Unformatted text preview: Massachusetts Institute of Technology  Physics Department Physics8.01 Fall 1999 Solutions for Assignment # 2
by Dru Renner Problem 2.1
(a)
g : We need to mathematically describe the motion of the rst stone. This is onedimensional motion, in the vertical direction, with a constant acceleration due to the Earth's gravity. Label the vertical direction with the axis pointing up from the surface of Earth and with the origin ( = 0) at the surface of Earth. If we set 0 = 15 0 m/s and = 9 8 m/s2 and choose the time = 0 to coincide with the throwing of the stone, then the motion is = 0 1 2 2 Now we can ask, when is the rst stone at the height = 11 0 m? This question leads to the equation
x x x v : t x v t gt h : h = v0 t 1 2 gt 2 which is a quadratic equation in with the two solutions
t = v0 q t v0 g 2 2 gh =) t = 1:22
x t s and
v t+ = 1 84 s
: 0 If we label the quantities for the second stone 0 , 0 , and 0 , then similarly
x 0 = v 0 t0 1 gt02
0 2 But you must be careful here: 0 is the amount of time elapsed since the second stone was thrown and the second stone is thrown 1.00 s after the rst stone is thrown. So there is a delay of = 1 00 s between the throwing of the stones, and thus we can write the relationship between and 0 as
t t : t t t 0 =t
t t Now at the times = , or 0 = , the two stones are required to hit, implying that 0 = . This leads to the equation
t t t t x h 1 h = 0( v
0 t t ) 1 ( 2
g t t )2 with the solution
v0
0 = h 1 +2 ( g t t t )2
v
0 t Problem 2.2
d gt So for the two possible values , there are two possible velocities 0 : = 1 22 s with 0 = 51 1 m/s = 114 mph and + = 1 84 s with 0 = 17 2 m/s = 38 5 mph The speed of 114 mph would be a very good fast ball and almost as good for a tennis serve ( which could be nearly 140 mph ), therefore the plausible answer is 17.2 m/s = 38.5 mph. (b) Now you must wait 1.30 s before throwing the second stone. The equation above, 1 )2 ) ( ) still works, but now = 1 30 s. The rst thing you 0 = ( + 2 ( should notice is that = 1 22 s is less than = 1 30 s. So the rst ball reaches the height , for the rst time, before you even throw the second ball. So the only possibility is to strike the rst ball at the second time + = 1 84 s with 0 = 23 0 m/s = 51 mph
t t : v
0 : t : v 0 : : v 0 h g t t = t t t : t : t : h t : v 0 : If an object experiences free fall for a length of time then the distance it falls is given by = 1 2. Thus measuring both and provides a value for : 2
t t d g g = 22
t d g d If the error for the measurement of is and the error for the measurement of is then, by our simple method, the error for is
d t t 2( + ) ( )2
d d t t g where the rst quantity represents the largest value of consistent with and . The data from lectures, the resulting values for g, with the experimental error, and the consistency with the value of = 9 80 m/s2 for Boston are given below. ( Both lectures make the identical measurement for the 3.000 m drop. )
g d t g : 2 Distance Dropped d = 3:000 0:003 m d = 1:500 0:003 m d = 1:500 0:003 m Time of Flight t = 0:781 0:002 s t = 0:551 0:002 s t = 0:550 0:002 s Experimental Value for g g = 9:84 0:06 m/s2 g = 9:88 0:09 m/s2 g = 9:92 0:09 m/s2 Consistency yes yes no Problem 2.3 The last value for g is not consistent with the value of g = 9:80 m/s2. This is most likely due to an underestimate of the error for d. The way the apple hangs and rotation of the apple as it falls are factors that are hard to account for, so the error of 0.003 m is probably too small. ~ ~ ~ ~ ~ The sum A + B + C is graphically performed by: (1) drawing A, (2) `sliding' B so ~ lies on the tip of A, (3) `sliding' C so that the tail of C lies on the tip ~ ~ ~ that the tail of B ~ ~ ~ ~ ~ of the `new' B , and (4) drawing the vector, A + B + C , from the tail of A to the tip of ~ the `new' C . ~ ~ The vector C is drawn along the direction opposite to C with the same length as ~ ~ ~ ~ ~ C . Then, the sum A + B C is graphically performed as above but with the vector C ~. used in place of C
N N C A+B+C
B C
E W C B A B C B A
W A+BC
E C Problem 2.4 ~ The vector A has the components: Ax = 5:0, Ay = 3:0, and Az = 1:0. The magnitude is given by p jAj = Ax + Ay + Az = (5:0)2 + ( 3:0)2 + (1:0)2 = 35:0 = 5:9
2 2 2 q q The angle x ~ between A and the xaxis can be found from the relation Ax = jAj cos
3
x which gives
x = cos1 Ax = cos1 p5:0 = 32:3 jAj 35:0 Ay = cos1 p 3:0 = 120 jAj 35:0
! ! ! ! ! ! Similarly,
y = cos1 and
z = cos1 Az = cos1 p1:0 = 80:3 jAj 35:0 Problem 2.5 You are given the vector The magnitude is jvj = 49 = 7, so the vector p ~ = 3x 6y + 2z v ^ ^ ^ ~ 1v v0 = jvj ~ = 3=7x 6=7y + 2=7z ^ ^ ^ has the same direction as ~ but magnitude jv0j = 1. Finally the vector v ~ ~ ^ ^ v00 = 2v0 = 6=7x 12=7y + 4=7z ^ Problem 2.6 ~ has the same direction as v0, which has the same direction as ~ , but now has magnitude v 00 j = 2. jv
You are given the distances to Venlo (31 km) and Eindhoven (39 km), but you are not given any information concerning the directions to these cities. The greatest distance between the two cities is 31 km + 39 km = 70 km. This occurs when the two cities are in opposite directions. The shortest distance between the two cities is 39 km  31 km = 8 km. This occurs when the two cities are in the same direction. The actual distance is 47 km, which is greater than 8 km and less than 70 km. 4 Problem 2.7
you must satisfy You are given the vectors Problem 2.8 You are given the vectors ~ ^ ^ ^ and B = 2x + 1y 3z The calculations for parts (a), (b), and (c) are all done `componentwise.' ~ A = 5x 3y + z ^ ^ ^ (a) ~ ~ A+B = (
= 5 + 2 )x + ( ^ 3x 2y 2z ^ ^ ^ 3 + 1 )y + ( 1 + 3 )z ^ ^ ~ ~ ~ ~ In fact it is true that A B = B A for any two vectors: (e) ~ ~ A B = ( ( 3)( 3) (1)(1) )^ + ( (1)(2) ( 5)( 3) )y + ( ( 5)(1) ( 3)(2) )^ x ^ z = 8x 13y + 1z ^ ^ ^ ~ ~ B A = ( (1)(1) ( 3)( 3) )x + ( ( 3)( 5) (2)(1) )y + ( (2)( 3) (1)( 5) )^ ^ z ^ = 8x + 13y 1z ^ ^ ^ = (8^ 13y + 1z x ^ ^) ~ ~ ~ ~ In fact it is true that A B = B A for any two vectors:
5 ~ ~ B A = (2)( 5) + (1)( 3) + ( 3)(1) = 16 (d) ~ ~ A B = ( 5)(2) + ( 3)(1) + (1)( 3) = 16 (c) ~ ~ 2A 3B = ( 2(5) 3(2) )^ + ( 2(3) 3(1) )y + ( 2(1) 3(3) )z x ^ ^ x ^ ^ = 16^ 9y + 11z (b) ~ ~ A B = ( = 5 2 )x + ( ^ 7x 4y + 4z ^ ^ ^ 3 1 )y + ( 1 3 )z ^ ^ =) =) ~ ~ A B AxBx + Ay By + Az Bz 2( 1) + ( 3)(a) + 0( 5) a = 0 = 0 = 0 = 2 3 ~ ^ ^ ~ A = 2x 3y and B = 1x + ay 5z ^ ^ ^ ~ ~ Now you must choose a value of a that makes A and B perpendicular. Mathematically Problem 2.9 We are given the vectors ~ ^ ^ A = 2x 3y
and we want to nd a vector and ~ B = x + 4y 5z ^ ^ ^ ~ ^ ^ ^ V = ax + by + cz ~ ~ ~ ~ that has unit length ( jV j = 1 ) and is perpendicular to both A and B ( V A = 0 and ~ B=0) ~ V
=) =) =) =) =) =) ~ ~ V A 2a 3b a = 0 = 0 = 3b 2 ~ ~ V B = 0 a + 4b 5c = 0 3 5c = 0 ( using a = 3 b ) 2 b + 4b 2 b = 2c a = 3c ( using a = 3 b ) 2 ~ V = 3cx + 2cy + cz ^ ^ ^ ~ So now we know that V has the form ~ Now imposing the condition that V is normalized to unity gives the equation
=) =) =) ~ V 9c + 4c2 + c2 14c2 c
2
j j = = = = 1 1 1 p 1= 14 So this nally gives just two vectors ~ V= p 1 (3x + 2y + z ) ^ ^ ^ 14 ~ ~ ~ ~ Alternatively, we could form A B which is perpendicular to both A and B . Then we could normalize this new vector and include the two possible directions this would ~ give V = jA 1 Bj A B . You can check that this expression is identical to the result ~ ~ ~ ~ above.
6 Problem 2.10
(a) The position as a function of time is given as
~ = (6 2t)x + (3 + 4t 6t2 )y (1 + 3t 2t2 )z r ^ ^ ^ The velocity vector at any time t is given by = = = q p jvj = ( 2)2 + ( 32)2 + ( 9)2 = 1109 = 33:3 So the speed is 33:3 m/s. (c) The acceleration vector at any time t is given by = = = a ~ So in particular at t = 3 a The magnitude of ~ at t = 3 is So the magnitude of acceleration at t = 3 is 12:6 m/s2. 7 q p jaj = (0)2 + ( 12)2 + (4)2 = 160 = 12:6 ~= a (b) The speed at t = 3 is given by the magnitude of ~ at t = 3 which is v = v d~ dt d d x ( 2)^ + dt (4 dt (0)^ + ( 12)y ^ x 12y + 4z ^ ^ 12t)y ^ z ( 4)^ d (3 dt ^ 4t)x 12y + 4z ^ ^ ~ v So in particular at t = 3, ^ z = 2x + (4 12(3))y (3 4(3))^ ^ = 2x 32y + 9z ^ ^ ^ v ~ = r d~ dt d ^ d ^ d ^ (6 2t)x + dt (3 + 4t 6t2)y dt (1 + 3t 2t2 )x dt (2)^ + (4 6(2)t)y (3 2(2)t)z x ^ ^ ^ 2x + (4 12t)y (3 4t)z ^ ^ Problem 2.11
z t Choose the axis pointing vertically up the axis pointing horizontally along the direction of the car and choose the  origin to be the corner of the ramp. Also choose the origin to be the moment when the car is at the corner of the ramp, and let 0x be the unknown speed of the car at that moment. The subsequent motion of the car is projectile motion. The vertical motion of the car is given by
x x z v z = 1 2 gt 2 So we can ask the question when does the car fall the distance = 2 0 m? This gives the equation
h : = 1 2
h gt 2 with solution t u 2 = u 2(2 0m) = 0 64 s ? t = (9 8m s2 )
h : g : : = s v The horizontal motion of the car is given by
x = v0x t so at = ? the horizontal position is
t t x ? ? = v0x t ? For the stunt driver to avoid crashing, So =) =) =)
x x must be larger than the distance = 24 0 m.
d : ? ? > > > > d v0x t v0x v0x q d t? d
2h d g ( using ? =
t q 2h
g ) Therefore, to clear all the cars
v0x > > 37 6 m s
: = p 24:0 22:0 9:8 8 Problem 2.12 We will need the same equation three times, so consider the case of arbitrary ball speed, v0 , arbitrary angle, , and arbitrary range d. The motion will be projectile. The usual quantities v0x and v0z are given by
v0x = v0 cos and v0z = v0 sin It is simplest to choose a coordinate system with the origin on the ball and the usual labels for vertical and horizontal. The vertical motion is 1 z = v0 sin( )t gt2 2 We want to know the time t? when the ball strikes the ground. This gives the equation 1 0 = v0 sin( )t? 2 gt?2 which has the solutions t? =0 and t? = 2v0 sin( )=g of which the nonzero solution corresponds to the ball striking the ground. The horizontal motion is given by
x So the range is given by
d
2 = v0 cos( )t? = 2v0 cos(g ) sin( ) Solving that equation for v0 gives
v0 = 2 cos( gdsin( ) ) s (a) Now we want to determine v0 when = 14 and d = 240 m. v u u t v0 = 240(9:8) = 70:8 m=s 2 cos(14) sin(14) We want to determine d when = 14 and v0 = 70:8 m/s +0:6 m/s = 71:4 m/s. (b) Suppose the ball speed was greater by 0:6 m/s. = v0 cos( )t 9 d 2 = 2(71 4) cos(14) sin(14) = 244 1 m 98
: : : : : The ball will travel 244 1 m 240 0 m = 4 1 m farther. (c) Suppose the angle was greater by 0 5 . We want to determine when = 14 +0 5 = 14 5 and 0 = 70 8 m/s.
: : d : : v : d 2 = 2(70 8) cos(14 5) sin(14 5) = 248 0 m 98
: : : : : : : : Problem 2.13 The ball will travel by 248 0 m 240 0 m = 8 0 m farther. The motion of the skier is identical to the motion of the car 3in problem 2.11, if you choose a similar coordinate system, with 0 = 110 km/h = 11010 m = 30 6 m/s. So the (60)(60)s motion is given by
v : x = v t 0 and z = 1 2 gt 2 (a) The equations above descibe the motion of the skier, but the equations do not know where the ground is. We must supply that information. Normally we choose the coordinates so that the ground corresponds to = 0, but that is not true in this case. The gound is really a hill. It starts at = 0 and = 0, but then it slopes down at an angle of 45, i.e. it is a line through the origin with slope 1. The mathematical description of the line is
z x z zground = xground The skier landing on the slope is mathematically indicated by the intersection of the curve of the skier and the curve of the ground. Suppose this intersection occurs at the horizontal position because this is a point on the ground we know that the correspondingly = So if the skier lands at time = ?, then
x z x t t x x z =) =) = = 0? 1 = 2 ?2
z v t gt
v t 0 ? = = = 10 1 ?2 2 gt 2v0
g t t ? ? 0 or of which the nonzero solution is the one that corresponds to the skier landing. At that time,
x = v0 t ? 2 6m 2 = 2 0 = 2(30 9 8 s) = 191m
v : = g : and So the distance down the slope is
d z = = 191m s
x = = p x 2 + z 2 = 270m Problem 2.14 (b) The skier attains large speeds which make considerations of air resistance necessary. Air resistance makes the actual distance shorter than our calculated distance. This will be explored in more detail in lecture. If the radius of the orbit is = 1 50 1011 m, then the circumference of the orbit is
R : C =2 R = (2)(3 14)(1 50 1011m) = 9 42 1011m
: : : The Earth travels this distance once in the time 3 15 107 s. So it's speed is
: v = 1 year = (365)(24)(60)(60) s =
: = = C =2 R 11 = 9342 10 7m = 2 99 104m s 15 10 s
: : So the centripetal acceleration is given by
R R a Problem 2.15 In problem 2.14 we actually calculated the centripetal acceleration in terms of the radius of the orbit and the period of the orbit. That result, = 4 2 2 , is true for all the planets. This result does assume that all planets have circular orbits, but most orbits are elliptical, so the radius we use is the average distance of the planet to the sun. This was referred to as in the lecture supplement for 9/17/99. Elliptical orbits will be discussed later in the course.
a R= R on the web site http://ask.com and by using the formula above for .
a (a) Here is a table of information found by asking \what is the distance to out planets?" = v 2 = 2 R 2 =4 2 R 2 4 s)2 = (2199 10 mm = 5 97 103m s2 50 1011
: = : : = 11 Planets Mean Radius (m) Mercury 5.791010 Venus 1.081011 Earth 1.501011 Mars 2.281011 Jupiter 7.781011 Saturn 1.431012 Uranus 2.871012 Neptune 4.501012 Pluto 5.911012 Period (s) Centripetal Acceleration (m/s2) 7.60106 3.96102 1.94107 1.13102 7 3.1610 5.93103 5.94107 2.55103 3.74108 2.20104 9.29108 6.54105 9 2.6510 1.61105 5.20109 6.57106 7.84109 3.80106 (b) Below is a loglog plot of the centripetal acceleration versus the orbit radius. The dots show the data points, and the lines connect consecutive points. The axes of your plot may be di erent depending on the units chosen, but the slope of the curve should be the same. ( I expressed the radii in units of 1010 m and the acceleration in units of 106 m/s2. )
Acceleration versus Radius for the Planets  log 5 Mercury
10 log10 Plot 4.5 6 m/s ) log10 ( Centripetal Acceleration/ 10 4 2 3.5 3 2.5 2 1.5 1 Pluto 1 1.5 2 10 log10 ( Radius of Orbit Around Sun/ 10 m ) 2.5 3 0.5 0.5 Later you'll learn how to determine the best curve for a set of data points, but these points seem to lie on such a straight line that we will just assume that the relationship is linear. Now we will use the slope between any two points to make a guess about the true log log 3 slope. For my guess, I used the points for Mercury and Pluto: slope log 539600log 591 80 79 00 2 00.
: : : : 12 You should notice that all the planets, independent of their mass, lie on this curve. This indicates a relationship between radius and acceleration: log = 2 00 log + 2 , where and are constants independent of the mass of which implies that = the planet. ( As was shown in lecture. ) Notice that the expression from problem 2.14 related , , and , but this expression is independent of indicating that there must be additional relationships between and . Here are a few lines of code that will produce graphs similar to the one above. First you need to nd Mathlab. Log onto an Athena workstation and either (1) choose the `Numerical/Math' option from the top, then choose `Analysis and Plotting' and `MATLAB' or (2) at the command line type `add matlab matlab'. Then at the prompt, which should look like ` ', type the following lines.
a : R C a K =R C K a R R >> radii = acc = 5.79, 10.80, 15.00, 22.80, 77.80, 143.00, 287.00, 450.00, 591.00] 39600, 11300, 5930, 2550, 220, 65.4, 16.1, 6.57, 3.80] Problem 2.16
w plot( log10(radii), log10(acc), 'o', log10(radii), log10(acc), '' ) Let be the speed of the wind blowing from A to B be the speed of the plane measured relative to the air and be the distance from A to B. For the trip from A to B, the ground speed (speed of the plane measured relative to the ground) would be
v d v + w and the time it would take to travel that distance would be
t A!B = +
d v w For the trip from B to A, the ground speed would be
v = w and the time of ight would be
t B!A v d w At this point you should notice that we could make B!A be very large by making , which already indicates that the round trip with wind will be longer. The total time of ight would be ! 1 1 = 2 1 + w = A!B + B!A = + 1 2 2
t w v t t t d d v w v w v w =v 13 The travel time without the wind (w = 0) is
t0 = 2d=v so we can write tw as
tw = t0 1 w2=v2 1
! 1 ! For jwj < v, 1 w 2 =v 2 >1 so tw > t0, and the round trip with wind is longer. But for jwj > v ! 1 1 w2=v2 < 0 which would make tw < 0 ! So clearly we should be careful here. The special points w = v cause problems: the above expression for tw becomes in nite. In fact for w = v the ground speed for the B to A portion is v w = 0, indicating that the plane will never be able to leave point B. This means that someone standing on the ground at B will see the plane oating still in the air. ( Where as a bird in the wind will see the plane as moving forward. ) So all considerations of the trip from B to A are irrelevant: the plane never makes that portion of the trip. This is revealed in the expression for tB!A which becomes in nite for w = v. So the analysis above breaks down: we can not talk about the round trip because it doesn't occur. In fact for w > v the ground speed is v w < 0, and the plane continues to be blown further in the A to B direction. If the wind were to blow from B to A instead, the above results would work by considering w < 0. And when w = v (when the wind blows from B to A with speed v), the ground speed from A to B would be v + w = 0. The plane would never even begin its trip since it couldn't leave A. And for w < v the plane immediately is blown further away from B. So for this case the plane never even completes the rst half of the round trip. 14 ...
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 Acceleration, Orders of magnitude, ax, Ay

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