#$&* course phy 201 020. `query 20 *********************************************
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Given Solution: `a** If we add the x components of the original two vectors we get the x component of the resultant. If we add the y components of the original two vectors we get the y component of the resultant. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qExplain how we get the components of a vector from its angle and magnitude. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x = magnitude * cos(angle) y = magnitude * sin(angle) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** To get the y component we multiply the magnitude by the sine of the angle of the vector (as measured counterclockwise from the positive x axis). To get the x component we multiply the magnitude by the cosine of the angle of the vector (as measured counterclockwise from the positive x axis). ** NOTE REGARDING THE CIRCULAR DEFINITION OF TRIGONOMETRIC FUNCTIONS VS. THE RIGHT-TRIANGLE DEFINITIONS: Students with a background in trigonometry often use the right-triangle definitions of sine and cosine (sine and cosine defined in terms of opposite and adjacent sides and hypotenuse), as opposed to the circular definition (using a coordinate system, with angles measured counterclockwise from the positive x axis--the definition used in this course). The two definitions are pretty much equivalent and completely consistent. The circular definition is a bit more general for two reasons: The circular definition can be applied to positive or negative angles, and to angles greater that 180 degrees, whereas triangles are limited to positive angles less than 180 degrees. The circular definition can yield positive or negative components, whereas the sides of triangles are all positive. In most applications it is your choice which definition you use. Some applications are easier if you use the right-triangle definition, others are easier of you use the circular definition, and some simply require the circular definition. In developing this course I chose to express all trigonometric solutions in terms of one of the definitions, in order to avoid confusion for students with a weak background in trigonometry. If only one of the definitions is to be used, it must be the more general circular definition with its four simple rules (x coordinate = magnitude * cos(angle), y coordinate = magnitude * sin(angle), magnitude = sqrt ( (x coordinate)^2 + (y coordinate)^2 ), angle = arcTan ( y coord / x coord), plus 180 deg or pi rad if x coord is negative). The circular definition is sufficient for Principles of Physics or General College Physics.. However General College Physics students are to have completed a year of precalculus or equivalent, which includes trigonometry, and it is expected that these students can reconcile the circular and right-triangle definitions and approaches, and understand the right-angle trigonometry in their text. University Physics students are of course expected to already be familiar with trigonometry and the use of vectors (though in reality some refreshing is usually required, and is provided in the first chapter of the University Physics text). However students at the level of University Physics should encounter no serious obstacle with the trigonometry. In a nutshell, here is a summary of how the right-triangle definitions are related to the circular definitions: On a circle of radius r centered at the origin, any first- or second-quadrant angle gives us a triangle in the upper half-plane having that base angle. The hypotenuse of this triangle is r, the adjacent side is the x coordinate r cos(theta), and the opposite side is the y coordinate r sin(theta). Thus adjacent side / hypotenuse = r cos(theta) / r = cos(theta), opposite side / hypotenuse = r sin(theta) / r = sin(theta), and opposite side / adjacent side = r sin(theta) / ( r cos(theta) ) = sin(theta) / (cos(theta)) = tan(theta). The definitions of the cosecant, secant and cotangent functions are then made in the usual manner. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qprin phy, gen phy 7.02. Const frict force of 25 N on a 65 kg skiier for 20 sec; what is change in vel? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: -25 * 20 = -500 N
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Given Solution: `aIf the direction of the velocity is taken to be positive, then the directio of the frictional force is negative. A constant frictional force of -25 N for 20 sec delivers an impulse of -25 N * 20 sec = -500 N sec in the direction opposite the velocity. By the impulse-momentum theorem we have impulse = change in momentum, so the change in momentum must be -500 N sec. The change in momentum is m * `dv, so we have m `dv = impulse and `dv = impulse / m = -500 N s / (65 kg) = -7.7 N s / kg = -7.7 kg m/s^2 * s / kg = -7.7 m/s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:
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Given Solution: `a**STUDENT SOLUTION: Momentum conservation gives us m1 v1 + m2 v2 = m1 v1' + m2 v2' so if we let v2' = v, we have (.023kg)(230m/s)+(2kg)(0m/s) = (.023kg)(170m/s)+(2kg)(v). Solving for v: (5.29kg m/s)+0 = (3.91 kg m/s)+(2kg)(v) .78kg m/s = 2kg * v v = 1.38 kg m/s / (2 kg) = .69 m/s. INSTRUCTOR COMMENT: It's probably easier to solve for the variable v2 ': Starting with m1 v1 + m2 v2 = m1 v1 ' + m2 v2 ' we add -m1 v1 ' to both sides to get m1 v1 + m2 v2 - m1 v1 ' = m2 v2 ', then multiply both sides by 1 / m2 to get v2 ` = (m1 v1 + m2 v2 - m1 v1 ' ) / m2. Substituting for m1, v1, m2, v2 we will get the result you obtained.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q**** Univ. 8.75 (11th edition 8.70) (8.68 10th edition). 8 g bullet into .992 kg block, compresses spring 15 cm. .75 N force compresses .25 cm. Vel of block just after impact, vel of bullet? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The spring ideally obeys Hook's Law F = k x. It follows that k = .75 N / .25 cm = 3 N / cm. At compression 15 cm the potential energy of the system is PE = .5 k x^2 = .5 * 3 N/cm * (15 cm)^2 = 337.5 N cm, or 3.375 N m = 3.375 Joules, which we round to three significant figures to get 3.38 J. The KE of the 1 kg mass (block + bullet) just after impact is in the ideal case all converted to this PE so the velocity of the block was v such that .5 m v^2 = PE, or v = sqrt(2 PE / m) = sqrt(2 * 3.38 J / ( 1 kg) ) = 2.6 m/s, approx. The momentum of the 1 kg mass was therefore 2.6 m/s * .992 kg = 2.6 kg m/s, approx., just after collision with the bullet. Just before collision the momentum of the block was zero so by conservation of momentum the momentum of the bullet was 2.6 kg m/s. So we have mBullet * vBullet = 2.6 kg m/s and vBullet = 2.6 kg m/s / (.008 kg) = 330 m/s, approx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!