- #1
riseofphoenix
- 295
- 2
There MUST be steps to solving problems dealing with Conservation of Energy! No?
There just HAS to be... like Step 1) What does the problem give you...
Step 2) Was does the law of Conservation of Energy state
Etc Etc...
For example...16. A 66.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 31.0 m/s. A second person, with a mass of 55.0 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 2.30 m/s, and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged? Disregard the friction between the skates and the ice.
Thrower: ______ m/s
Catcher: ______ m/s
––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
This practice test my teacher gave our class has a LOT of problems dealing with the ONE single, SEEMINGLY STRAIGHTFORWARD, concept of Conservation of Energy! There HAS to be a certain amount of steps I have to take in order to approach ANY problem dealing with Conservation of Energy, right?? Can anyone list steps I have to take in order to approach problems like these? Because my approach almost NEVER works!
This is what I did - the fact that I just CAN'T seem to fully UNDERSTAND these types of problems is SO frustrating that I'm literally fuming and I can't think straight. And I'm such a lOGICAL thinker, which explains why I NEED to know that there is some kind of pattern in these problems!
To find speed for thrower:
1) Energy system of thrower vs. Energy system of catcher (?)
PEA + KEA = PEB + KEB
0 + (1/2)mv2 = 0 + (1/2)mv2
(1/2)(66)v2 = (1/2)(55)(2.30 - 0)2
v2 = 145.475/33
v = √(4.4083)
v = 2.09 m/s
"INCORRECT" Correct answer is - Thrower: 2.28 m/s
??
akkajsnflkajdsnflkjadnflkjdnsa
F. My. Life.
-.-
Help anyone?
There just HAS to be... like Step 1) What does the problem give you...
Step 2) Was does the law of Conservation of Energy state
Etc Etc...
For example...16. A 66.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 31.0 m/s. A second person, with a mass of 55.0 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 2.30 m/s, and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged? Disregard the friction between the skates and the ice.
Thrower: ______ m/s
Catcher: ______ m/s
––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
This practice test my teacher gave our class has a LOT of problems dealing with the ONE single, SEEMINGLY STRAIGHTFORWARD, concept of Conservation of Energy! There HAS to be a certain amount of steps I have to take in order to approach ANY problem dealing with Conservation of Energy, right?? Can anyone list steps I have to take in order to approach problems like these? Because my approach almost NEVER works!
This is what I did - the fact that I just CAN'T seem to fully UNDERSTAND these types of problems is SO frustrating that I'm literally fuming and I can't think straight. And I'm such a lOGICAL thinker, which explains why I NEED to know that there is some kind of pattern in these problems!
To find speed for thrower:
1) Energy system of thrower vs. Energy system of catcher (?)
PEA + KEA = PEB + KEB
0 + (1/2)mv2 = 0 + (1/2)mv2
(1/2)(66)v2 = (1/2)(55)(2.30 - 0)2
v2 = 145.475/33
v = √(4.4083)
v = 2.09 m/s
"INCORRECT" Correct answer is - Thrower: 2.28 m/s
??
akkajsnflkajdsnflkjadnflkjdnsa
F. My. Life.
-.-
Help anyone?