Aldo Bautista, Homework 3 Due: Sep 20 2005, 4:00 am Inst: Maxim Tsoi This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 3) 10 points Consider the setup of a gun aimed at a target (such as a monkey) as shown in the figure. The target is to be dropped from the point A at t = 0, the same moment as the gun is fired. The bullet hits the target at a point P, which is at the same horizontal level as the gun. Let the initial speed of the bullet be v0 , let the angle between the vector v0 and the horizontal (x-) direction be , and OP= L and AP=h. The gravitational acceleration is g. Denote the time taken to hit the target by T. The acceleration of gravity is 9.8 m/s2 .
A
1
Solution: The time T is the time it takes for the monkey to fall from A to P, a distance h. For any object falling a distance h from rest (neglecting air friction), we have from (1): 1 0 - h = 0 + (-g)T 2 2 or T = 2h g
v0
O
002 (part 2 of 3) 10 points Find the initial speed v0 (magnitude of the vector v0 ) which allows the projectile to meet the target at location P. (Hint: T defined in part 1 is also the time taken for the bullet to travel, following the projectile trajectory, from O to P). Let the distance OP be L = 2.91 m, the angle = 46.1 , and the time T = 0.785577 s. (Given g = 9.8 m/s2 ). Correct answer: 5.34219 m/s. Explanation: We divide v0 into components: vx = v0 cos vy = v0 sin ()
P
This time is given by 1. T = 2. T = 3. T = 4. T = 5. T = 6. T = 3 7. T = 2 gh. 2h . g h . g 2h . correct g h . 2g gh. gh. or
The x-velocity is unchanged (there is no horizontal acceleration) so we simply use (1) where a = 0 and the bullet is traversing a distance L: L - 0 = v0x T + 0 v0x = L/T Now we use () to find the speed v0 : v0 = v0x L = cos T cos
With the given values we find v0 = 2.91 m m/s 0.785577 s cos(46.1 ) = 5.34219 m/s
Explanation: Basic Concepts: Constant acceleration: 1 x - x0 = v0 t + at2 2 v = v0 + at (1) (2)
003 (part 3 of 3) 10 points
Bautista, Aldo Homework 3 Due: Sep 20 2005, 4:00 am Inst: Maxim Tsoi Now the same setup is to take place at some planet where the gravitational acceleration is g = g/4. Keep v0 , and h to be the same as before. Find the new height, i.e. the y-coordinate of the new point of collision. (Hint: you should convince yourself that for this new case, the time taken for the bullet to travel from O to the new point of collision P' should still be T ). h 4 h 2. y = 2 h 3. y = 3 1. y = 4. y = 2h 5. y = 3h correct 4 2h 6. y = 3 h 7. y = 5 3h 8. y = 5 9. y = h 10. y = 3h Explanation: If g = g/4, we would expect the bullet to hit its target at a higher point, since the target will not fall quite as fast. However, as noted in the hint, the x-motion is still unaccelerated so T will still be the same. So the only difference is that the event, while taking the same amount of time to transpire, will occur at a height yP instead of yP = 0. 1 yP = h - g T 2 2 But h is the height the target fell in the previous setup: 1 h = gT 2 2 so 3h h 1 g T2 = h - = yP = h - 2 4 4 4
2
Alternative solution: If one found the right answer in Part 1, it can be utilized in the following, completely equivalent solution: T = is the same as T = 2h g 2(h - yP ) g
Thus, setting these expressions equal and squaring both sides, 2h 2(h - yP ) = g g (the negative root is unphysical since we know yP < h). But g = g/4 so we have, cancelling the 2's, h 4(h - yP ) = g g or 4h - 4yP = h so the solution is yP = 3h 4
004 (part 1 of 2) 10 points A ship cruises forward at vs = 5 m/s relative to the water. On deck, a man walks diagonally toward the bow such that his path forms an angle = 13 with a line perpendicular to the boat's direction of motion. He walks at vm = 3 m/s relative to the boat.
vm
vs
At what speed does he walk relative to the water? Correct answer: 6.38346 m/s.
Bautista, Aldo Homework 3 Due: Sep 20 2005, 4:00 am Inst: Maxim Tsoi Explanation:
3
vm
v
vs
When you complete the parallelogram, the resultant velocity v with respect to the water is the side of the triangle opposite the obtuse angle, which has a measure of = 90 + . Let vs be the velocity of the ship, vm be the velocity of the man, and v be the resultant velocity of the man relative to the water (Earth). By the law of cosines
2 2 v 2 = vm + vs - 2 vm vs cos 2 2 v = vm + vs - 2 vm vs cos 1/2
005 (part 2 of 2) 10 points At what angle to his intended path does the man walk with respect to the water? Correct answer: 49.7471 . Explanation: The law of sines can be used to compute the requested angle , which is the angle opposite the ship's path and velocity. sin sin = vs v vs sin = sin v vs = arcsin sin v 5 m/s = arcsin sin(103 ) 6.38346 m/s = 49.7471 . Alternate Solution: Using vector components from Part 1, we have vy tan = vx vy = arctan vx 2.92311 m/s = arctan 5.67485 m/s = 27.2529 . Therefore the angle between vm and v is = 90 - - = 90 - (13 ) - (27.2529 ) = 49.7471 . 006 (part 1 of 2) 10 points A car travels 29.1 km due north and then 36.2 km in a direction = 45.6 west of north.
N S B R A x W E y
= (3 m/s)2 + (5 m/s)2
-2(3 m/s)(5 m/s) cos(103 )]1/2 = 6.38346 m/s . Alternate Solution: We can analyze the vector addition using the components of the vectors. Note: v = v s + vm , or vx = vs + vm sin = (5 m/s) + (3 m/s) sin(13 ) = 5.67485 m/s and vy = vm cos = (3 m/s) cos(13 ) = 2.92311 m/s Thus the speed of the man with respect to the water is v= =
2 2 vx + vy
(5.67485 m/s)2 + (2.92311 m/s)2
= 6.38346 m/s .
Bautista, Aldo Homework 3 Due: Sep 20 2005, 4:00 am Inst: Maxim Tsoi Find the magnitude of the car's resultant displacement. Correct answer: 60.2605 km. Explanation: The magnitude of the resultant displacement vector, R, can be obtained from trigonometry as applied to the obtuse triangle. Since = 180 - 45.6 = 134.4 and R = A + B - 2 A B cos , we find that for C1 = -2AB cos = -2 (29.1 km)(36.2 km) cos 134.4 = 1474.08 km2 , R= (29.1 km)2 + (36.2 km)2 + 1474.08 km2 = 60.2605 km . 007 (part 2 of 2) 10 points Calculate the direction of the car's resultant displacement, measured counterclockwise from the northerly direction. Correct answer: 25.4169 . Explanation: The direction of R measured from the northerly direction can be obtained form the law of sines from trigonometry: sin sin = . B R Thus sin = B sin R 36.2 km sin 134.4 = 60.2605 km = 0.429201 .
2 2 2
4
A camel sets out to cross the desert, which is 48.6 km wide in the north-south direction. The camel walks at the uniform speed 1.65 km/hr along a straight line in the direction 46.7 north of East (only the camel knows why he chose that particular direction).
N W S E
How long will it take the camel to cross the desert? Correct answer: 40.4722 hr. Explanation: Let the total distance walked by the camel be L; then the camel's northward displacement is LN = L sin while his eastward displacement is LE = L cos . To cross the desert, the camel needs LN = desert width = 48.6 km and hence L= LE sin (48.6 km) = sin(46.7 ) = 66.7791 km ,
and the time he needs to walk this distance is simply T = L v (66.7791 km) = (1.65 km/hr) = 40.4722 hr .
Thus, = arcsin 0.429201 = 25.4169 . 008 (part 1 of 2) 10 points
009 (part 2 of 2) 10 points Calculate the camel's eastward displacement while he crosses the desert. Correct answer: 45.7983 km. Explanation: As we already mentioned, LE = L cos = (66.7791 km) cos(46.7 ) = 45.7983 km .
Bautista, Aldo Homework 3 Due: Sep 20 2005, 4:00 am Inst: Maxim Tsoi 010 (part 1 of 1) 10 points A river flows with a uniform velocity v. A person in a motorboat travels 1.54 km upstream, at which time a log is seen floating by. The person continues to travel upstream for 82 min at the same speed and then returns downstream to the starting point, where the same log is seen again. Find the flow velocity of the river. Assume the speed of the boat with respect to the water is constant throughout the entire trip. (Hint: The time of travel of the boat after it meets the log equals the time of travel of the log.) Correct answer: 0.156504 m/s. Explanation: Let u be the boat's speed in still water and let be the distance traveled upstream in t = 82 min and T be the time of return. Then, for the log, d = v (t + T ) , and for the boat, = (u - v) t + d = (u + v) T . Combining the above gives +d d = + . v u-v u+v Substituting = (u - v) t from the equation above yields the equation for v (u - v) t + d d =t+ . v u+v Simplifying this equation, we obtain v= d 2t 1540 m = (2) (4920 s ) = 0.156504 m/s . 1. 2. 3. 4. 5.
5
the same way by the water flow, the motion of the boat relative to you is independent of the water flow. In other words, if the boat is moving away from you with a speed say u, for a time t, it will be moving back toward you with the same speed as you see it. So the time interval for the entire trip, i.e. moving away from you and coming back to you is 2t. During this time, the distance which the log has d drifted is d. So the drift-speed of the log is 2t . 011 (part 1 of 1) 10 points The velocity of a projectile at launch has a horizontal component vh and a vertical component vv . Note: Air resistance is negligible. When the projectile is at the highest point of its trajectory, which of the following show the vertical and the horizontal components of its velocity and the vertical component of its acceleration in 3 columns? Vertical Velocity 0 vv vv 0 0 Horizontal Velocity vh vh 0 0 vh Vertical Acceleration g correct 0 0 g 0
Explanation: Basic concept: Newton's second law of motion F = ma The only force on the projectile is the gravitational force, which gives projectile a constant vertical acceleration of the magnitude g. There is no acceleration in the horizontal direction, which means at the highest point, the horizontal component of the velocity is the same as the initial value vh . One other obvious thing: the vertical component of the velocity is zero at the highest point, because from the trajectory, the projectile moves horizontally at the highest point.
Digression: The answer can be readily understood in the following way. Imagine you are at rest on the log observing the motion of the boat. You need to convince yourself that, since both you and the boat are affected in
Bautista, Aldo Homework 3 Due: Sep 20 2005, 4:00 am Inst: Maxim Tsoi 012 (part 1 of 1) 10 points Suppose that three balls are rolled simultaneously from the top of a hill along the slopes as shown below. 1 2 3 2 v 0 tf 3 6. R = 2 v0 tf 5. R =
6
Which one reaches the bottom first? 1. 2 and 3 2. 1 and 3 3. 2 4. 1 correct 5. 1 and 2
Explanation: The range R is given by the horizontal velocity times vx the flight time tf . The range is the maximum distance the missile with velocity v0 can travel so the launch angle is 45 v0 and vx = v0 cos(45 ) = . 2 014 (part 1 of 3) 10 points Neglect: Air friction. Your teacher tosses a basketball. The ball gets through the hoop (lucky shot).
18 m /s
73
013 (part 1 of 1) 10 points For anti-ballistic missile system, the time of flight tf is determined by the initial speed v0 of the missile and the maximum range R of the incoming missile. Find their relationship. 1 1. R = v0 tf correct 2 1 2. R = v0 tf 2 3. R = v0 tf 4. R = 3 v 0 tf
Figure: Not drawn to scale. How long does it take the ball to reach its maximum height? Correct answer: 1.75648 s. Explanation: Let : = 73 , v0 = 18 m/s , h1 = 2.397 m , h2 = 3.048 m , and ytop = maximum height of ball s trajectory . Note: The horizontal distance to the basket
2.397 m
Explanation: The one on the left. That ball gains speed quickly at the beginning, where the slope is steeper, so its average speed is greater even though it has less acceleration in the last part of its trip.
3.048 m
7. The balls reach the bottom at the same time.
6. 3
Bautista, Aldo Homework 3 Due: Sep 20 2005, 4:00 am Inst: Maxim Tsoi is superfluous for Parts 1 and 2. This problem has two distinct parts, vertical motion in Parts 1 and 2, then horizontal motion in Part 3. Basic Concepts:
2 v0 sin2 (2) 2g 1 ytop - h1 = v0 sin t - g t2 (3) 2 Solution: Using Eq. 1, the time t1 to reach the maximum height, (i.e., vytop = 0, the velocity at the top) is v0 sin(73 ) t1 = g (18 m/s) sin(73 ) = (9.8 m/s2 )
7
Or symbolically, we have t2 = = 2 [ytop - h2 ] g
1/2
v = v0 sin - g t
(1)
ytop - h1 =
2 v0 sin2 2 [h2 - h1 ] - g2 g 2 sin2 (73 ) (18 m/s) = (9.8 m/s2 )2
1/2
2 [(3.048 m) - (2.397 m)] - (9.8 m/s2 ) = 1.71824 s .
1/2
The total time of the ball's trajectory is t = t 1 + t2 = (1.75648 s) + (1.71824 s) = 3.47472 s .
= 1.75648 s . y1 = 1 2 gt 2 1 1 = (9.8 m/s2 ) (1.75648 s)2 2 = 15.1176 m , or v 2 sin2 = 0 2g (18 m/s)2 sin2 (73 ) = 2 (9.8 m/s2 ) = 15.1176 m , so = y1 + h 1 = (15.1176 m) + (2.397 m) = 17.5146 m .
016 (part 3 of 3) 10 points What is the horizontal length of the shot? Correct answer: 18.2864 m. Explanation: The horizontal length of the shot is = v0 cos t = (18 m/s) cos(73 ) (3.47472 s) = 18.2864 m .
ytop
015 (part 2 of 3) 10 points How long does it take the ball to reach the hoop? Correct answer: 3.47472 s. Explanation: The time t2 for the ball to decend from the top into the basket is t2 = 2 [ytop - h2 ] g
1/2
017 (part 1 of 4) 10 points Denote the initial speed of a cannon ball fired from a battleship as v0 . When the initial projectile angle is 45 with respect to the horizontal, it gives a maximum range R.
y 45 R/2 R x
2 [(17.5146 m) - (3.048 m)] = (9.8 m/s2 ) = 1.71824 s .
1/2
Bautista, Aldo Homework 3 Due: Sep 20 2005, 4:00 am Inst: Maxim Tsoi The time of flight tmax of the cannonball for this maximum range R is given by 2 v0 3 g v0 2. tmax = g v0 3. tmax = 4 g v0 4. tmax = 2 g 1 v0 5. tmax = 3 g v0 6. tmax = 2 correct g 1 v0 7. tmax = 2 g 1 v0 8. tmax = 4 g v0 9. tmax = 3 g 1 v0 10. tmax = 2 g Explanation: The cannonball's time of flight is 2 v0y 2 v0 sin 45 v0 t= = = 2 . g g g 1. tmax = 018 (part 2 of 4) 10 points The maximum height hmax of the cannonball is given by 1. hmax 2. hmax 3. hmax 4. hmax 5. hmax 6. hmax 7. hmax v2 = 2 0 g 1 v2 = 0 2 g v2 = 0 g v2 =4 0 g 2 1 v0 = 2 g 2 2 v0 = 3 g 1 v2 = 0 3 g
2 v0 g 2 1 v0 correct 9. hmax = 4 g v2 10. hmax = 3 0 g Explanation: Use the equation
8
8. hmax = 2
2 2 vy = vy0 - 2 g h .
At the top of its trajectory vy = 0. Solving for h yields h=
2 vy0
2g v 2 sin2 45 = 0 2g 2 1 v0 = . 4 g
019 (part 3 of 4) 10 points The speed vhmax of the cannonball at its maximum height is given by 1. vhmax = 2. vhmax 1 v0 4 = 3 v0
3. vhmax = 4 v0 1 4. vhmax = v0 3 5. vhmax = 2 v0 6. vhmax = 7. vhmax 1 v0 2 1 = v0 correct 2
8. vhmax = 2 v0 9. vhmax = v0 2 v0 3 Explanation: 10. vhmax =
Bautista, Aldo Homework 3 Due: Sep 20 2005, 4:00 am Inst: Maxim Tsoi At the top of the cannonball's trajectory, vy = 0. Hence the speed is equal to vx . 1 |v| = vx = v0 cos 45 = v0 . 2 020 (part 4 of 4) 10 points At a new angle, , the new range is given by R R = . 2 The corresponding angle, , which is greater than 45 , is given by 1. 72 < 74 2. 70 < 72 3. 74 < 76 correct 4. 78 < 80 5. 60 < 62 6. 64 < 66 7. 76 < 78 8. 66 < 68 9. 68 < 70 Explanation: The maximum range corresponds to when = 45 R= Thus for R =
2 v0 sin[2 (45 )] v2 = 0. g g
9
You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 3 m at your feet, then a horizontal shelf of 7 m , then another drop of 7 m to the bottom of the canyon, which has a horizontal floor. You kick a 0.33 kg rock, giving it an initial horizontal velocity that barely clears the shelf below. The acceleration of gravity is 9.8 m/s2 . Consider air friction to be negligible. v 3m 10 m 7m x What initial horizontal velocity v will be required to barely clear the edge of the shelf below you? Correct answer: 8.94614 m/s. Explanation: Let : y1 = 3 m , y2 = 7 m , x = 7 m. and y= 1 2 gt 2 2 y1 = g t 2 1 so
7m
10. 62 < 64
Basic concepts: Since all of the vertical motion is down, we can consider down to be positive, and deal with positive gravity, vertical velocities, and distances. Thus voy = 0, and vertical distances are defined by
R , we need 2 sin[2 ] = 1 . 2
There are two solutions for that satisfy the above = 15 or = 75 .
There is an initial horizontal velocity but aox = 0, so horizontal distances are defined by x = vox t Solution: Once launched, the vertical motion defines the time to drop to the first shelf
021 (part 1 of 2) 10 points
Bautista, Aldo Homework 3 Due: Sep 20 2005, 4:00 am Inst: Maxim Tsoi t1 = 2 y1 g
10
Thus the horizontal distance traveled is x = vox t1 x vox = t1 =x so
g 2 y1 (9.8 m/s2 ) 2 (3 m)
= (7 m)
= 8.94614 m/s . 022 (part 2 of 2) 10 points How far from the bottom of the second cliff x will the projectile land? Correct answer: 5.78019 m. Explanation: The time to drop the distance y = y1 + y2 to the bottom of the cliff is t= 2 [y1 + y2 ] , g
1 the speed when the maximum height is 5 1 projectile is at its maximum height. 5 What is the initial projection angle? Hint: First verify the following relation1 2 2 2 2 ships vx = vx0 + vy , where vy = vy0 - 5 1 2 h 2gh = (1 - )vy0 . 5 5 Correct answer: 79.6533 . Explanation: First find the y component of the initial velocity in terms of the maximum height, h, and gravity, g.
2 2 vy = vy0 - 2gh 2 0 = vy0 - 2gh
The y component of the velocity at 1/ 5 the maximum height is 1 2 2 vy(1/5h) = vy0 - 2gh 5 1 = 2gh - 2gh 5 vy(1/5h) = We are given, 2gh 1 - 1 5
vy0 =
2gh
and the total horizontal distance traveled in that time is x2 = vox t = vox 2 [y1 + y2 ] g 2 [(3 m) + (7 m)] 9.8 m/s2
= (8.94614 m/s) = 12.7802 m .
1 speedmax height = speed 1 height 5 5 At the maximum height the speed of the projectile is simply due to the x component of the velocity. Note that the x component of the velocity is constant. Hence vx = = 1 5 1 5
2 2 vx + vy(1/5h) 2 vx + 2gh 1 -
Thus the projectile will land a distance x = x2 - x = (12.7802 m) - (7 m) = 5.78019 m , from the bottom of the cliff. 023 (part 1 of 1) 10 points The speed of a projectile when it reaches its
1 5
Solving for vx yields vx =
1 2gh 1 - 5 (5)2 - 1
Finally, since we now have an expression for vx and vy0 , we can find the initial projection angle.
Bautista, Aldo Homework 3 Due: Sep 20 2005, 4:00 am Inst: Maxim Tsoi vy0 = vx 2gh (5)2 - 1 (5)2 - 1 (1 - 1 ) 5 vws
11
= tan
-1
2gh(1 - 1 ) 5
= 79.6533
024 (part 1 of 2) 10 points A river flows at a speed vr = 5.12 km/hr with respect to the shoreline. A boat needs to go perpendicular to the shoreline to reach a pier on the river's other side. To do so, the boat heads upstream at an angle = 40 from the direction to the boat's pier. Find the ratio of vb to vr , where vr is defined above and vb is the boat's speed with respect to the water. 1 vb = vr cos2 v 2. b = tan vr vb 3. = cos2 vr v 4. b = sin2 vr v 1 5. b = vr tan vb 1 = correct 6. vr sin 1 v 7. b = vr sin2 vb 1 8. = vr cos vb = cos 9. vr vb 10. = sin vr Explanation: 1. Let : vbs =? , boat relative to shore vws = vr , water relative to shore vbs = vb , boat relative to water . This is a problem about relative velocity. See the figure below.
The velocity of the boat relative to the shore is given by - =- +- . vbs v v ws bw It is easy to see from the figure above that vws , vbw and vbs form a right triangle if the boat moves northward relative to the earth. Therefore, vws = vbw sin v 1 = bw = . vws sin 025 (part 2 of 2) 10 points If the time taken for the boat to cross the river is 11.6 min, determine the width of the river. Correct answer: 1.17968 km. Explanation: By similar reasoning, we know relative to the shore, the velocity of the boat is vws vbs = = 6.10178 km/hr . tan Therefore the time taken for the boat to cross the river is given by L vbs L = vws tan L tan . = vws vws t = L = tan (5.12 km/hr) (11.6 min) (1hr/60 min) = tan 40 (5.12 km/hr) (0.193334 hr) = tan 40 = 1.17968 km . t=
v bw
vbs
= tan-1
L
Bautista, Aldo Homework 3 Due: Sep 20 2005, 4:00 am Inst: Maxim Tsoi 026 (part 1 of 1) 10 points A science student is riding on a flatcar of a train traveling along a straight horizontal track at a constant speed of v1 . The student throws a ball along a path that she judges to make an initial angle of with the horizontal and to be in line with the track. The student's professor, who is standing on the ground nearby, observes the ball to rise vertically. How high does the ball rise? 1. h = 2. h = 3. h = 4. h = 5. h = 6. h = 7. h = 8. h = 9. h = 10. h =
2 v1 tan2 g 2 v1 cos2 2g 2 v1 sin 2 2g 2 v1 sin cos 2g 2 v1 sin cos g 2 v1 tan2 correct 2g 2 v1 sin2 g 2 v1 cos2 g 2 v1 sin 2 g 2 v1 sin2 2g
12
where vbc = velocity of the ball relative to the car, vce = v1 , velocity of the car relative to the earth, vbe = velocity of ball relative to the earth. vbe = vbc + vce . Since is the initial angle, this diagram is only for the case where t = 0 . Since vbe has zero horizontal component, we have vbex = 0 = vbcx + vcex and the vertical components give vbey = vbcy + vcey , so vbcx = -vcex = -v1 , and vbcy = vbey = v0y . Denote the highest point the ball rises by h,
2 2 vy = v0y + 2 a y ,
2 0 = v0y - 2 g h
h=
2 v0y
2g
.
For the initial angle of the ball as observed by the student in the car is tan = so we get h= vbcy v0y = , vbcx -v1
2 v1 tan2 . 2g
Explanation: y v1 = vce
vbc
vbe
O
x