Phys 201A Homewok 5 Solutions 3. In each of the thee cases, you can find the changes in the velocity vectos by adding the second vecto to the additive invese of the fist and dawing the esultant, and by adding the thid vecto to the additive invese of the second and dawing the esultant. (a) 5 since minus is the same as plus which equals (b) Hee, finding the diffeence gives us a change in diection of the velocity vecto that points to the ight, which is choice 1. (c) In this case, the change in diection of the velocity vecto is staight downwad choice 7. 9. The point P is displaced vetically by 2R, whee R is the adius of the wheel. It is displaced hoizontally by half the cicumfeence of the wheel, o R. Since R = 0.450 m, the hoizontal component of the displacement is 1.414 m and the vetical component of the displacement is 0.900 m. If the x axis is hoizontal and the y axis is vetical, the vecto displacement is = ((1.414 m) ˆ i + (0.900 m) ˆ j ). (b) The displacement has a magnitude of = ( πr) 2 + ( 2R) 2 = R π 2 + 4 = 1.68 m and (b) an angle of tan 1 2R πr = tan 1 2 π = 32.5
above the floo. In physics thee ae no exact measuements, yet that angle computation seemed to yield something exact. Howeve, thee has to be some uncetainty in the obsevation that the wheel olled half of a evolution, which intoduces some indefiniteness in ou esult. 13. The diagam shows the displacement vectos fo the two segments of he walk, labeled A and B, and the total ( final ) displacement vecto, labeled Δ. We take east to be the +x diection and noth to be the +y diection. We obseve that the angle between A and the x axis is 60. Thus, the components of A ae A x = (250 m)(cos 60 ) = 125 m and A y = (250 m)(sin 30 ) = 216.5 m. The components of B ae B x = 175 m and B y = 0 m. The components of the total displacement ae Δx = A x + B x = 125 m + 175 m = 300 m and Δy = A y + B y = 216.5 m + 0 m = 216.5 m. (a) The magnitude of the esultant displacement is Δ =Δ = Δx 2 +Δy 2 = (300 m) 2 + (216.5 m) 2 = 370 m. (b) The angle the esultant displacement makes with the +x axis is Δy tan 1 Δx = tan 1 216.5 m 300 m = 36.
(c) The total distance walked is d = 250 m + 175 m = 425 m. (d) The total distance walked, d, is geate than the magnitude of the esultant displacement, Δ. The diagam shows why: A and B ae not collinea. 34. (a) The x-component of a is 3.00 m. (b) The y-component of a is 0 m. (c) The x-component of b is ( 4.00 m)cos 30 = 3.46 m. (d) The y-component of b is ( 4.00 m)sin30 = 2.00 m. (e) The x-component of c is ( 10.0 m )cos120 = 5.00 m. (f) The y-component of c is ( 10.0 m)sin120 = 8.66 m. (g) and (h) Since a has no y-component, q times the y-component of b must equal the y- component of c so q(2.00 m) = 8.66 m and q = 4.33. Also, p times the x-component of a plus q times the x-component of b must equal the x- component of c. pa x + qb x = c x p(3.00 m) + 4.33(3.46 m) = 5.00 m p = 6.66 m 3. We adopt the positive diection choices used in the textbook so that equations such as those in Table 5-1 ae diectly applicable. The initial velocity is hoizontal so that v 1 y = 0 m / s and v 1 x = v 1 = 161 km h. Conveting to SI units, this is v 1 x = 44.7 m/s.
(a) With the oigin at the initial point (whee the ball leaves the pitche s hand), the y-coodinate of the ball is given by y = 1 2 g( Δt) 2, and the x-coodinate is given by x = v 1 x Δt. Fom the latte equation, we have a simple popotionality between hoizontal distance and time, which means the time to tavel half the total distance is half the total time. Specifically, if x = (18.3 m)/2, then t = (9.15 m)/(44.7 m/s)= 0.205 s. (b) And the time to tavel the next 9.15 m must also be 0.205 s. It can be useful to wite the hoizontal equation as Δx = v 1 x Δt in ode that this esult can be seen moe clealy. (c) Fom y = 1 2 g( Δt) 2, we see that the y-coodinate of the ball is 1 2 ( 9.8 m / s 2 )( 0.205 s) 2 = 0.205 m at the moment the ball is halfway to the batte. (d) The ball s y-coodinate when it eaches the batte is 1 2 ( 9.8 m / s 2 )( 0.409 s) 2 = 0.820 m, which, when subtacted fom the pevious esult, implies it has fallen anothe 0.615 m. Since the value of y is not simply popotional to t, we do not expect equal time-intevals to coespond to equal height-changes; in a physical sense, this is due to the fact that the initial y-velocity fo the fist half of the motion is not the same as the initial y-velocity fo the second half of the motion. 11. Taking the y axis to be upwad and placing the oigin at the fiing point, the y-coodinate is given by y = v 1 Δt sin θ 1 1 2 g( Δt) 2 and the y-component of the velocity is given by v 2 y = v 1 sin 1 gt. The maximum height occus when v 2 y = 0. Thus, t = (v 1 /g) sin 1 and v 1g y max = v 1 sinθ 1 sinθ 1 1 2 g(v 1 sinθ 1 ) 2 g = (v 1sinθ 1 ) 2 2 2g. 21. We adopt the positive diection choices used in the textbook so that equations such as those in Table 5-1 ae diectly applicable. The coodinate oigin is at gound level diectly below the elease point. We wite 1 = 37 fo the angle measued fom +x, since the angle given in the poblem is measued fom the y diection. We note that the initial speed of the pojectile is the plane s speed at the moment of elease. (a) We use Eq. 5-7 to find v 1.
y 2 y 1 = (v 1 sinθ 1 ) Δt 1 2 g( Δt) 2 0 m 730 m = v 1 sin( 37 )(5.00 s) 1 2 (9.8 m / s 2 )(5.00 s) 2 which yields v 1 = 202 m/s. (b) The hoizontal distance taveled using Eq. 5-5 is x = v 1 cos 1 t = (202 m/s)cos( 37.0 )(5.00 s) = 806 m. (c) The x-component (hoizontal) of the velocity (just befoe impact) is v x = v 1 cos 1 = (202 m/s) cos( 37.0 ) = 161 m/s. (d) The y-component (vetical) of the velocity (just befoe impact) is v 2 y = v 1 sin 1 gt = (202 m/s) sin ( 37 ) (9.80 m/s 2 )(5.00 s) = 171 m/s. 23. We adopt the positive diection choices used in the textbook so that equations such as those in Table 5-1 ae diectly applicable. The coodinate oigin is at gound level diectly below impact point between bat and ball. (a) We want to know how high the ball is fom the gound when it is at x = 97.5 m, which equies knowing the initial velocity. Using the given maximum ange infomation and Eq. 5-5, we solve fo t and then substitute that expession into Eq. 5-7 to solve fo v 1 : Δx = v 1 cos θ 1 Δt Δt = Δx = 107 m v 1 cos θ 1 v 1 cos 45.
Δy = v 1 sinθ 1 Δt 1 2 gδt 2 Δx 0 m = v 1 sinθ 1 1 Δx 2 g v 1 cos θ 1 v 1 cosθ 1 sinθ 1 Δx ( Δx) 2 = cos θ 1 2 g 1 v 1 = () v 2 1 cos θ 1 ( ) 2 gδx 2sinθ 1 cos θ 1 = 2 ( 9.8 m / s 2 )107 m 2 sin 45 o ( ) ( )( cos 45 o ) = 32.4 m / s. Then, Eq. 5-5 with the actual distance to the fence tells us when it is above the fence: Δt = Δx = 97.5 m = 4.26 s. v 1 cos θ 1 ( 32.4 m / s)cos 45 At this moment, the ball is at a height (above the gound) of y 2 = y 1 + ( v 1 sinθ 1 )Δt 1 g( Δt 2 )2 = 1.22 m + ( 32.4 m / s) ( sin 45 )4.26 ( s) 1 2 9.8 m / s 2 ( )( 4.26 s) 2 = 9.88 m which implies it does indeed clea the 7.32 m high fence. (b) At t = 4.26 s, the cente of the ball is 9.88 m 7.32 m = 2.56 m above the fence. 32. The (x,y) coodinates of the points ae A = (15 m, 15 m), B = (30 m, 45 m), C = (20 m, 15 m), and D = (45 m, 45 m). The espective times ae t A = 0 s, t B = 300 s, t C = 600 s, and t D = 900 s. Aveage velocity is defined by Eq. 5-22. Each displacement Δ is undestood to oiginate at point A. The magnitudes of the thee displacements ae ( 15 m) 2 + ( 30 m) 2 = 34 m, 5 m, and ( 30 m) 2 + ( 60 m) 2 = 67 m. (a) The aveage velocity having the least magnitude is fo the displacement going fom A to C. v = ( 5 m) /( 600 s) = 0.0083 m / s at 0 (measued ccw fom the +x axis).
(b) The aveage velocity having the geatest magnitude fo the displacement going fom A to B: v = ( 15 m) 2 + ( 30 m) 2 300 s = 0.11 m / s at θ = tan 1 30 m 15 m = 297 (ccw fom +x) o 63 (which is equivalent to measuing 63 clockwise fom the +x axis). 43. Since the x and y components of the acceleation ae constants, then we can use Table 2-1 fo the motion along both axes. This can be handled individually (fo x and y) o togethe with the unit-vecto notation (fo ). (a) Since 1 = 0 m, the position vecto of the paticle is (adapting Eq. 2-17) = v 1 Δt + 1 2 a ( Δt) 2 = [( 8.0 m / s)ˆ j ]Δt + 1 2[ ( 4.0 m / s 2 )ˆ i + ( 2.0 m / s2)ˆ j ]Δt ( ) 2 = [( 2.0 m / s 2 )( Δt) 2 ]ˆ i + {[ ( 8.0 m / s )Δt ]+ [( 1.0 m / s 2 ) ( Δt ) 2 ]}ˆ j. Theefoe, we find when x = 29 m, by solving ( 2.0 m / s 2 )( Δt) 2 = 29 m, which leads to t = 3.8 s. The y-coodinate at that time is y = (8.0 m/s)(3.8 s) + (1.0 m/s 2 )(3.8 s) 2 = 45 m. (b) Adapting Eq. 2-11, the velocity of the paticle is given by v 2 = v 1 + a Δt. Thus, at t = 3.8 s, the velocity is v = [ ( 8.0 m/s )ˆ j ]+ [( 4.0 m / s 2 )ˆ i + ( 2.0 m / s2)ˆ j ]3.8 ( s) = ( 15.2 m / s)ˆ i + ( 15.6 m / s)ˆ j which has a magnitude (speed) of
v = v = v x2 + v y2 = ( 15.2 m / s) 2 + ( 15.6 m / s) 2 = 22 m / s. 49. We apply Eq. 5-35 to solve fo speed v and Eq. 5-33 to find the magnitude of the acceleation a. (a) Since the adius of Eath is 6.37 10 6 m, the adius of the satellite obit is 6.37 10 6 m + 640 10 3 m = 7.01 10 6 m. Theefoe, the speed of the satellite is ( ) ( ) = 7.49 10 3 m/s. v = 2π T = 2π 7.01 10 6 m ( 98.0 min)60 s / min (b) The magnitude of the acceleation is ( ) 2 a = v2 = 7.49 103 m/s 7.01 10 6 m = 8.00 m/ s 2. 56. We wite ou magnitude-angle esults in the fom ( R,θ). All angles ae measued counteclockwise fom +x, but we will occasionally efe to angles which ae measued counteclockwise fom the vetical line between the cicle-cente and the coodinate oigin and the line dawn fom the cicle-cente to the paticle location (see in the figue). We note that the speed of the paticle is v = 2/T whee = 3.00 m and the peiod T = 20.0 s so v = 0.942 m/s. The paticle is moving counteclockwise in Fig. 5-39. (a) At t = 5.00 s, the paticle has taveled a faction of t T = 5.00 s 20.0 s ev = 1 4 ev
aound the cicle (stating at the oigin). Thus, elative to the cicle-cente, the paticle is at φ = 1 4 (360 ) = 90 measued fom vetical (as explained above). Refeing to Fig. 5-39, we see that this position (which is the 3 o clock position on the cicle) coesponds to x = 3.00 m and y = 3.00 m elative to the coodinate oigin. In ou magnitude-angle notation, this is expessed as ( 4.24 m, 45 ). Although this position is easy to analyze without esoting to tigonometic elations, it is useful (fo the computations below) to note that these values of x and y elative to coodinate oigin can be gotten fom the angle fom the elations x = sin and y = cos. Of couse, = x 2 + y 2 and comes fom choosing the appopiate possibility fom tan 1 (y/x) (o by using paticula functions of vecto-capable calculatos). (b) At t = 7.50 s, the paticle has taveled a faction of ( 7.50 s) /20.0 ( s)= 3/8 of a evolution aound the cicle (stating at the oigin). Relative to the cicle-cente, the paticle is theefoe at = 3/8 (360 ) = 135 measued fom vetical in the manne discussed above. Refeing to Fig. 5-39, we compute that this position coesponds to x = (3.00 m)(sin 135 ) = 2.12 m and y = 3.00 m (3.00 m)(cos 135 ) = 5.12 m elative to the coodinate oigin. In ou magnitude-angle notation, this is expessed as (R, ) = (5.54 m, 67.5 ). (c) At t = 10.0 s, the paticle has taveled a faction of (10.0 s)/(20.0 s) = 1/2 of a evolution aound the cicle. Relative to the cicle-cente, the paticle is at = 180 measued fom vetical (see explanation, above). Refeing to Fig. 5-39, we see that this position coesponds to x = 0 m and y = 6.00 m elative to the coodinate oigin. In ou magnitude-angle notation, this is expessed as ( R, θ)= ( 6.00 m, 90.0 ). (d) We subtact the position vecto in pat (a) fom the position vecto in pat (c): ( 6.00 m, 90.0 ) ( 4.24 m, 45 )= 4.24 m, 135 using vecto-capable calculatos). If we wish instead to use unit-vecto notation, we wite ( ) using magnitude-angle notation (convenient when Δ = (0 m 3.00 m) ˆ i + (6.00 m 3.00 m) ˆ j = ( 3.00 m) ˆ i + ( 3.00 m) ˆ j
which leads to Δ = 4.24 m and = 135. (e) Fom Eq. 5-22, we have v = Δ Δt whee Δt = 5.00 s which poduces ( 0.600 m / s) ˆ i + ( 0.600 m / s) ˆ j in unit-vecto notation o (0.849 m, 135 ) in magnitude-angle notation. (f) The speed has aleady been noted (v = 0.942 m/s), but its diection is best seen by efeing again to Fig. 5-39. The velocity vecto is tangent to the cicle at its 3 o clock position (see pat (a)), which means v is vetical. Thus, ou esult is ( 0.942 m / s, 90 ). (g) Again, the speed has been noted above (v = 0.942 m/s), but its diection is best seen by efeing to Fig. 4-37. The velocity vecto is tangent to the cicle at its 12 o clock position (see pat (c)), which means v is hoizontal. Thus, ou esult is ( 0.942 m, 180 ). (h) The acceleation has magnitude v 2 / = 0.296 m/s 2, and at this instant (see pat (a)) it is hoizontal (towads the cente of the cicle). Thus, ou esult is 0.296 m / s 2, 180 ( ). (i) Again, the acceleation has a magnitude v 2 / = 0.296 m/s 2, but at this instant (see pat (c)) it is vetical (towads the cente of the cicle). Thus, ou esult is 0.296 m / s 2, 270 ( ). 57. To calculate the centipetal acceleation of the stone, we need to know its speed duing its cicula motion (this is also its initial speed when it flies off). We use the kinematic equations of pojectile motion fom Table 5-1 to find that speed. Taking the +y diection to be upwad and placing the oigin at the point whee the stone leaves its cicula obit, then the coodinates of the stone duing its motion as a pojectile ae given by x = v 1 x Δt and y = 1 2 a y Δt 2 = 1 2 gδt 2 (since
v 1 y = 0 m / s ). It hits the gound at x = 10 m and y = 2.0 m. Fomally solving the second equation fo the time, we obtain Δt = 2y / g, which we substitute into the fist equation to find the x- component of it: v 1 x = x g 2y = 10 m ( ) 9.8 m / s 2 2( 2.0 m) Theefoe, the magnitude of the centipetal acceleation is = 15.7 m / s. a = v2 ( 15.7 m/ s) 2 = 1.5 m = 160 m/ s 2.