Copy of PhysPracticeProblems

PDF

School

University of Alabama *

*We aren’t endorsed by this school

Course

MISC

Subject

Physics

Date

Nov 24, 2024

Type

PDF

Pages

Uploaded by MajorFlyMaster764

2. Given the following three displacement vectors A, B, and C, find vector D such that the sum of all four vectors is zero. A = 72.4 meters at 32° east of north B = 57.3 meters at 36° south of east C = 17.8 meters south (I SEL PN ) Michelson obtained: ¢ = 186,285 +/- 2.5 mi/s Current value: 186,282.025 mis (using miles units). The speed of light was finally known accurately. What a big number! ) 3. The most distant manmade object is the Voyager 1 spacecraft, launched on 05 Sept., 1977. It is currently ~14.5 billion km from earth, Problem 1. (a) Given that ¢ = 300,000 km/s (rounded), what was the minimum rotation frequency (in rotations per second) required for Michelson to observe light in the detector and, thus, measure the speed of light? (b) At this frequency, what would the detector _ traveling at 61,500 km/h. Anyway, at this speed, how many years would it observe? A continuous stream of light or flashes of light? At other frequencies, what would take to reach the nearest star system (a-Centauri, around 4.5 light years it observe? (c) Estimate the accuracy with which Michelson had to know the distance ? between he mountain peaks. away)? 7. Two people are pushing on a 6 kg block of ice, as shown below. The 4. A small object moves along the x-axis with acceleration: friction is negligible. (a) what is the acceleration of the ice? (b) what is a (t) = _(0 032 m/s’)(lS Os — t) the magnitude of the normal force? H g g R At t = 0, the object is at x = —14.0 m and has a velocity of 8.0 m/s. What is 13N 11N the x-coordinate of the object when t = 10 s? (from Y&F) 60~ 30° < 00; - Xo = -14 m £ -0 e vo=8m/s g -0.2 t=10s 8. A truck tilts its bed upward to allow a 95 kg crate to slide off. Given a -0 3 coefficient of static friction of 0.429, at what angle will the crate start g 5 We want x, sliding? 0O 2 4 6 8 10 12 14 tme (s) From the notes p.7, we have: x, =X, +I(: v.(t)dt and v, =v, + I‘:a,(t )dt A truck company image 9. A basket of flowers is suspended from two cables as shown. The mass 5. You throw a rock down a 100 meter deep well, with an initial vertical is 6.2 kg. What is the tension in each wire? velocity of 10.0 m/s downward. (a) What is the rock’s time of flight? T 40° 2 T (b) What is the rock’s speed just before it hits the bottom of the well? 10. Atwood’s machine is a mass and pulley design as shown. 6. From Y&F: In the middle of the night you are standing a horizontal i e e distance of 14.0 m from the high fence that surrounds the estate of your o, e, CHecK ATWEOD rich uncle. The top of the fence is 5.00 m above the ground. You have TERSLES taped an important message to a rock that you want to throw over the fence. The ground is level, and the width of the fence is small enough to be ignored. You throw the rock from a height of 1.60 m above the ground and at an angle of 56° above the horizontal. (a) What minimum initial speed must the rock have as it leaves your hand to clear the top of the fence? (b) For the initial velocity calculated in part (a), what horizontal distance beyond the fence will the rock land on the ground? Giventhat M1 =3.1kgand M2 =4.4 kg, find the acceleration of the two masses. Note: we can treat the cable and pulley as very light compared to the masses. 13. A racing cart accelerates around a horizontal circular track with a radius of curvature 11. A Bugatti Veyron has a minimum curb weight of 1838 kg. When a roadway is banked of 200 meters. During the time it is in the curve, the cart increases its speed uniformly ; i y at the proper angle, a car can get around a curve of radius 85 m without needing any from 36 km/: t0 30 km/h ovfr: period Of, 2 secrnds.. L 5 friction from the tires. What is this bank angle for the Bugatti traveling at its maximum (a) What is the magnitude of the tangential acceleration over this interval? speed of 407 km/h? (b) What is the magnitude of the centripetal acceleration and the total acceleration when the cart has reached the maximum speed? (c) What is the angle between the direction of the total acceleration and the radial acceleration at this moment? (d) What is the angle through which the cart turned during the 20 second interval, and the length of the race track it covered. 14. A spring having a spring constant of 150 N/m is hung vertically. When unstretched, the spring has a length of 25 cm. A 2 kg mass is then hung from the spring. What is th ilibri . f th ith h . it of 12. From YF: A 3.0 kg box that is several hundred meters above the earth’s surface is (a) What s the equilibrium position of the mass with respect to the suspension point o suspended from the end of a short vertical rope of negligible mass. A time-dependent the spring? upward force is applied to the upper end of the rope and results in a tension in the rope (b) If the mass is set into oscillation, determine the angular frequency ® and the of T(t) = (36 N/s)-t (T is tension, t is time in seconds). The box is at rest initially. The only frequency f of the oscillation. forces on the box are the tension in the rope and gravity. (a) What is the velocity of the (c) Is the motion linear or angular? Why is there an angular frequency? box at (i) 1.0 s and (ii) 3.0 s ? (b) What is the maximum distance that the box descends below its initial position? (c) After what time does the box return to its initial position?

15. A rope is laid out straight on a flat horizontal table, with one end hanging off the 18. Knowing that g = 9.8 m/s? at the earth’s surface, make a plot showing the acceleration due to gravity as a function of distance, starting from the center of the earth and going out to a distance of 4 times the Earth’s radius. If you are using a calculator instead of Python, then you can just calculate 10 or 20 points and plot them. Show the equations you use. Treat the earth as a uniform sphere. 19. From YF: Two stars in a binary star system orbit in circular orbits around their common center of mass. The first star, Alpha, has an orbital speed of 36 km/s. The second start, Beta, has an orbital speed of 12 km/s. The orbital period is 137 days. What are the masses of the two stars? table. The coefficient of static friction between the rope and the table is p.. What maximum fraction of the length of the rope can hang over the edge without the rope slipping? White the answer in terms of pu_. Plot your results using Python, Matlab, or Mathematica. 16. A 260 gram mass is attached to a vertical spring and put into motion having a period of 1.12 seconds. (a) by how much does the mass stretch the spring when it is in equilibrium? (b) if this experiment were repeated on the surface of the moon, where the gravitational acceleration is about 1/6 of what it is on earth, then how would the period change? 17. United Airlines Flight 826, “On December 28, 1997, United Airlines Flight 826, a Boeing 747-100, was flying from (now Narita International Airport) to Honolulu International Airport, Hawaii. Two hours into the flight, at an altitude of 31,000 feet (9,400 m), the plane received reports of severe clear-air turbulence in the area and the seat belt sign was turned on. Moments later, the aircraft suddenly dropped around 100 feet (30 m), seriously injuring 15 passengers and 3 crew members. The plane turned around and landed safely back in Tokyo, but one passenger, a 32-year-old Japanese woman, died.” — Wikipedia. The maximum “g-force” was about 1.8g. Treating the planes motion as harmonic (cosine-like), what was the plane’s oscillation period and what was its maximum vertical speed? 20. YF 6-28. A block of ice with mass 2.00 kg slides 1.35 m down an inclined plane that slopes downward at an angle of 36.9° below the horizontal. If the block of ice starts from rest, what is its final speed? Ignore friction. 21. YF 6-94. An airplane in flight is subject to an air resistance force proportional to the square of its speed But there is an additional resistive force because the airplane has wings. Air flowing over the wings is pushed down and slightly forward, so from Newton's third law the air exerts a force on the wings and airplane that is up and slightly backward. The upward force is the lift force that keeps the airplane aloft, and the backward force is called induced drag. At flying speeds, induced drag is inversely proportional to v2 so the total air resistance force can be expressed by F,. =V + b/, where a and b are positive constants that depend on the shape and size of the airplane and the density of the air. For a Cessna 150, a small single- engine airplane, a = 0.3 Ns?/m? and b = 350,000 Nm?/sZ. In steady flight, the engine must provide a forward force that exactly balances the air resistance force. (a) Calculate the speed at which this airplane will have the maximum range (that is, travel the greatest distance) for a given quantity of fuel. (b) (b) Calculate the speed (in km/h) for which the airplane will have the maximum endurance (that is, remain in the air the longest time). 24. YF 13-79. Interplanetary Navigation (hard). The most efficient way to send a spacecraft from the earth to another planet is to use a Hohmann transfer orbit (Fig. P13.79). If the orbits of the departure and destination planets are circular, the Hohmann transfer orbit is an elliptical orbit whose perihelion and aphelion are tangent to the orbits of the two planets. The rockets are fired briefly at the departure planet to put the spacecraft into the transfer orbit; the spacecraft then coasts until it reaches the destination planet. The rockets are then fired again to put the spacecraft into the same orbit about the sun as the destination planet. (a) Find the difference in the orbital energy of a small spacecraft of 900 kg mass, ignoring air resistance and assuming 100% efficiency? (b) How long does a one-way trip from the earth to Mars take, between the firings of the rockets? (c) To reach Mars from the earth, the launch must be timed so that Mars will be at the right spot when the spacecraft reaches Mars’s orbit around the sun. At launch, what must the angle between a sun—Mars line and a sun—earth line be? Orbit of Mars Information: M., =2.0x10% kg Rearth-sun = 150 million km R nars-sun = 228 million km Mspacecvaft =900 kg Tmars = 687 earth days 26. YF 8-42. A 5.0 g bullet is fired horizontally into a 1.2 kg wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.20. The bullet remains embedded in the block, which is observed to slide 0.31 m along the surface before stopping. What was the initial speed of the bullet? How much kinetic 22. YF 6-84 modified. Can profs fly? All birds, independent of their size, must maintain a minimum power output of 10 watts per kilogram in order to fly by flapping their wings. (a) The Andean giant hummingbird (Patagona gigas) has mass 70 g and flaps its wings 10 times per second while hovering. Estimate the amount of work done by such a hummingbird in each wingbeat. (b) A 75 kg prof can maintain a power output of 1.4 kW for no more than a few seconds; and his personal FTP (functional threshold power, the average power he can maintain over 60 minutes) is 292.5 W. If he attached fake wings and flapped his arms, could he fly? Explain. energy was dissipated in the impact? 27. YF 8-49. Canadian nuclear reactors use heavy water moderators in which elastic collisions occur between the neutrons (mass = 1 amu) and deuterons (deuterium) of 23. YF 6-24. You throw a rock with a weight of 3.0 N vertically into the air from ground level. You observe that when it is 15 m above the ground, it is traveling at 25 m/s upward. Use the work—energy theorem to find (a) the rock’s speed just as it left the ground and (b) its maximum height. mass = 2 amu. (a) What is the speed of a neutron, expressed as a fraction of its original speed, after a head-on, elastic collision with a deuteron that is initially at rest? (b) What is its kinetic energy, expressed as a fraction of its original kinetic energy? (c) How many such successive collisions will reduce the energy of a neutron to 1/59,000 of its original value?

28. From YF 13-65. This is a real world raindrop problem. It is hard, so | don’t really expect people to solve it, but hopefully everyone will understand the answer. So when it rains, the initial little tiny droplets fall through the cloud, starting from rest at their moment of formation and gaining more and more mass (starting from zero) as they fall. While a droplet falls, obviously it also gains speed. At first, we will ignore wind resistance. (a) Write down the force-momentum equation for this situation. (b) Simplify the above equation as much as possible, given that the mass increase can be written as m = kx (in other words, the mass increases linearly with the distance fallen). (c) Using the equation for (b), find the acceleration of the raindrop. Remember, it starts from rest. Why is it smaller than g? (d) How far does the raindrop fall after the first 3 seconds since it started to fall? (e) What is the mass of the raindrop after the first 3 seconds after it started to fall, if k = 0.02g/m? (f) What would the equation in (b) look like if we included wind resistance? (g) What if the raindrop didn’t start from rest — in other words, it already had some initial velocity upon formation? 29. YF 10-36. An airplane propeller is 2.08 m in length (from tip to tip) and has a mass of 117 kg. When the airplane’s engine is first started, it applies a constant net torque of 1950 Nm to the propeller, which starts from rest. (a) What is the angular acceleration of the propeller? (b) What is the propeller’s angular speed after making 5.0 revolutions? (c) How much work is done by the engine on the propeller during the first five revolutions? (d) What is the average power é output of the engine during the first 5.0 revolutions? (e) What is the instantaneous power output of the motor at the instant that the propeller has turned through 5.0 revolutions? 30. YF 10-48. Suppose that a dense iron asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and become distributed uniformly within the earth’s outer crust. What would have to be the mass of this asteroid, in terms of the earth’s mass, for the day to become 25.0% longer than it presently is as a result of the collision? Assume that the asteroid contributes negligibly to the total volume of the earth, and that the earth is a uniform sphere. 31. An astronaut on the moon is asked to confirm that the acceleration due to gravity on the moon’s surface is actually 1.62 m/s2, as predicted. To do this, one can roll a solid cylinder down an inclined plane (this can be done in the moon-lander) and Answers: Lecture 2 Problem 2. Find vector components: Sketch the problem: g A4 =724c0s(90-32)=384m| A, =724sin(90-32)= 61.4m| B, =573cos(-36) = 46.4m B, =57.3sin(-36)=-33.Tm C, =0 C,=-17.8m since A, +B,+C,+D, =0and A +B +C +D =0, D,=-A -B, -C,=-384-464=-848 D,=-A,-B, -C,= -61.4+33.7+17.8 = -99 We must be careful! The angle is positive, but both components are Problem 1 Considering every other distance very small, the total distance light must travel is 70 km. The time to do this is 70km / (300,000 km/s) = 2.33 x 10“*s. To get flashes of light in the detector, the mirror must rotate at least 1/8 of a rotation in this time. So, the total time for a complete rotation is 8 x 2.33 x 104s = 1.87x103s. The rotation frequency is just inverse of this, or 536 Hz. The actual measurements had to be good to at least five significant digits (since the final accuracy was to a few parts in 100,000. So he had to know the distances between the mountains down to around 10 centimeters! This required an amazingly accurate survey to be conducted. At other frequencies, the detector would observe nothing — only darkness. At the right frequency, it would get flashes of light appearing every 1/8 of a turn (at 8 times the frequency of the mirror rotation). What if the mirror were 4 sided? Then it would have to rotate at twice the frequency, but there would only be half the number of flashes per cycle. Interestingly, the flash frequency is inverse to the light travel time only! 33. YF 37-43. A space probe is sent to the vicinity of the star Capella, which is 42.2 light-years from the earth. The probe travels with a speed of 99.3% of the speed of light. An astronaut recruit on board is 19 years old when the probe leaves the earth. What is her biological age when the probe reaches Capella? 34. YF 37-5. The negative pion is an unstable particle with an average lifetime of 26 ns, when the pion is almost stationary. (a) If the pion is made to travel at very high speed relative to a laboratory, its average lifetime is measured to be 420 ns. Calculate the speed of the pion expressed as a fraction of c. (b) What distance, measured in the laboratory, does the pion travel during its average lifetime? 35. YF 37-31. What is the speed of a particle whose kinetic energy is equal to (a) its rest energy and (b) five times its rest energy? 36. YF 37-50. The distance to a particular star, as measured in the earth’s frame of reference, is 7.11 light-years. A spaceship leaves the earth and takes 3.35 years to arrive at the star, as measured by passengers on the ship. (a) How long does the trip take, according to observers on earth? (b) What distance for the trip do passengers on the spacecraft measure? rd 37. YF 37-21. Two particles in a high-energy accelerator experiment approach each other head-on with a relative speed of 0.89c¢. Both particles travel at the same speed as measured in the laboratory. What is the speed of each particle, as measured in the laboratory? 38. YF 37-45. A spaceship moves past the earth at a speed of 0.80c toward the west. Another spaceship approaches moving to the east. Both spaceship crews measure their relative speed of approach to be 0.98c. What mass would the crews of both spaceships measure for the standard kilogram, kept at rest on the earth, (a) according to classical physics and (b) according to the special theory of relativity? Problem 3 It’s just units conversion, but the answer is a bit scary. Many ways to do it. time = distance/speed distance 45LY speed 61,500km/h converting years into seconds 300,000km \ ( 60s | [ 60min | ( 24h | [ 365day 45years | ——— | — || —— || =— | - i s min h day year | 4.26x10"km 61,500km/ h " 61,500km / h =692,254h Converting this back will give you about 79,024 years.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Answers: Lecture 2 Problem 2. Find vector components: Sketch the problem: A, =T724c0s(90-32)=38.4m B, =57.3cos(—36) = 46.4m C.=0 C,=-17.8m since A, +B,+C,+D, =0and A +B +C +D =0, D,=-A -B -C,=-384-464= -848 D,=-A;-B,-C,= -61.4+33.7+17.8= -9.9 From slide 5 of the Sec 1 notes: R=1(-84.8m) +(~9.9m)" =85.4m tan@ =-9.9m/-85.4m 0=6.61 quadrant! Problem 3 It's just units conversion, but the answer is a bit scary. Many ways to do it. time = distance/speed dis tan ce 45LY speed — 61,500km /[ h converting years into seconds 300,000km \ ( 60s | [ 60min \ [ 24k | ( 365day 4.5years | ——— || — || ——|:| =—|:| —=——= : 2 s min h day vear ) 4.26x10"km 61,500km/ h " 61,500km / h =692,254h Converting this back will give you about 79,024 years. A, =724sin(90-32) =61.4m B, =57.3sin(-36) =-33.Tm We must be careful! The angle is positive, but both components are negative! So we are pointing to the 3¢ Problem 7. This should be pretty straightforward. You should draw an FBD. Let’s find the horizontal net force and the acceleration: 2F_ =ma_=13cos(60)—11lcos(30)=6a, a_=—0.5m/s> Soitaccelerates to the left. As for the normal force, it will be upward: XF, =0=-13sin(60)—11sin(30) —mg + N N =75N Problem 8. Again, the FBD: This will be the max angle at which we have equilibrium. Even one arcsecond higher and it N slides. £ Parallel: ZF,=mgsin@—u N =0 “~mgsin0 Perp: / I XF =N-—-mgcos@=0 » mgcosO mg So, XF,=mgsin@— umgcos =0 then g, =tan€ @ =tan '(0.429)=23.2" Problem 9. We should know the drill now. T,sin40 Horiz: ZF, =T, -T,c0os40=0 Vert: IF, =T,sin40-mg =0 Plug in numbers Substitute: T, =72.5N ZF,,=T,——.m—g—cos40=O T, =94.6N sin 40 Problem 10. Here we have a pulley. Multiple pulley problems can be very complex, but this single pulley is pretty easy. It simply turns the direction but it does not change anything else. The accelerations are the same and the tension is constant in the cable. FBDforM1 FBDforM2 XF,, =T-M,g=Ma LFy,=M,g-T=Ma Pulley just turns the direction of straight cable. These eqns can l be solved by summation (or by substitution) IF,, +ZF,, =(M:‘M1)g=(M1+M:)“ Mg M,g a=(_M!__M'.).g=|_7,,,/S2 Mzg (Ml + M: ) a| T T la If we choose to do it by substitution, and ignore the direction change of the pulley (which implies that the tension is in the same direction!), then we get these equations: The - sign is due to the downward acceleration in this case. It doesn’t just “drop out” as a2 = -al because if you try to keep al and a2 separate, there will be too many variables. To solve it you need the extra information, that a2 = -al. LF,=T-Mg=Ma IFy,=T-M,g=-Ma Letting T = M, g + M,a from the first eqn and subbing it into the second equation leads to the same result as before. Problem 11. The Bugatti. Draw FBD Horiz: N |

ne =3Ign is due 10 the downwara acceieralion in this case. ZE,,, =T- Mlg = Mla It doesn’t just “drop out” as a2 = -al because if you try to keep al and a2 separate, there will be too many variables. ZF:‘, 3= T-M 8= -M ,a To solve it you need the extra information, that a2 = -al. Letting T = M,g + M,a from the first eqn and subbing it into the second equation leads to the same result as before. Problem 11. The Bugatti. Draw FBD Horiz: LF, =Nsinf=ma, Vert: LF, =Ncos@-mg=0 gV (113.05m/s) = > Sol 0':86.30 gr 9.8Im/s 85m Set equal to centripetal acceleration and substitute as usual: 2 13(a): the final speed is 90 km/h which is 25 m/s, and the initial one is 10 m/s. a,(t):%%’-: 25m/s—10m/s = 0.75m/ ¢ s p b t)m L == 3.13m/ s (b) a.(r) =200 mls The total acceleration is the vector sum of a_and a,. Note that a_is center pointing and a, is tangential, so they are at an angle of n/2. So, (d) 0=co[,t+—l-art2 2 vy vy =—‘:19t+—l— r '3=lOm/s20s+125m/s—lOm/s r 2 t 200m 2 200m 20s =1.75rad S=rf=200-1.75=350m 14a. Just balance the spring with an FBD. kx = mg, so x = mg/k =2*9.8/150 = 0.131 m. That's the distance of the stretching. But the distance from the hanging point is then 0.131 +0.25 = 0.381 m. 14b. w=k/m=8.66rad /s ,'=;=l-38H: 14c. The motion is entirely on a line, it is linear. We think of an angular frequency because the motion follows a cosine function. It looks exactly the same as the “x component” of an object going in uniform circular motion. In other words, the x- component of an object going in UCM (with by definition a constant angular frequency) is given by x = Acos(0). Note that this implies that Vk/m: is unitless. You should check this. The only part of the rope contributing to the friction 1s the fraction on the table. The fraction off the table does not contribute to friction, 15. It looks like this: 5 = y—£=% 0 - ‘ L X [ fe—1-1—s| };F':ng—,u'Nzo - | got this plot: . X L-x 208 ' I £=M=M=/, (l—i) Eo“ / w L mg mg L 02 / X u, R R T T L 1+ i 12. The forces on the box are mg downward, and there is a time-dependent tensional force upward. This means that there will also be a time-dependent acceleration, and we need to find position and velocity. We have seen this before. It will involve integration. a,(!)=2—F-’4ﬂ ZF,(t)=T(t)—mg =ma (1) s a,(!)=%—g m Therefore, a,(f)= SOV (st -g=12m/s’ t-9.8m/ s’ 3kg We have to integrate this to find Av = v, = v, = v (since v, = 0): J121° v (0)=[120-9.8= iy OB = 61* —9.81 (a) Att=1s,v,(t) =—3.8 m/s and at 3s v,(t) = 24.6 m/s We have to integrate again to find Ay =y, =y, =y, (since y, can be set to 0): 6r' 9.8’ g f———= 2 -4.9¢* 3 2 At the lowest position, v, will be zero, so 6t? = 9.8t, for t = 1.63s (b) At t = 163s, y(t) = 2(1.63)* - 4.9(1.63)2 =~ 4.36 m (c) Setting y = 0 and solving for t yields 2t* = 4.9t? and solve fort = 2.45 s (1) =J.(:612 -9.8& = _4r'm 4r°0.26kg ™ (1.12s) =31.2cm 16a. T = 27r~/m/k So, _mg 02698 ko 818 =8.18kg / s 16b. The period shows no dependence on the gravitational acceleration (see formula). So the period on the moon is the same as on earth. 17. a,, =40’ W= 2—” =qa, /A4 d 2 2 T = e = e = 8,28 Janm 14 Jl.8~9.8m/s“ 30m Vo = A0 = axd = b =23m/s T 8.2s That was the exact plane g inside ~ This tells us that the acceleration is linear with radius as you go into the earth. _ GM Outside it goes as a (r)? Pet(str, 31, (r, 8, 9), Bontds S A PlotTheme « “nice”, Framelabel = (“earth radit®, "¢ (w/s')7}] R* function. Knowing that g, ;... = 9.8 m/s?, we can _ finalize a piecewise function: "‘E 98~ r<R > R g=128 r>R 18. Recall that the gravity inside a uniform spherical shell is zero. Therefore, within the earth, you will feel a force only from the part of the ball remaining below you. The solution to this is therefore a piecewise function —in other words, the equation is different below and above the surface. Below the surface: 4 s GM GEIIR P 4 —_ = ~ =—'GIZR R R 3 P | got the following using Mathematica. You can use python. slr.,2) l-'l«ulul{{o.li, . ¢n}, {‘.:_).'.‘ . ..}”

19. v,= 36 km/s v =12 km/s m, T=137 days We want the masses. Strategy: 1. Use v = 2nR/T to find the orbital radius of each star. 2. The magnitude of force on each star must be equal. We write an expression for the force: 2 2 P This tells us that R,M, = R,M,, because each one feels the same centripetal force. 3. Set this force equal to gravity force: _ GMM, 4x’RM, 4n’RM, £ (Rl + R: )2 T3 TZ 4. Solve these for M; and M,: Mo M N o M Can plug in numbers here or... GT* . GT* 5. Add the masses: 47° (R +R,) (M, +M,)= —orr 6. Putting in numbers gives M, + M, = 3.12 x 103° kg 7. Solve. From step 2, R, M =M, | So, '-x Q‘Jx ,=3.12:10% Using the orbital radii from step 1, replace subscripts 1 with o and 2 with 3 and you get M, R p,=3.12.10% s A R/' M= 7.8 x 10 kg Mg=2.34 x 10*°kg 20. This should be pretty easy: just find the amount of PE the block lost, let that equal to the amount of KE it gained, and then find the speed. h=1.3551n(36.9) = 0.81m PE = mgh = 2kg x9.8m/ s* x0.81m =15.9J = 28 - M =3.99m/s d m 2kg 21a. At first glance, we probably have no idea what to do. So we better boil it down to the simple things. Basically, we know that the airplane produces some constant forward force, and wind resistance is opposing it. Therefore, the distance it can travel will be greatest when the opposing force of wind resistance is the smallest. Maybe we should take a look at how this strange wind resistance equation looks. So | simply plotted it: | noticed that it has a minimum around 30 m/s. To find the exact value, 0.3 5000 therefore, | set the derivative equal to —~ 4000 Sl % 3000} %:“““Z—@:zav—zbv-’ =0 2000¢ 1000 Solve it for v: 0 20 40 60 80 109=(—b-) =(M) =329m/s=118km/h a v (m/s) 21b. Here, we need the plane’s engine to be working for the longest time before it runs out of fuel. This means that we must find the speed at which the least amount of energy is lost to air resistance per unit time. In other words, at what speed will air resistance do the least work per unit time? W, d=F,, -vt=(av'+b/v)t So: W, /t=av'+b/v alr m We might suspect a similar kind of solution. There will be some minimum value of W, /t, the work done by air resistance per unit time. Set the derivative equal to zero: {(av’ +bv' ‘__(‘"I;_) =0=3av =b/V Solving this for v as before : av 14 14 v =(i) =(M) =25m/s=90km/h 3a 0.9 This is the speed at which the airplane stays in the air for the longest time, although it doesn’t get as FAR as it would have at the 118 km/h. 22a. We know the power per kg and the power, so we can calculate the power, and then the energy per flap: 0.7J s X —x——=0.07J/ fla s 10 flaps Jep L4 x0.07kg =0.7W kg 22b. We just have to see if a human bird is possible, using myself as the example. 292.5W Sadly, to be a bird | need about 10 W/kg to fly, as 75k >0 /k8 explained in part (a). In a sprint, | can do 1.4k kW for about 5 seconds. This corresponds to 18.6 W/kg. So, with fake wings as good as a hummingbird’s wings, | could fly for about 5 seconds (at least, assuming | could maintain the same power with my arms as | can with my legs)! 23. This is an easy example of energy conservation. E= mgh+ my” —3le5m+5—kgx(25m/s) =140.66J . ,25 ’2xl4066_303m/s 3/9.8 _E_140.66J b) h,.. = =46.9m 3N Problem 5

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

25. This one is about the conservation of mechanical energy. Set equal and solve for v, v, = J‘.j _ﬂ+m—M =6.83x10"m /s L ¥y 26. Start by finding the amount of work done by friction on stopping the block. This will let us know the total original mechanical energy of the system. W, =Fd=puNd=(0.2)(1.205-9.81)(0.31)=0.733J This must be the amount of KE the bullet-block had originally just after impact. So: Vi = oo —2XO'733-IIOn1/v o \lm \} 1.205 ' ’ The collision was inelastic (they stuck not bounce) so by conservation of momentum: Note: you can’t just assume that the work done by m,v,, 1.205-1.10 friction was the same as the kinetic energy of the =t = ——————=265m/ s bullet originally, because the collision is inelastic, In other words, a bunch of the bullet’s energy is lost of] sound, heat, and deformational energy in the block. My e The amount of KE lost in the impact was: 1 g A 5 1 g ; AKE =2 My Vius =5 MaViy = 0005265 ~21.205-1.10° =1757 27. This problem is an elastic collision problem. From page 18 of the notes: _| m—m, F : A _|1=2 _ Ve a) v, = m,+m, Vio since v, is zero. This gives: v, = :2‘ Vio = 3 The neutron has 1/3 of its original speed. b) Since KE goes as v?, it means that 1/3 of the speed is 1/9 of the KE. " (l)"_ 1 (l)"_ ) 9 59000 9 59000 Log (9") = Log (59000) = nLog (9) _ Log(59000) ~ Log(9) SO 9" =59000 28. Slow and steady. In this problem, we see that the momentum of the raindrop is changing, because BOTH its speed AND its mass are increasing. Therefore: a) F=d—p=m‘£+vﬂ dt dt dt d(k ; s b) F=kxﬂ+"—(—v)=k.vd—‘+\'k£=hd—‘+k\*'=mg=kxg dt dt dt dt dt dv 2 dv 2 kx; +kv® =kxg Dividing both sides by k: xj: +Vv =xg c) To do this, we need v = at and x = (1/2)at? I 5 sog A e N V- 1, - —ata+at =—at' g d) \’—Eﬂf —ggl =14.7m 2 2 3 22 _ 1 7 g% "p® e) m=kc=0.02x14.7 =0.294g g s a =5 dv 2 2 2 f) kx —+ kv = kxg —cv” 1 dt a=—g Itis %maller Wi & Bacause it pick You would have to include the v, terms in part c. But this would imply all these s(alionafy water mlolleczlles.ns . 8) that the raindrop had an initial velocity when it had zero mass, which seems 31. The easiest way to do this one is by energy conservation this time: by Vo g 51 Y mgh, = KE,, ,+KE,, , = Emr; + Elm,' = Emv; +E(5mr‘)[7'J Solve this forvz: 1? = -4-gh So a plot of v2vs h has a slope of 4g/3. The 3 slope would thus be 4(1.62)/3 = 2.16 m/s2. C v A Obviously, either choice will decrease his net weight by the same amount, However, the moment of inertia of the wheels is very important. The tires contribute majorly to this, because they are on the wheel rims, By creating a larger moment of inertia, a great deal more effort and energy has to be expended on bringing the wheels up to speed when accelerating on the bike. Similarly, even decelerating would be harder as well - the brakes have to work harder to stop the large angular momentum of the wheels. https://www.schwalb com/technology-fag/rolling o/ 33, L, =42.2ly I I }/ = = = = = 8 47 JI-vi/et J1-0.993 t,=t/y=50y Age is thus 19 + 5 = 24 years Problem 6 -v. kdczl ém 1. =0 =) pa- \j; i O x,‘:'ILf‘x. |.6m oy =19 8 m/s '———'N#m_'\,_y|.._. Ax = O VeV, +at wL.en x=/4m then Y= S5m d=vt+3at’ V‘I s ‘V.z ‘20()' "v.:—) 8:56‘ As ofden +he case, hime 15 the “C\g We have info at+ +he fence x=Vt Y= Vgt -4t = 5-16 =34 So, d34=V.unS6. _ 1% _ 49 __ )4 VoCos 56 V,* cos™ 56 Sunplqu J% tan 56-3. 4 = 49 ~|4? Vi cos'5é solve for Ve Vo =| 49~y = 133 m/s cos*56 (1ytan 56 -3 L&) That's +he answer +o part a. NOW L)) we L‘now Voy »Qy and A'.j_ So vse d=v,t ‘;‘_a‘l’.!L agan, Ay ==lbm =V t-49t = 1335056t - 49t I+ q'uao{(a‘hc. -4qt*+ |llot *1.6=0 Solve this +o get t=238s or -013s So #inally , x= Vit =133¢c0s56~2.38 = 17.8m and +he distance Lcsonal the fence 1s 178-140=3.8m 25. This one is about the conservation of mechanical energy. E, = lmv(f " GMm E;u I—m ’ _ GMm 2 7, 2 r, Set equal and solve for v;. v, = [v —M+M =6.83x10"m/s % ¥y

24a. The kinetic energy of a satellite (earth or mars) in a circular orbit is given by KEml = l sat = 2 2 R 2R GM, The gravitational potential energy of a satellite is given by PE = — Rm GM GM GM Orbital energy at the earth’s distance: £, . = s _ LU RN P 2 R.-.uvh sn Rnu-lh—mu 2 R«-unh sun GM_m Orbital energy at Mars’ distance: E. .m —T"'—‘“‘ The difference in energy: £, —E, = CM e OM aMoae _ ) 36.10° satk - 2R mars - sun 2R carth—sun In reality, the rocket must overcome the earth’s gravity also, but will get some of that energy back from the gravity of mars, just not as much because Mars is smaller. The rocket must also provide the orbital energy we just calculated. 24b. From the diagram, we see that the Hohmann orbit has a semi-major axis equal to (1/2)(Ruah-sun + Rmarssun) = 189 million km. And we see that it takes one half period to go there in this orbit. 2 1.89-10")" g, - 227 = - ) — = 2.24x10’s = 259days 2JGM,,, \/(6-67'10 )(2.0-10") About 8.5 months 24c. During these 259 days, Mars will pass through a certain angle, 6, in its orbit. We can look up the orbital period of Mars or calculate it with the information given (satellite orbit centered on the Sun), giving a Martian year equal to 687 earth days. 259days 687days 0=2n =237rad about 135.8 degrees In the same time, the spacecraft completes a half elliptical orbit as shown, for a total angle of rad or 180 degrees. This means that the angle we seek is 180 — 135.8 = 44.2 degrees. So it is like this: 29. This is more of a routine question to get up to speed with the variables. The main thing to think about is which moment of inertia to pick. The propellor is most like a rod rotated about its center, so: ;- — ;7?2 12 T 1950 Nm a) a=—= 1 = I 5 117kg (2.08m)’ =46.2rad | s* b) @’ =W +2a0 Since @, =0, w=2a0 =2.46.2rad / s* 10xrad =53.9rad | s c) W=10=1950-107r =6.13x10*J 6.13x10"J d) P ZALL = w = - tw (@, —@,)/a (539rad/s—0)/46.2rad /s avg =52.5kW e) Here we need the power at a particular moment. Remember our linear formulas: W = Fd = Fuvt W/t=P=Fy W, =10 = rart P = 1w =1950Nm -53 9rad / s = 105kW 30. This is about the change n the Earth’s moment of inertia, while the angular momentum is conserved. The asteroid is being treated as a point mass (no extra volume). Consider the Earth: 2 > Ls=1w Loy = gfwmnl.R;mm Tosren =27 @y If the asteroid becomes distributed uniformly throughout the earth’s crust, it would form a thin spherical shell, so: 2 2 lu.\l - ——MuuR‘:.u 3 By conservation of angular momentum: 2 7 2 2 2 2 g M earth Rmrlh Dorino = (g M earth Rmrlh + 3 M ast me. )(Urul'lh\ f Solve for M,,,, setting 1.25@,,,pf = @egreno 5 129@ix S s g'anhR- rth —L = EMe.uth:unh 2 'Z_M.u:R;m/- 5 = wmulh. 4 5 3 25 i B o 2 . Yy R —2rm R 52 Mopra R — 5 M R = EM 2t Rares S cce il R B wiopiibes s, — M 2 2 ass 3 3 Rmrlh M, = %Mmm. To increase the earth’s day by 25%, the asteroid would have to be 15% of the earth’s mass. Treating the asteroid as a point mass would have given 10%.

34. a) We can find gamma: Then the speed: v=c\/l—l/y2 =c\1-1/16.15" =0.998¢ b) d=vi=0.998¢x420-10"s =126m 35. Notes slide 35: a) £y =moc , Setting these equal gives y= 2. Then we need to solve for KE=(}/—l)m0C' speed. 1 V= [ 50 = ,,1_1/ 2 J1=vi/e? P 4 v =0.866¢ b) Smyc® =(y—1)myc® Thistime y=6sov=0.986¢c 36. a) L =7.”ly 3.35y ; t=701y/v=pt, = =t ’0 - 3.35-‘) \‘l - v' /(" 2 7:1 lI_V : 29,3 Solving this for v (rememberc=1ly/y): V' = K (l—v /¢ ) 35y v =4.504(1-1") v =0.905ly/ y = 0.905¢ Y= —= 2.35 t=At, =787y J1-0.905 . i i b) L=—==——=3.03LY ¥ 35 o, m NP 1-w,/c* We'll assign particle 1 the S’ system and the lab is S. Then: g v,” is the speed of particle 2 with respect to particle 1. v, is the speed of particle 2 measured in the lab. Q— —Q v is the speed of particle 1 measured in the lab. S vV, =V V_,—(—V,) _2v, 2 v, i T R 2 g 2 2 =0-89(' l=w_ /c¢® 1=v (-v,))/c" 1+v./c" 1+v /c° This is a quadratic, see as follows: [ _2v =0.89¢ Solve this quadratic with 1+v2/¢c? a=0.89 Lo b=-2c 2v, =(1+v.; /c‘)0.89c c=0.89c2 2v, =0.89¢ +0.89v) /¢ , bt Jb? —4ac X 2a 0.89¢% - 2cv, + 0.89vf ={) L _2c£v4c -4-0.89-0.89¢7 _ 2¢+0.912¢ ' 1.78 1.78 This yields two solutions of which only one is possible. v, = or 1.64c. Alternatively, sy, v v, = i t= —=— =(.89¢c s S l=w /" _1+v. /c” @ 0.89¢> +2cv, +0.89v> =0 v, =or —-1.64c —Q S 38. For this one, we will need the speeds of both the spacecraft. a) Classically, the speed doesn’t affect the mass so they would observe 1 kg. b) For spaceship A, it is moving at 0.8c relative to the earth, so y= 1.67 m=ym,=1.6Tkg For spaceship B, we have to figure out its velocity, like in earlier questions. We'll assign spaceship A to the S’ system and the Earth is S. Then: v,” is the speed of Ship B with respect to Ship A. -0.98c v, is the speed of Ship B with respect to the earth (we want this). v is the speed of S’ relative to earth. 0.8¢ — ®— = v -f-v _ —0.98¢ +0.8¢ __0.833c Yol+w, /¢ 1+(0.8¢)(-0.98¢)/ ¢ so y=1.81and m=1.81kg

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Copy of PhysPracticeProblems

Related Documents