QUIZ 3 (ANSWERS) |
Sample Test 1 |
READ ALL EXAMPLES. SOME PROBLEMS ARE JUST LIKE AN EXAMPLE. |
CH. 2 & 3 |
CH. 2 - Multiple choice 9, 10, 14, 15 and exercise/problems FREE FALL (READ EXAMPLES 2.10 AND 2.11 !) *48, *51, *53, *56, *59, *60, 73*, 75*, 79*, 83*; RELATIVE MOTION 61*, 64* (See example 2.12 !) |
CH. 3 - INTRODUCTION TO PROJECTILE MOTION (MORE SUCH PROBLEMS ASSIGNED IN QUIZ 4 COMING UP) Multiple choice 1, 2, 3, 4, 5, 6 (note- 4 and 5 are 1D questions exploring the y-motion occurring in 2-D projectile motion) exercises/problems 12*, 18*, 56* . |
* DISCUSSIONS PROVIDED. |
SELECTED DISCUSSIONS TO exercises/problems
BELOW.
ALL PROBLEMS ARE DUE EVEN THE ONES WITHOUT DISCUSSIONS, WHICH ARE DESIGNED TO HELP YOU DO RELATED EXERCISES. |
For problems below, unless otherwise stated, assume up is positive, and thus the downward directed acceleration component is negative: a_{y} = -g. |
48. (a) Use V_{2y}^{2} = V_{1y}^{2}
- 2*g*(change in y) and set the second velocity, at the top, equal to
zero. Find the first velocity. (b) Find the total time by using change in y = V_{1y} *t - (1/2)*g*t^{2}. Set change y = 0 = V_{1y} *t - (1/2)*g*t^{2} = t*[ V_{1y} - (1/2)*g*t ] and solve V_{1y} - (1/2)*g*t = 0 for the total time. Note you can also get the total time by getting the
time to arrive at the top of the path by using V_{2y}
= V_{1y} - g*t and setting the second
velocity = 0 and solving for t. The total time of flight is twice this
value !! |
51. (a) The total time is 8.5 seconds, so
the time to get to the top (where the velocity is zero) is 4.25 seconds.
You can find the initial velocity using : V_{2y} = 0 = V_{1y} - 0.379g*t and plugging t = 4.25 seconds. Find V_{1y} . Then use V_{2y}^{2} = 0 = V_{1y}^{2} - 2*0.379g*(change in y) and find change in y. (b) see part (a) (c) See figure 2.28. The section of the graph you are interested in is between 0 and 3 seconds. In the current problem you replace 3 seconds by 8.5 seconds and g by 0.379g. |
53. (a) Use V_{2y}
= V_{1y} - g*t
and change in y = V_{1y} *t -
(1/2)*g*t^{2} for the two times given. Remember to
add 40.0 m to change in y. Note the first velocity is +5.00 m/s. (b) Use change in y = - 40.0 m = V_{1y} *t - (1/2)*g*t^{2}. and solve the quadratic equation for t. Note the first velocity is +5.00 m/s. (c) Use V_{2y} = V_{1y} - g*t and the time t from the previous part. You can also use V_{2y}^{2} = V_{1y}^{2} - 2*g*(change in y), where change in y = - 40.0 m, in order to get the second velocity. Note take the negative square root. The first velocity is +5.00 m/s. (d) Use V_{2y}^{2} = V_{1y}^{2} - 2*g*(change in y) and set the second velocity, at the top, equal to zero. Find change in y. Remember to add 40.0 m to your answer. (e) See figure 2.28 as a guide. |
56. (a) Use V_{2y} = V_{1y}
- g*t . Note the initial ( first) velocity is negative even
though the speed is given: V_{1y}= - 15.0 m/s. (b) change in y = V_{1y} *t - (1/2)*g*t^{2}.
Note the initial ( first) velocity is negative. |
59. (a) From #48, we know the total time of
flight = 2* V_{1y} / g, where V_{1y } is the
initial velocity. Solve for the initial velocity of the faster stone and then the initial velocity of slower stone. Then use time of flight = 2* V_{1y} / g again to get the time of flight of the slower stone. (b) Use V_{2y}^{2} = V_{1y}^{2} - 2*g*(change in y) and set the second velocity, at the top, equal to zero. Note if we are dealing with the slower stone, the change in y = H. Solve for the initial velocity of the slow stone in terms of H. Then find the initial velocity of the faster stone in terms of H by using the relationship between the initial velocity of the fast stone and the initial velocity of the slow stone. |
60. In this problem you will encounter the
square root of 2. (a) Let the first coconut (A) be dropped from H and the second coconut (B) dropped from H/2. V_{A}^{2} = 2gH and V_{B}^{2} = 2g(H/2) are the squares of the speeds when each hits the ground. Take the ratio, then square root to get the relationship between the two speeds at impact with ground. Note |V_{A} | = V. Write the B coconut speed in terms of V. (b) |V_{B} | = gT and |V_{A} | = g*time. Find the time of the A coconut in terms of T using the relationships you found in part (a). Your answer will involve a square root of 2. |
73. SOLVE A VERTICAL FREE FALL
EQUATION FOR TIME OF FLIGHT OF EGG. THEN SUBSTITUTE INTO PROFESSOR'S
UNIFORM HORIZONTAL MOTION EQUATION. In particular, we can set the downward to be the positive y direction. In this case, change in y = V_{1y} *t + (1/2)*gt^{2}. Since the egg is dropped, V _{1y} = 0. Solve for the time t. Note the change in y is positive but is less than 46.0 m since the egg hits the head of a 1.80 m tall professor. Thus you must subtract 1.8 m from 46 m to get change in y. The plug time t into the formula for the horizontal displacement of the professor: x = V*t, where V is the horizontally directed l speed of professor as she moves to the right. |
75. (a) Use V_{2y} = V_{1y}
- gt and find the the requested time t given V_{2y} = + 20
m/s. (b) Repeat for V_{2y} = - 20 m/s . (c) V_{2y} = - 40 m/s (d) V_{2y} = V_{1y} - gt . Set the second velocity equal to ZERO and find t. (e) See figure 2.28. The section of the graph you are interested in is between 0 and 3 seconds. |
79. To begin this problem find the
height and velocity the rocket goes before shutting off the
engine. change in y = V_{1y} *t + (1/2)*(2.50 m/s^{2})*t^{2}. and V_{2y} = V_{1y} + (2.50 m/s^{2} ) t, where the first velocity is ZERO since the rocket begins from rest. t = 20.0 seconds. (a) Use V_{By}^{2} = V_{Ay}^{2} - 2*g*(change in y). Set the speed at point B, the highest point, equal to ZERO. Find the change in y and add it to the value of change of y you got when the engine was on. Note: The V_{Ay }= V_{2y} you found above. (b) At the highest point the velocity is zero but the acceleration is not zero. (c) The time can be found in two pieces: 1. while the engine is on and 2. while it rocket is rising and falling in free fall (engine shut off) . 1. The first time is 20.0 seconds 2. You know the total change in y from shut off to when it hits the ground. It equals the negative of the vertical displacement between launch and shut off. You calculated the vertical displacement between launch and shut off at the beginning the problem above: change in y = (1/2)*(2.50 m/s^{2})*t^{2}, where t = 20.0 s. Solve the following quadratic equation for t', the time the rocket is in free fall. -change in y = V_{2y } *t' - (1/2)gt^{'2} . The total time is 20.0 s + t'. You can get the velocity when it hits the ground this way: V_{3y} = V_{2y } - gt^{'} . There are other ways to solve this problem without a quadratic equation which will be discussed in class. |
83. Let t_{1 }= time to hit the
bottom of cliff and t_{2} = time the sound takes to get from
bottom of cliff to your ear at the top. Let H be the unknown cliff height. We know H = (1/2)*g*t_{1}^{2}. Also t_{2} = H/(330 m/s) and t_{1} = 10.0 s - t_{2} . Substitute and solve for H = (1/2)*g*( 10.0 - H/330)^{2}. Multiply out the right hand side and solve for H. Solve a quadratic equation for H. Clearly if you neglected the speed of sound, you would have overestimated the cliff height. |
61. See example 2.12 |
64. Hint: If no wind is blowing, the trip is simply 2*L/V, where L = 2,000 mi and V = 600 mi/h. If the wind is blowing west to east, relative to the ground the plane has larger speed than on the way to Chicago, and relative to the ground the plane has a smaller speed from Chicago to SF. Thus on the way to Chicago, the plane takes less time and on the way back the plane takes more time. |
CH. 3 Intro to Projectile Motion |
USE EQUATIONS 3.12, 3.13, 3.14 AND 3.15
PAGE 77. AND SEE FIGURE 3.11 WHICH WE REVIEWED IN CLASS. We assume up is the positive direction, such that the acceleration is straight down and has component a_{y} = -g. THERE IS NO ACCELERATION IN THE X DIRECTION. THAT IS WHY THE X-COMPONENT OF VELOCITY IS CONSTANT. |
12. See example 3.3. THIS JUST
LIKE THE SIMULATION LAB. Referring to the equations mentioned above, we
see for the x-motion, x = V_{o}*cos0*t = V_{o}*t since the launch angle is zero. Thus V_{x} = V_{o}. Y-MOTION: y = -(1/2)g*t^{2} since sin0 = 0. V_{y} = -g*t since sin0 = 0. y = -9.00 m = -(1/2)*gt^{2}. Solve for t. x =1.75 m = V_{o}*t. Solve for V_{o}. |
18. This is admittedly a sophisticated
problem. It may be helpful to do #56 first, which simulates figure 3.11.
For the stone, the initial velocity components are: Vox = V_{o}*cos theta = 15 m/s Voy = -Vo sin theta = - 20 m/s. Note the y-component of velocity is negative. Also note we do not need the initial launch angle theta in order to find the requested info. X-MOTION: x = (15 m/s)*t. Y-MOTION V_{y} = -20 m/s - g*t and y = -20 m/s*t - (1/2)*g*t^{2}. (a) You are given the time t = 4.00 s. Plug this time into the y-motion equation for the vertical displacement to get y = -20 m/s*t - (1/2)*g*t^{2}. We are assuming y = 0 at the initial level of the balloon. Take the absolute value of the answer to get the initial height H of the balloon: H = |y| (b) x = (15 m/s)*t (c) This is a little tricky. The distance of the rock from the balloon is given by the Pythagorean Theorem. You need to get the x displacement of the rock from the balloon first which is given by x = (15 m/s)*t, where you plug in t = 4.0 seconds. Now you need to get the y component of displacement of the rock from the balloon. That simply is the difference between the height you computed in part (a) and the vertical distance moved by the balloon in 4.0 seconds: 20.0 m/s*t - H. The distance D between the rock and balloon is given by D^{2} = x^{2} + (20 m/s*t - H)^{2}. Take the square root of both sides to get D. |
56. Study examples 3.4. and 3.5. We will review this in class. V_{x} will be constant for the entire trip so the plot V_{x }vs t will be a straight horizontal line (zero slope) . The plot of V_{y} vs t will look like that of figure 2.28 (b), where the slope = -g. |