: Acceleration
12/2/2010 1) Instantaneous speed - speed at a certain instant Average speed - Speed of the entire distance  2) a. V = 1km/15s V = .06m/s b. V = 84m/6s V = 14m/s c. V = 9.6km/2h V = 4.8km/h d. V = 400km/4.5h V = 89km/h 3) a. Positive Acceleration b. Positive Acceleration  c. No Acceleration  d. Negative Acceleration  e. No Acceleration  f. No Acceleration  4) a.A,D b. B c. A  d. C  5. a. 7 b. 10.5 c. 11.6 d.12.25 e.12.6 f. 12.8 g. 13 h. 13.125 i. 13.2 j. 13.3 k. 13.36 l. 13.5 6) a. A = (0-45km/h)/9s A = -5 m/s b. negative value  7) a. constant velocity b. speeding up c. speeding up, slowing back down d. slowing down, speeding back up 8) V = 100mi/2h V = 50mi/h  9) No this is the average speed over the entire distance 10) 11. 20m/s  12. Yes, they can move 10m

=Section 3=

What Do You Think?
11/29/2010 A force is a push or pull that acts on an object. The force will affect a bowling ball differently because the weight of the bowling ball is much greater than that of a tennis ball. The more weight an object has, the more force it takes to push it. The tennis ball will accelerate much faster than the bowling ball.

Investigate: Newton's Third
11/29/2010 2a. The car just keeps accelerating

3a. The cart speeds up even though the same amount of force is applied 3b. The smaller bend means that there is less force exerted and that means the acceleration is less than that of a ruler with a large bend (more force exerted) 3c. The car speeds up faster if the bend is larger 3d. The greater the constant force pushing on an object, the greater the acceleration of the object

4a. The greater the mass of the object, the slower it accelerates (if the force is the same as the force being applied to a smaller-massed object)

5a. 5b. When equals amounts of a constant force are used to push objects that have different masses, the more massive object accelerates slowly
 * 200g cart || 500g cart || 1000g cart ||
 * 3.84 seconds || 5.68 seconds || 6.6 seconds ||

6a. I would have drawn the conclusion that the large objects needed more force to be able to go at the same pace as the smaller object.

8a. The ruler bends more with the coin

9a. The ruler bends more and more as there are more coins taped to it 9b. 2 pennies represent a small force, 5 pennies represent a large force 9c. The mass of the pennies, and gravity.

Physics Talk: Newton's Second
12/3/2010 The more massive an object, the more force it takes to accelerate it. Newton's second law of motion is: the acceleration of an object is directly proportional to the unbalanced force acting on it and is inversely proportional to the object's mass. Newton's second law can be written as an equation: acceleration = force/mass. There are a lot of different everyday forces. Newton's second tells you that accelerations are caused by unbalanced forces. Force makes lighter objects accelerate a great deal faster than heavier objects. The acceleration due to gravity is 9.8m/s^2. Weight is the force of gravity acting on an object, and it depends on the mass of the object and the acceleration due to gravity. You can calculate the weight of an object by using F(of gravity) = m*a(of gravity). The mass is in kilograms, and the g is the acceleration due to gravity (9.8). When two forces act at the same time, the direction as well as the magnitude of the forces determine the motion of the object. If the forces are in the same direction, then the sum of the forces will cause a larger acceleration than either force alone. Likewise, if they are in opposite directions, then the net force could be zero and there would be no acceleration. A free-body diagram can determine the net force. A free body digram is used to show the relative size and direction of all forces acting on an object.

Checking Up: Newton's Second
12/3/2010 1. The acceleration of an object is determined by its weight and the amount of force acting on it 2. The effect of increasing the weight of an object is: the object will accelerate at a slower speed than the lighter object 3. 1 N basically = 1 KG, plus the amount of force already acting on the object, which is 1g. So you multiply the 30n x 9.8m/s^2 to find the weight of the object. 4. Weight would increase, mass would increase

Physics To Go: Newton's Laws
12/3/2010 1. 2a. Long jumper in flight and shot put ball in flight because both of the objects are in the air 2b. Yes, the acceleration is negative. If the force is negative, so is the acceleration 2c. Yes because the running back is running in one direction, and the person tackling the running back is moving in the other direction with greater force. 3. F = ma. The force is 42 = 0.3kg*a = 140 m/s^2 4. F = 0.04*20 = 0.8N 5a. Catching and throwing a baseball will look like the objects are not really moving 5b. The bowling ball will accelerate slower than the baseball, and the force of catching the bowling ball will be greater then the force of catching a baseball 6. F = 0.1(9.8m/s^2) = 0.98N 7a. F = (150)(4.38) = 657N 7b. 657N = m(9.8m/s^2) m = 67.04 kg 8. The tug of war would basically see which side is stronger than the other. If the rope goes more towards one side, that side is exerting more force than the other side. 9. The force of my hand will always act on the ball until the ball comes to a complete stop, such as another force acting on it. 10. The net force of the object is 90N 11. The combined force is 800N 12. 125N = .7kg(a) a = 178.57m/s^2
 * Newton's Second Law || F= || m* || a ||
 * Sprinter beginning 100-m dash || 350N || 70kg || 5m/s^2 ||
 * long jumper in flight || 800N || 80kg || 10m/s^2 ||
 * shot-put ball in flight || 70N || 7kg || 10m/s^2 ||
 * ski jumper going downhill before jumping || 400N || 80kg || 5m/s^2 ||
 * hockey player "shaving ice" while stopping || -1500N || 100kg || -15m/s^2 ||
 * running back being tackled || -3000N || 100kg || -30m/s^2 ||

=Section 4=

Part A
11/8/2010 1a. Yes, they both hit the floor at the same time. 2a. The coins sound like they are both hitting the floor at the same time. 3a. The speed does not affect whether or not the coins hit the floor at the same time, because they always do. 3b. Yes, the distance between the two coins increases as the speed of the coin being flicked increases 3c.

4a. The height of the coins does not affect whether or not they hit the floor at the same time 4b. The distance the horizontal coin traveled was greater at a bigger height than at a smaller height.

Part B
1a. The ball creates an arc, when it is thrown up in the air, the ball moves diagonally since there is a force pushing it both up and forward. 2a. The faster the chair traveled, the bigger the arc the ball created in the air. The ball, obviously, went diagonally and fell diagonally, but when it was at its apex, it had to slow down a little before it came back down, creating a rounded top. 2b. How fast the ball was launched and how fast the person was going affect the range of the ball.

Active Physics Plus: Velocities
12/9/2010 1a. Vy = 30m/s, Vx = 10m/s. Vx does not change. 1b.
 * t || Vy ||
 * 0 || 30 ||
 * 1 || 20 ||
 * 2 || 10 ||
 * 3 || 0 ||
 * 4 || -10 ||
 * 5 || -20 ||
 * 6 || -30 ||

2a. Vertical = 25m/s, Horizontal = 43.3m/s 2b. Horizontal = 43.3m/s, Vertical = 15.2m/s 2c. Horizontal = 43.3m/2, Vertical = -24m/s 2d. 50 m/s.

Physics To Go: Motions
12/9/2010 1. Lab Diagrams up top 2. Lab Diagram 3.

9a. 12 9b. 24 10a. 8.5 10b. 4.25

=Section 5=

What Do You See?
12/13/2010 There's a soccer game going on, where the ball makes a parabola hitting another players head which in turn makes another parabola that goes into the goal.

What Do You Think?
12/13/2010 1. The bigger the angle the object is thrown, the smaller the parabola is. The smaller the angle, the bigger the parabola it makes is. 2. The greater the force, the greater the distance. If an object is launched at a higher velocity, it will obviously travel more, and therefore will increase the range of the ball.

Investigate: Projectiles
12/14/2010 Experiment 5: Acceleration due to Gravity Average Velocity = distance/time
 * Time || Distance || Speed || Acceleration ||
 * 0.000000 || 0.0000 || 0 || 0 ||
 * 0.053508 || 0.0500 || 0 || 0 ||
 * 0.089924 || 0.1000 || 1.373022 || 0 ||
 * 0.119426 || 0.1500 || 1.694800 || 10.90698 ||
 * 0.144898 || 0.2000 || 1.962939 || 10.52681 ||
 * 0.167650 || 0.2500 || 2.197609 || 10.31425 ||
 * 0.188406 || 0.3000 || 2.408941 || 10.18173 ||
 * 0.207604 || 0.3500 || 2.604437 || 10.18314 ||

V average = (d2-d1)/(t2-t1) V = (.05 - 0) / (0.053508 - 0) V average = (d3-d2)/(t3-t2) V = (0.1 - 0.05) / (0.89924 - 0.053508)

Acceleration = v2 - v1 / t2-t1

Physics Plus pg. 190
12/16/2010 A ball rolls off a table and lands on the floor.

Description of Motion in the Vertical Direction: As soon as the ball leaves the table it begins to fall downward a distance of "y" at a rate of 9.8 m/s^2 starting from rest. (Vyi = 0). The time it takes to reach the floor is "t".

Vertical Variables: y = Distance, Vyi = Velocity Initial a = Acceleration t = Time Vyf = Velocity Final

Vertical Equations: y = 1/2(gt)^2 + Vyi(t) g = (Vyf - Vyi) / t y = 1/2(Vyi + Vyf) t Vyf^2 = Vyi^2 + 2gy V = d/t does not apply in the vertical because a is not equal to 0, there is a change in speed

Description of Motion in the Horizontal Direction: The ball moves at a constant speed as it moves a distance of "x". It moves horizontally at a time of "t", same as in the vertical direction. a = 0. Vxf = Vxi.

Horizontal Variables: x = Distance t = time Vx = Velocity

Horizontal Equations: Vx = x/t

Examples: 1. A ball is pushed such that it rolls across a table at a constant speed of 5.0m/s. The ball rolls off the table, which is 75 cm tall. The ball lands on the floor. Organize the given info. Calculate "t". -0.75 m = 1/2(-9.8)t^2 + 0 t = 0.39 seconds
 * Variable || x || y ||
 * d || ? __= 1.95 meters__ || -75 cm = -0.75 m ||
 * t || ? __= 0.39 seconds__ || ? __= 0.39 seconds__ ||
 * Vi || 5.0m/s || 0m/s ||
 * Vf || 5.0m/s || ? __= -3.822 m/s__ ||
 * a || 0 || -9.8m/s ||

Vx = x/t x = Vxt = 1.95 meters

a = (Vyf - Vyi)/t -9.8 = (Vyf - 0) /0.39 seconds Vyf = -3.822 m/s

2. What is the final velocity of the ball?(Not Vyf or Vxf, but Vf). Use Pythagorean Theorem

Vertical and Horizontal Problems
12/17/2010

Sample Problem
Given: y distance = 1.6m x speed = 6.0m/s Solve for t: 1.6 = 1/2(9.8)t^2 t = 0.57 seconds

Solve for x distance: V = d/t d = vt d = (6.0)(.57) d = 3.42 m

Physics #1-2 pg 192
1. Given: y distance = 1.7 m x speed = 7.0 m/s Solve for t: 1.7 = 1/2(9.8)t^2 t = 0.59 seconds

Solve for x distance: d = vt d = (7)(.59) d = 4.13 m

2. Given: y distance = 1.5 m x velocity = 45 m/s

Solve for t: 1.5 = 1/2(9.8)t^2 t = 0.55 seconds

Vi = 5.4m/s

Solve for x distance: d = vt d = (45)(.55) d = 49.8 meters

Physics to Go #1-10
1. 45 degrees.

2a. The time the ball would be in the air would be longer 2b. The time the ball would be in the air would be shorter

3a. 60 3b. 75

4. The angle is less because the long jumper has to run really fast horizontally, therefore the long jumper requires more energy to jump at a higher angle.

5. The higher the horizontal velocity, the higher the hypotenuse (which is used to determine the distance of where a long jumper lands). Apparently he can also jump pretty high.

6a. The direction of acceleration for the ball is downwards. 6b. The direction of velocity of the ball is eastward.

7a. Given: x velocity = 5.0 m/s t = 3 seconds Horizontal distance = 15 meters Vertical velocity = a = (vf - vi) t 9.8 = (vf)/3 vf = 29.4 m/s 7b. Remains at 5m/s cause the horizontal speed does not change 7c. d = 1/2(9.8)(3)^2 d = 44.1 meters

8. The ball will travel the furthest if theta was 45 degrees

9. I don't know what this question is even asking

10a. The direction of acceleration is downwards 10b. 100 = 1/2(9.8)t^2 t = 4.51 seconds 10c. 20 = d/4.51 d = 90.35 meters

=Section 6= =What Do You See?= 1/3/2011 A guy is pushing off the wall, the wall is concave at first, then it becomes convex as soon as he's moving away. In Order to Jump you must: Bend your knees and straighten them out really fast while also lifting your heel, pushing yourself up

Part A: Push, Push back, Pull, Pull back
1a. The person only accelerates for the first second. He accelerates only till his hand is off the wall, and accelerates in the opposite direction of where he's pushing. 1b. As soon as the person pushes off, he's traveling at a constant speed if there were no friction. You would technically travel on forever if there was no motion stopping it. 1c. The object that is pushing you is the wall, and the direction of the push is the opposite of the direction you exerted your force into. 1d. I push on a grocery cart when I'm at the grocery store. I usually push it ahead of me. 1e. The force I exert on the wall is less than the force the wall exerts on me because I am the one who's moving, not the wall.

2a. Student A went backwards. 2b. The force of student B's push caused the motion of student A 2c. Student B also went backwards 2d. The force of student A, likewise caused the motion of student B.

3a. The forward force comes from the ground. Every time I step on the ground and exert force through my feet, there is another force that is exerted that accelerates me forward. The backward force exerted by my shoe is less than the forward force, since I'm moving forward. 3b. You can walk or run on a slippery surface in terms of force, but in terms of friction, it would be incredibly hard. I don't see how in the world it is possible to describe this scenario in terms of force, but I guess this answer is sufficient.

4a. One person goes 2000, the other goes 1000. The person who goes 2000 accelerates faster than the person who goes 1000, but they crash into each other eventually anyways.

Part B: Observing a Meter Stick Pushing Back
3a. The force of the coin isn't strong enough to negate the force of the books. 4a. 4b. The more the weight is on the ruler, the more force it applies, therefore bending the meter stick further 4c. Yes, the meter stick is deflecting, just on a way smaller scale then with the weights. 4d.
 * Weight || Deflection ||
 * 1N || 5 mm ||
 * 2N || 8 mm ||
 * 5N || 15 mm ||
 * 10N || 23 mm ||

=Ball and Cup Lab:= (Was not able to upload the file)

**Ball and Cup Trajectory Predictions** Daniel Hwang, Lexie Halperin, Active Physics, Period 4, Group 4 January 6, 2011

__Objective:__ What is the distance the ball will travel horizontally off the table? __Theory & Hypothesis:__ The methods needed to solve this problem come from the x/y table, and also some equations to find the final velocity of the ball off the ramp. To find the final velocity off the ramp, you have to time how long it takes the ball to reach the end, then find the length of the ramp. Using the equation d = 1/2(Vinitial + Vfinal)t, you can solve for Vfinal because Vinitial is 0. Once the final velocity is determined, the horizontal constant speed is also obtained. Using multiple equations to solve for variables of the x/y table, such as d = 1/2at^2 + Vit and V = d/t, the question can be easily solved.

__Materials & Methods:__ To understand how far the ball traveled, many essential variables are needed to be solved for. As mentioned above, the final velocity off the ramp, height of the fall, and the acceleration due to gravity are all key variables. Once these variables are solved for, an x/y table can be created to easily determine the other variables needed to answer the question, which are the times of x and y, which can be used to solve for the distance the ball travels in the x direction. A stop watch is needed to time how long it takes for the ball to reach the end of the ramp, and once the measurement for how far the ball will travel in the x direction is obtained, then a ruler is needed to ensure a precise location for the cup.

__Observations & Data:__

__Data Table__
 * Variables || x || y ||
 * d || ? || 0.92 m(92 cm) ||
 * a || 0 || 9.8m/s^2 (Due to gravity) ||
 * Vf || 162.16 cm/s || ? ||
 * Vi || 162.16 cm/s || 0 ||
 * t || ? = 0.4333 seconds || ? = 0.4333 seconds ||

__Equations and Explanations__

The velocity in the x direction can be solved for by using the equation: d = 1/2(Vf + Vi)t. The time it took for the ball to reach the end of the ramp was: 1.48 seconds, and the distance it traveled was 120 cm. Therefore, plugging the variables into the equation (120 = 1/2(Vf+Vi)1.48), the Vf turns out to be 162.16cm/s.

Once the horizontal speed is determined, the next variable needed is the time the ball is in the air. The time is the same for both the x and the y directions. To solve for the time, the y axis variables are needed to be used. Since the distance and acceleration are both available for the y axis, the appropriate equation to use is d = 1/2at^2. The distance is 0.92m, and acceleration is 9.8m/s^2. Plugging the variables into the equation, 0.92m = 1/2(9.8)t^2, the time comes out to be around 0.4333 seconds. Once the essentials are solved for, the big question can be answered: how far will the ball travel horizontally? Using the simple equation of V = d/t, both the time and the velocity is available, therefore the distance can be easily solved using simple algebra. D = vt, so D = (162.16m/s)(0.4333 seconds), D = 70.215 cm.

__Analysis & Discussion:__ The one thing that was missing from the calculations was the height of the cup. The variables that were solved for were extremely precise, since the ball hit the upper lip of the cup and bounced out. This proves that the distance that the ball traveled would be precise, if the cup had not been there, of course, and if the ball was simply to land on the ground. Neglecting the cup height was the only err in this experiment, however, the calculations were very, very precise. There were not many difficulties that were encountered during the actual experiment, it all went pretty smoothly.

__Conclusion:__ The distance the ball travels horizontally is 70.215 cm, with the absence of the cup, of course.

= Section 7: =

What do you see?
Friction is less on the ice, therefore it is easier to move the shoe. On more ribbed landscapes, the shoe is harder to move, as demonstrated by the girl. Some sports require special shoes to traverse the ground easily.

Investigate: Friction
1a. Newbalance: 600 grams + 1000 gram weight. 1.6 N. 1b. The textures of the floor or table, texture of the soles, and the weight of the shoe 2. We will able to conclude the friction between the shoe and the surface. The amount of force necessary to move the shoe across the surface. We will need a spring scale to determine the amount of force needed. 3a. The surface of the table smooth 3b. 0.8N 3c. 0.5 4a. 1.6N 4b. 0.5 4c. No, the coefficient is not affected because it is always proportional to original weight/force. 4d. It would not affect the coefficient because it always remains the same no matter what, of course, if the shoe and the surface are the same. 5a. 5b. The coefficient is 1.7 5c. The coefficient is different because the floor has more friction than the table does. 5d. No, it would not make a difference because the coefficient remains the same if the object's texture and the surface remain the same.

Checking Up: Friction
1/7/2011 1. The force of friction is equal to the force that is read on the scale because that is the amount of grams required to move the object at a constant speed. 2. It has no units because it is a force divided by a force, therefore canceling out the units 3. The coefficient of friction is determined by dividing the force of friction by the perpendicular force exerted by the surface on the object (aka normal force)

Physics To Go: Friction
1/7/2011 1. A soccer player on a wet playing field will have to wear bigger cleats to gain more traction onto the ground. 2. Hockey players need skates to reduce friction, in order to "glide" around the ice. Skates are more effective then cleats on ice because cleats need to be bigger to actually be able to run across the ice, so instead a small blade is more efficient. 3. No, because the other courts may have different textures. Let's say that the basketball player's home court is a wooden floor, there can be other courts with gravel floors, therefore changing the friction shared between her shoes and the court. 4. Tennis players may have different shoes for different surfaces, because some surfaces are more rugged than others. Tennis players need traction to keep balanced, and if a certain pair of shoes slide off the court, then another pair is needed. 5. 18N.

=Section 8=

What Do You Think?
1/14/2011 There is not enough restoring force in the 11 m long pole to make the person go 12 m up. The factors that affect the height available to be attained is the height and elasticity of the pole, and of course, the ability of the pole vaulter himself.