AP Physics 2. Norton High School Teacher: Mr. Murray

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1 AP Physics 2 Norton High School Teacher: Mr. Murray craigmurray@norton.k12.ma.us Course Description: AP Physics 2 is an algebra-based, introductory college-level physics course. The material covered is typical of a second semester college level physics class (ex. Physics 102). Students explore principles of fluids, thermodynamics, electricity, magnetism, optics, and topics in modern physics. The course is based on seven Big Ideas, which encompass core scientific principles, theories, and processes that cut across traditional boundaries and provide a broad way of thinking about the physical world. The following are the Big Ideas: Objects and systems have properties such as mass and charge. Systems may have internal structure. Fields existing in space can be used to explain interactions. The interactions of an object with other objects can be described by forces. Interactions between systems can result in changes in those systems. Changes that occur as a result of interactions are constrained by conservation laws. Waves can transfer energy and momentum from one location to another without the permanent transfer of mass and serve as a mathematical model for the description of other phenomena. The mathematics of probability can be used to describe the behavior of complex systems and to interpret the behavior of quantum mechanical systems. Textbook: Giancoli, Physics, 6 th Edition Summer Assignment: All AP Physics students will read, study, and complete the following packet. This material should be review material used in previous math, science, and physics classes. It is full of basic skills that you will be required to know, understand, and utilize throughout the year. All material will be due upon the first day of class in September. On that day you will be given an answer packet, and an opportunity to ask any questions that you may have. A test will be given on the second day of class to ensure you have mastered the skills.

2 Scientific Notation Convert the following numbers into scientific notation: 1) 3,400 2) ) 101,000 4) ) ) 1,000,000 7) ) 4.50 Convert the following numbers into standard notation: 9) 2.30 x ) 1.76 x ) x ) 8.65 x ) 9.11 x ) 5.40 x ) 1.76 x ) 7.4 x 10-5

3 Significant Figures Significant figures are a very important part of physics and engineering problems. The number of significant figures attaches a level of precision to that number. Saying that a person has a mass of 50 kg vs. 47 kg vs kg are all very different answers and should be interpreted differently. Identify the number of Significant Figures: 1) 5,300 9) ) ) 12, ) ) ) 125,521 12) ) 3,458,000 13) 100 6) ) 1,200 7) 5,700,00 15) ) 10 16) 9,400,000

4 Addition and Subtraction of Significant Figures Significant figures are a very important part of physics and engineering problems. It is important to carry the correct number of significant figures throughout a problem. This will ensure that your answer has the correct level of precision. Complete the following problems, answers shall have the correct number of significant figures. SET # = = = = = = SET # = = = = = = SET # = = = = = = SET # = = = = = =

5 Multiplication and Division of Significant Figures Significant figures are a very important part of physics and engineering problems. It is important to carry the correct number of significant figures throughout a problem. This will ensure that your answer has the correct assumption of precision. Complete the following problems, answers shall have the correct number of significant figures. SET # x 2.5 = x 3.58 = x 727 = / = / 9.0 = / 55 = SET # x 2.00 = x 3.45 x 7.42 = x = / 7 = / 9.0 = / 5 = SET # x 5.6= x 10.6= x 6.95= 4. 80/0.675= /106= /0.356= SET # x 5.2 = x 8.57 = x 538 = / = / 7 = / 38 =

6 Unit Conversions When converting from one a unit to another it is useful to use conversion factors. Conversion factors are fractions that equal 1 because the numerator (top) equals the denominator (bottom), but they use different units. In algebra you learned that multiplying any quantity by 1 does not change the quantity s value, so we can use conversion factors to change the units of a measurement without changing the value of a measurement. For instance, since 1 km = 1000 m, 1 min = 60 s, and 1 hr = 60 min, we can create the conversion factors below: 1 km _ = 1 _1 min_ =1 _ 1 hr =1 and _1000 m_ = 1 _60 s_ =1 _60 min_ = m 60 s 60 min 1 km 1 min 1 hr By using all 3 of these conversion factors you can convert 15 km/hr into m/s. Be careful to use the conversion factor that cancels the units you are trying to convert out of. If you can cancel (reduce to 1) those units you did it correctly. 15 km 1000 m 1 hr 1 min 1 hr 1 km 60 min 60 s = 4.2 m/s Carry out the following conversions. If you have trouble, start by writing out the conversion factors you ll need year (non-leap year) into seconds km into cm g into kg kg into mg 5. 5 m/s into mi/hr

7 6. 60 L into ml cm/min into m/s 8. The speed of light (in a vacuum) is 3.0 x 10 8 m/s. Convert this speed into km/day and mi/hr kg = g cm= dm MHz = Hz m/s = mi/hr ft/s 2 = m/s x 10 5 N/m 2 = N/cm g*cm 2 /s 2 = kg*m 2 /s kg*km/min 2 = kg*m/s 2

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