| Chapter 2: Do listed problems up to
#27 for Quiz 1. Do listed problems from #15 to #67 for Quiz 2 Ch. 2--CQ means conceptual question |
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| ANSWERS | |||||||||||
| Quiz 1 (Note the addition of #12) | |||||||||||
| CQ 6 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 27 | ||
| Go to masterinphysics.com for assignment 1 on "Motion Part 1" from Quiz 1. | |||||||||||
| Quiz 2 (Note the addition of #12 to Quiz 1) | |||||||||||
| 15 | 16 | 19 | 35 | 45 | 46 | 49 | 50 | 51 | 57 | 66 | 67 |
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Measurement Lab Links: |
| Helpers!! Click this link then click the links "Practice Questions" and "Practice Problems" in the left hand column. |
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for 3,000 Solved Problems in Physics (Schaum's Solved Problems Series) |
| CQ 6. Reviewed in class. |
| 5. Find the the slope of each of the three line segments. Remember Vx = slope of x. For example the slope of the first segment, between 0 and 20 seconds, is (100 - 50)m/(20 - 0) s = 2.5 m/s. |
| 6. (a) Find the slope of each line of the three segments. Remember Vx = slope of x. For example the slope of the third segment, between 3 and 4 seconds, is (0 - 20)m/(4 - 3) s = - 20 m/s. (b) Turning points occur when Vx changes sign, indicating a change in direction. See example 2.9 for a general description. The turning point occurs at t = 3 seconds, when x = 20 m. |
| 7. (a) Again, turning points occur when Vx changes sign, indicating a change in direction. Example 2.9 gives a general description. Indeed that example shows you everything you need to know ! The turning point occurs at t = 1 second. But you have to find x = x(1) at that time. Here's how you find that position: x(1) = x(0) + (area under the Vx -curve between 0 and 1 second). x(0) = x0 is given. The triangular area is negative. The base of the triangle is 1 second, and the height is -4 m/s. (b) Use xf = xi + (area under the Vx -curve between ti and tf) repeatedly. Here's two examples: (i) x(2) = x(0) + (area under the Vx -curve between 0 and 2 seconds). Clearly the area of interest is zero, since you are adding the areas of a negative triangle and a positive triangle of equal magnitudes. x(0) = x0 is given. (ii) x(3) = x(0) + (area under the Vx -curve between 0 and 3 seconds). Clearly the area of interest is positive, since you are adding the areas of a negative triangle and a positive triangle of unequal magnitudes. The area of the negative triangle has been described in part (a) . For the positive triangle, the height is 8 m/s and the base is 3 s -1 s = 2 seconds. Again, x(0) = x0 is given. (iii) Find x(4) using the same method. |
| 8. ax = slope of Vx. Thus between 0 and 2 seconds, the acceleration is zero . Between 2 and 4 seconds, find the slope of the line segment using rise/run. |
| 9. x(0) = x0 = 0. (Starts
at Origin.) (a) ax = slope of Vx. 3 seconds is in the interval [0,8]. As in the previous problem, between 0 and 8 seconds, find the slope of the line segment using rise/run. Note: For t > 8 seconds the acceleration is zero. (b) (i) You can use the qualitative method to plot x vs t., using Vx = slope of x. Between 0 and 8 seconds, we see the x-curve is a concave up parabola. To find the vertex you need x(4) = x(0) + (area under the Vx -curve between 0 and 4 seconds) The triangular area is negative, with base 4 seconds and height -2 m/s. The vertex is ( 4, x(4) ). Note: At t = 0, the slope of the tangent is - 2 m/s. At t = 8 seconds, the tangent slope is + 2 m/s. For t > 8 seconds, x is a straight line with slope +2 m/s and joined to the parabola at t = 8 seconds. (ii) ax = slope of Vx. Between 0 and 8 seconds, find the slope of the line segment using rise/run. For t > 8 seconds, the acceleration is zero. |
| 10. (a) x(2) = x(0) + (area under Vx -curve between 0 and 2 seconds) (b) Look at the graph ! (c) ax = slope of Vx |
| 11. See detailed class notes. |
| 12. (a) Use equation 2.22. The initial and final velocities and position change xf - xi are all given. Solve for ax . (b) Is you answer reasonable? Does ax seem to large or too small? |
| Check back soon for more hints ! |
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27.
Consider the section of the problem dealing with the 35 m distance. You want the total time it takes to reach that distance. You can find the second time interval i. e, the period beyond the time when the car accelerates by slowing down. Call this second time interval t. During the first time interval of 2 seconds, the car has zero acceleration. The second time interval t gives the displacement we can call xf = xi + Vit + (1/2)*at2. Get a by finding the difference ratio a = (Vf - Vi )/t, where t is the value of the second time interval. Vi = the velocity of the car when it has zero acceleration. Vf = 0. xf = 35 m. From a rectangular computation, xi = 20 m. Solve for the two solutions for t. The two solutions represent instants above and below the moment when the velocity is zero---instants on either side of the turning points. Then add the solved t values to the first time interval. In other words, add 2 seconds to t. |
| Free Fall review using i , f notation: Vfy = Viy + ay(tf - ti) (2.19) Yf - Yi =Viy(tf - ti) + (1/2)*ay(tf - ti)2 (2.20) Vfy2= Viy2+ 2ay(yf - yi) (2.22) VAV = (Vfy + Viy )/2 (my 2.23) Yf - Yi = VAV(tf - ti) (my 2.24) If up is the positive y direction, then ay = -g. If down is the positive y direction, then ay = g. Note these equations are useful for any kind of motion in which the acceleration is constant ! You could have used them on Quiz 1 by changing y to x if desired. |
| More hints posted later--check back soon !!! |
| 15. The time interval is given. You
can conveniently let ay = g. (a) Use equation (2.20) to get Yf - Yi . The ball bearing is dropped. (b) Use equation (2.19) or (2.22) to find Vfy . Note you have also used my 2.23 and my 2.24 to get Vfy . |
| 16. See detailed class notes. We let ay = -g. |
| Incline plane review: Use equations (2.19) , (2.20) , (2.22) , (my 2.23) , and (my 2.24) above. Replace y with x. If up the incline is the positive x direction, then ax = -gsin(theta). If down the incline is the positive x direction, then ax = gsin(theta). |
| 19. See example 2.33. |
| 35. See example 2.19 for a demonstration of
how to construct the graphs. Assume t = 0 when the object with velocity V0x
is at the
left edge of the first horizontal segment
. As in example 2.19, we assume the object reaches the
third segment. But in this problem, on segment 3 the
object has a smaller velocity than on the first. # First segment: On the lowest horizontal segment, we have constant velocity Vx , zero acceleration, and position x increasing with positive slope linearly . The shapes are identical to velocity, acceleration and position plots segment 1 of example 2.19. # Second segment: On this segment, our x-axis is parallel to the incline. Vx decreases linearly according to equation (2.19) with ax = -g*sin(angle): Vx = Vix - g*sin(angle)*(t - ti). Here ti corresponds to the moment the object reaches the incline bottom. Vix = V0x . (Compare with the second segment in example 2.19 where Vx increases linearly.) When the object reaches the bottom of the incline, position x becomes a concave down parabola. (Please compare with example 2.19's segment 2 where x transforms into a parabola that's concave up. ) The acceleration is constant and negative: ax = -gsin(angle). (Please compare with segment 2, example 2.19, where acceleration +gsin(angle). #Third segment: On the highest horizontal segment, once again Vx is constant, acceleration zero, and position x increases with positive slope linearly. The graphs are identical to the velocity, acceleration and positions of segment 3, example 2.19. But here the steady velocity is less than the velocity of the first segment. Also in contrast to example 2.19: Position x increases linearly with positive slope less than than that of segment 1. |
| Kinematics review using i , f notation: Vfx = Vix + ax(tf - ti) (2.19) xf - xi =Vix(tf - ti) + (1/2)*ax(tf - ti)2 (2.20) Vfx2= Vix2+ 2*ax(xf - xi) (2.22) VAV = (Vfx + Vix )/2 (my 2.23) xf - xi = VAV(tf - ti) (my 2.24) |
| 45. The velocity has the same shape as figure ex2.11. See problem, 11--Quiz 1. But here Vx is constant = 20 m/s until t = 0.50 seconds when the car slows down with acceleration ax = -6.0 m/s2. Thus the slope of the second segment in figure ex2.11 would be -6.0 m/s2 not -2.0 m/s2 . We want to know how far the car travels before Vx = 0. If that distance is greater than 50 m, then the car will collide with the obstacle. First find the displacement between ti = 0 and tf = 0.50 seconds: (2.20) xf - xi =Vix(tf - ti) + (1/2)*ax(tf - ti)2 becomes xf - xi = 20(0.50 - 0) + 0, since the acceleration is zero between 0 and 0.50 s. Thus xf - xi = 10 m. Now use (2.22) with Vfx = 0, corresponding to ti = 0.50 seconds, xi = 10 m, Vix = 20 m/s, ax = -6.0 m/s2 and tf = time when car stops: Vfx2= Vix2+ 2ax(xf - xi). Solve for (xf - xi) and add to 10 m to get the total displacement. Note that you can also just solve for xf . Compare with 50 m to check for a collision. |
| 46. The problem is similar to the
previous. Here the total displacement of interest is 110 m not 50 m. (a) First find the displacement between ti = 0 and tf = 0.50 seconds: (2.20) xf - xi =Vix(tf - ti) + (1/2)*ax(tf - ti)2 becomes xf - xi = 20(0.50 - 0) + 0, since the acceleration is zero between 0 and 0.50 s. Thus xf - xi = 10 m. From this find the distance between the car and the intersection using your common sense and simple arithmetic. (b) Now use (2.22) with Vfx = 0, corresponding to ti = 0.50 seconds, xi = 10 m, Vix = 20 m/s, tf = time when car stops, and xf = 110 m: Vfx2= Vix2+ 2ax(xf - xi). Solve for ax. (c) The first time interval of motion is 0.50 seconds representing the period between seeing the red light and actually applying the brakes. This is the reaction time. (This time interval varies from person to person, and is an average. In general, the reaction time increases in drivers who are legally drunk, under the influence of mind-altering drugs or sleep derived. Stay alive, don't imbibe and drive !) You can easily find the second time interval (tf - ti) during which the brakes are engaged using (2.19) or (my 2.23 ) and (my 2.24). Add the two intervals for the total time. |
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49. Let up be the positive y direction. (a) (i) First find the displacement the object moves upward when the acceleration is ay = + 30 m/s2. Use Yf - Yi =Viy(tf - ti) + (1/2)*ay(tf - ti)2 where the time interval is given. The initial velocity is zero: Viy = 0. It will be helpful to find Vfy = Viy + ay(tf - ti) for this time interval since you can use it in the next part just below. (ii) Second, when the rocket runs out of fuel, ay = -g. Use Vfy2= Viy2+ 2ay(yf - yi) to find the second displacement. Note that when the object reaches the maximum height the velocity is zero: Vfy = 0. You can find the initial velocity Viy for this interval using information mentioned above in (i): Viy = 0 + (30 m/s2)(30 s). In other words, the initial velocity for this interval is the final velocity for the first interval in (i). Add the two displacements to get the maximum height from (i) and (ii)
to get the maximum height. Add all three time intervals to compute the total time of
flight. |
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50. This problem is just like 49, only different things are
sought. Same problem, just different givens and unknowns. For the first 16
seconds, the rocket has ay > 0. After 16 seconds , ay
= -g. Thus between 16 seconds and 20 seconds the rocket is in free
fall. However, the problem seems challenging, at least for me. Let us use A, B, C notation. A is at t = 0, B is at t = 16 s, and C is at t = 20 s. In other words, tA = 0, tB = 16 s, and tC = 20 s. Let the acceleration during the first 16 s be ay = a > 0. (a) VB = VA + a(tB - tA). We know (tB - tA) = 16 s and VA = 0. We want to solve for a. But we need to know VB to find a. For this we can use our definition of average velocity for constant acceleration (my 2.23) and (my 2.24) and the given information which says that (YC - YB) + (YB - YA) = 5100 m. Now, (YB - YA) = [(VA +VB )/2]( 16 s), where VA = 0. Also, (YC - YB) = [(VB +VC )/2]( 4 s), where VC = VB + ay(tC - tB) = VB - g(4 s) because the object slows down as it moves upward in free fall ! We can write the original equation in terms of VB , solve for VB and then find a. Here we go: (YC - YB) + (YB - YA) = 5100 m can now be re-written as [(VB +VC )/2]( 4 s) + [(VA +VB )/2]( 16 s) = 5100 m [(VB + VB - 4g)/2](4) + [VB /2]( 16) = 5100 [(2VB - 4g)/2]*(4) + [VB /2]*(16) = 5100. Here I have dropped the units . Solve this for VB and then solve for a using VB = a(tB - tA) ---------------------------------------------------------------------------------------- The time differences are all known: (tB - tA) =
16 s and (tC - tB) = 4 s |
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51. This problem is much easier than the previous !
Let down be the positive y direction. |
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57. Assume t = 0 when the object with velocity V0x
is at the
left edge of the first horizontal segment
. As in example 2.19, we assume the object reaches the fifth segment.
In this problem, on segment 5 the
object has zero acceleration and will have the same velocity as the first. # First segment: On the lowest horizontal segment, we have constant velocity Vx , zero acceleration, and position x increasing with positive slope linearly . The shapes are identical to velocity, acceleration and position plots segment 1 of example 2.19. # Second segment: On this segment, our x-axis is parallel to the incline. Vx decreases linearly according to equation (2.19) with ax = -g*sin(angle): Vx = Vix - g*sin(angle)*(t - ti). Here ti corresponds to the moment the object reaches the incline bottom. Vix = V0x . (Compare with the second segment in example 2.19 where Vx increases linearly.) When the object reaches the bottom of the incline, position x becomes a concave down parabola. (Please compare with example 2.19's segment 2 where x transforms into a parabola that's concave up. ) The acceleration is constant and negative: ax = -g*sin(angle). (Please compare with segment 2, example 2.19, where acceleration +g*sin(angle) #Third segment: On the highest horizontal segment, once again Vx is constant, acceleration zero, and position x increases with positive slope linearly. The graphs are identical to the velocity, acceleration and positions of segment 3, example 2.19. But here the steady velocity is less than the velocity of the first segment. Also in contrast to example 2.19: Position x increases linearly with positive slope less than than that of segment 1. #Fourth Segment: On this segment, our x-axis is parallel to the incline. Vx increases linearly according to equation (2.19) with ax = +g*sin(angle): Vx = Vix+- g*sin(angle)*(t - ti). Here ti corresponds to the moment the object reaches the right end of the third segment. Vix =constant velocity on the third segment. (Compare with the second segment in example 2.19 where Vx increases linearly.) #Fifth Segment: On the last horizontal segment, we have constant velocity Vx , zero acceleration, and position x increasing with positive slope linearly . The shapes are identical to velocity, acceleration and position plots segment 1 of example 2.19. Note: the velocity should be equal to the velocity on the first segment. |
| 66. Tina's x vs t curve is a parabola xT = (1/2)*at2 , vertex at the origin. a = her given acceleration. David's x vs t curve is a straight line xD = Vt going through the origin. V = his given speed. Set the xT = xD and solve for t > 0. With t you can answers both parts to this problem. |
| 67. Cat's x vs t curve is a concave up parabola xC = xC0 + (1/2)*aCt2 . aC = its given acceleration. xC0 = 1.5 m. Dog's x vs t curve is a concave down parabola xD = V0t + (1/2)*aDt2 going through the origin. V0 = its given initial velocity. aD = its given negative acceleration. Set the xC = xD and solve for t > 0. If you get a real answer then the dog catches up to the cat. For real t values, you must choose the smallest value of t giving you an x less than or equal to 3.0 m. Assume the dog does not collide with the cat when it first passes by the cat. The other value of t would then correspond to the dog passing by the cat while moving in the opposite direction from the first. |
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Measurement Lab Links: http://wise.fau.edu/~dchen/1.pdf http://phoenix.phys.clemson.edu/tutorials/measure/index.html http://en.wikipedia.org/wiki/Caliper http://www.physics.ccsu.edu/LEMAIRE/genphys/virtual_physics_labs.htm http://www.upscale.utoronto.ca/PVB/Harrison/Micrometer/Micrometer.html |
| error log #1: Note that a correction has been made to #27 hint. xf = 35 m. |
| error log #2: #19 refers to example 2.33. |
| error log #3: On #45, #46, the acceleration is zero between 0 and 0.50 seconds. |
| error log #4: On free fall review a correction has been made to (2.22): x has been replaced by y. |
| More hints later--check back soon. |