1. 150J
2. the same
3. 3.0kg-m/sec
4. 7.5 m/s West
5. 24J
6. 5J
7. 8 x 10^2 Joules
8. momentum
9. 48 kg-m/sec north
10. greater
Wednesday, November 28, 2007
Wednesday, November 21, 2007
Conservation of Momentum Answers
1/2: 16.9 m/s
3/4. 5 m/s
5. 7 cm/s
6. 1.15 m/s
7. 58 cm/s
8. 1.25 km/hr right
9. 1.14 m/s E
3/4. 5 m/s
5. 7 cm/s
6. 1.15 m/s
7. 58 cm/s
8. 1.25 km/hr right
9. 1.14 m/s E
Thursday, November 15, 2007
Corrections to Solutions
It appears my copy of the review sheet was different than yours. I have put the solutions below in the order the questions appear on your copy. Email me with any questions.
1. 0 Degrees (180 degrees is minimum resultant, 0 degrees is maxiumum)
2. Arrow up and to the right (use the parallelogram method)
3. 100N (use change in velocity - 10m/s, over change in time - 1.5 s to find acc., then multiply by mass to find the force)
4. 44m (use d = Vit + 1/2at^2, where a = 9.8, t = 3, Vi = 0)
5. decrease (remember because F = Gm1m2/r^2, the distance r plays a greater factor)
6. F (Force of friction only relies on Normal force, and coefficient of friction)
7. Arrow from the box, down and to the left, parallel to the plane (friction always opposes motion)
8. DE (this is where the slope of distance v. time is NOT constant, so velocity must be changing)
9. 10 m/s^2 (use v^2/r, NOT mv^2/r, because we are finding acceleration, not force)
10. copper (its coefficient is the smallest value).
11. 2.0s (Find the average acceleration (Vf + Vi)/2 , then use d = V*t to find t)
12. 1 x 10^ -2 N (mass of insect x velocity of insect)
13. 40m (use Vf^2 = Vi^2 + 2ad)
14. .2h (use distance = average velocity * time)
1. 0 Degrees (180 degrees is minimum resultant, 0 degrees is maxiumum)
2. Arrow up and to the right (use the parallelogram method)
3. 100N (use change in velocity - 10m/s, over change in time - 1.5 s to find acc., then multiply by mass to find the force)
4. 44m (use d = Vit + 1/2at^2, where a = 9.8, t = 3, Vi = 0)
5. decrease (remember because F = Gm1m2/r^2, the distance r plays a greater factor)
6. F (Force of friction only relies on Normal force, and coefficient of friction)
7. Arrow from the box, down and to the left, parallel to the plane (friction always opposes motion)
8. DE (this is where the slope of distance v. time is NOT constant, so velocity must be changing)
9. 10 m/s^2 (use v^2/r, NOT mv^2/r, because we are finding acceleration, not force)
10. copper (its coefficient is the smallest value).
11. 2.0s (Find the average acceleration (Vf + Vi)/2 , then use d = V*t to find t)
12. 1 x 10^ -2 N (mass of insect x velocity of insect)
13. 40m (use Vf^2 = Vi^2 + 2ad)
14. .2h (use distance = average velocity * time)
Wednesday, November 14, 2007
Solutions to Midterm Review Sheet - Reminder Tutoring
Tutoring
Room 203 - Period 5 - With Ms. Page
Room 606 - Period 6 - With Ms. Pellegrino (while lab is in session)
1. Arrow up and to the right (use the parallelogram method)
2. 44m (use d = Vit + 1/2at^2, where a = 9.8, t = 3, Vi = 0)
3. 0 Degrees (180 degrees is minimum resultant, 0 degrees is maxiumum)
4. 100N (use change in velocity - 10m/s, over change in time - 1.5 s to find acc., then multiply by mass to find the force)
5. DE (this is where the slope of distance v. time is NOT constant, so velocity must be changing)
6. 40m (use Vf^2 = Vi^2 + 2ad)
7. .2h (use distance = average velocity * time)
8. 1 x 10^ -2 N (mass of insect x velocity of insect)
9. F (Force of friction only relies on Normal force, and coefficient of friction)
10. 10 m/s^2 (use v^2/r, NOT mv^2/r, because we are finding acceleration, not force)
11. 2.0s (Find the average acceleration (Vf + Vi)/2 , then use d = V*t to find t)
12. Arrow from the box, down and to the left, parallel to the plane (friction always opposes motion)
13. decrease (remember because F = Gm1m2/r^2, the distance r plays a greater factor)
14. copper (its coefficient is the smallest value).
Room 203 - Period 5 - With Ms. Page
Room 606 - Period 6 - With Ms. Pellegrino (while lab is in session)
1. Arrow up and to the right (use the parallelogram method)
2. 44m (use d = Vit + 1/2at^2, where a = 9.8, t = 3, Vi = 0)
3. 0 Degrees (180 degrees is minimum resultant, 0 degrees is maxiumum)
4. 100N (use change in velocity - 10m/s, over change in time - 1.5 s to find acc., then multiply by mass to find the force)
5. DE (this is where the slope of distance v. time is NOT constant, so velocity must be changing)
6. 40m (use Vf^2 = Vi^2 + 2ad)
7. .2h (use distance = average velocity * time)
8. 1 x 10^ -2 N (mass of insect x velocity of insect)
9. F (Force of friction only relies on Normal force, and coefficient of friction)
10. 10 m/s^2 (use v^2/r, NOT mv^2/r, because we are finding acceleration, not force)
11. 2.0s (Find the average acceleration (Vf + Vi)/2 , then use d = V*t to find t)
12. Arrow from the box, down and to the left, parallel to the plane (friction always opposes motion)
13. decrease (remember because F = Gm1m2/r^2, the distance r plays a greater factor)
14. copper (its coefficient is the smallest value).
Thursday, November 8, 2007
Helpful Hints for the Exam Tomorrow
Forces:
Gravitational Force: F = (Gm1m2/r^2)
You should be comfortable drawing free body diagrams, identifying tension in a pulley system, and using the big 5 motion equations for horizontal and vertical motion.
- If there is a net force, that means an object is ACCELERATING.
- If there is NO net force (equilibrium), the object can still be moving, but there can't be a change in speed
- If something is moving at CONSTANT speed, then the sum of all the forces must be 0
- If something is moving at CONSTANT acceleration, then the NET FORCE must be constant.
- On an inclined plane, the parellel force is the weight * sin of angle.
- On an inclined plane, the perpendicular force is the weight * cos of angle
- When in doubt, use the equation Fnet = m * a
- As you make the angle of an incline greater, the parallel force increases (the object is more likely to slide down), and the perpendicular force decreases (the normal force decreases, also causing the object to slide).
- The direction of friction is always opposite the direction of motion.
Gravitational Force: F = (Gm1m2/r^2)
- If I double the mass of one object, I double the gravitational force between the two objects
- If I triple the distance between two objects, the force becomes 1/3^2 = 1/9 as big
- If I double mass, and distance, the force will still decrease because distance factors more than one mass
You should be comfortable drawing free body diagrams, identifying tension in a pulley system, and using the big 5 motion equations for horizontal and vertical motion.
And the Winner Is...
Wednesday, November 7, 2007
Monday, November 5, 2007
Homework: Week of 11/5-11/9
Monday: TPS p47 109-112, Start Packet
Wednesday: Work on packet, study for Exam
Friday: EXAM 4, start studying for midterm!
Wednesday: Work on packet, study for Exam
Friday: EXAM 4, start studying for midterm!
5 points BONUS!
The first person to post the correct answers to the BACK side of that handout will receive 5 extra points on the exam Friday. You can post your answers as a comment on this post. You have until WEDNESDAY, which is when I will post the solutions.
Solutions to Exam 3 Questions
Sunday, November 4, 2007
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