 Phyllis Fleming Physics Physics 107
 Answers - Two Dimensional Motion
 1 You find the resultant of two vectors graphically by (a) the parallelogram method  or (b) the polygon method. In the parallelogram method, complete the parallelogram, as shown by the dashed lines in Fig. 1a below. The resultant s is the diagonal. In the polygon method, move the tail of one of the vectors, for example s1, to the head of s2. The resultant s goes from the tail of s2 to the head of s1 as in Fig. 1b below of the statement of the problem. 2 For vectors along a straight line, you can assign a positive sign for a vector to the right and a negative sign for a vector to the left. Letting 8 m be to the right and 6 m to the left, 8 m - 6 m = 2 m. This is also shown in a vector diagram in Fig. for #2a above. Because the two vectors lie along the same straight line, I have drawn the vector for -6 m below the vector for +8 m. Remember two vectors are equal if they have the same magnitude and direction even if they do not start at the same point. Now let both 8 m and 6 m be to the right: 8 m + 6 m = 14 m. This is also shown in a vector diagram in Fig. for #2b above. Note that (82 + 62)1/2 = 10. If the vectors are at a right angle then the resultant vector has a magnitude of 10 m, as shown in a vector diagram in Fig. for #2c above.
 3 First find the components of A. Ax = A cos 37o = 10 cm (0.8) = 8 cm Ay = A sin 37o = 10 cm (0.6) = 6 cm. We are given that Bx = 8 cm and By = - 2 cm. For C = A + 1/2 B, Cx = Ax + Bx/2 = (8 + 4) cm = 12 cm                                 and  Cy = Ay + By/2 = (6 - 1) cm = 5 cm. C = (122 + 52)1/2 cm = 13 cm.  tan Θ = Cy/Cx = 5/12 = 0.42.  Θ= 26o. Fig. for #3 is shown above.
 4 The vector r1 is totally in the +Y-direction. The component of r1 along the X-axis, r1x = r2 cos 900 = (7.00 m)0 = 0. The component of r1 along the Y-axis, r1y = r1 sin 900 = 7.00 m (1) = 7.00 m. The component of r2 along the X-axis, r2x = r2 cos 450 = 7.07 m (0.707) = 5.00 m. The component of r2 along the Y-axis, r2y = r2 sin 450 = 7.07 m (0.707) = 5.00 m. The component of r along the X-axis, rx = r1x + r2x = 0 + 5.00 m = 5.00 m. The component of r along the Y-axis, ry = r1y + r2y = 7.00 m + 5.00 m = 12.0 m. r = rxi + ryj = (5.00i + 12.0j) m. The magnitude of r = r = (rx2 + ry2)1/2 = (5.002 + 12.02)1/2 m = 13.0 m. tan Θ= 12/5. Θ= 67.40.
 5 For motion in the horizontal direction along the X-axis, t = x/vx = 60 ft/(132 ft/s) = 0.454 s. For motion in the vertical direction along the Y-axis, the vertical drop y = 1/2 at2 = 1/2 (32 ft/s2)(0.454 s)2 = 3.3 ft.
 6 For horizontal motion of the rock, x = 24 m/s t           (Equation 1) For horizontal motion of the car, 90 m - x = 1/2(4.0 m/s2)t2          (Equation 2) Substituting Eq. 1 into Eq. 2: 90 m - 24 m/s t = 1/2(4.0 m/s2)t2   or t2 + 12s t - 45s2 t2 = 0   or  (t - 3s)(t + 15s) = 0  or  t = 3 s. Distance fallen by the rock = y = 1/2 at2 = 1/2(10 m/s2)(9s2) = 45 m.
 7 For the vertical motion of the ball, y = 1.0 m = 1/2(10 m/s2)t2 and t = 0.45 s. For the horizontal motion, x = vxt = (4.0 m/s)(0.45 s) = 1.8 m.
 8 The expressions for the quantities t, ymax, vy(t), and x will all remain the same. This is exactly what the principle of superposition is telling you. The object is subjected to two different influences, an initial velocity and a continuing acceleration due to gravity, and the object responds to each without altering its response to the other. The fact that the object has an initial horizontal velocity does not alter its vertical motion or the time it takes the object to hit the ground. The fact that the object is subjected to a vertical acceleration does not change its constant horizontal velocity motion. As a result you may separate a two-dimensional problem of a projectile into two one-dimensional problems.
 9 At the initial position i, vix = vo cos 370 = 25 m/s (0.80) = 20 m/s and viy = vo sin 370 = 25 m/s (0.60) = 15 m/s. At any time, the horizontal component of the velocity vx = vix + axt = vix + (0)t = 20 m/s. There is no acceleration in the x-direction and the object continues to move with a constant horizontal velocity. At the highest point, the vertical component of velocity is zero. It does not continue to move upward. It momentarily stops moving upward and then moves down. It does continue, however, to have the same constant horizontal velocity. Notice that this velocity at h is tangent to the path, as it should be. At f, the horizontal velocity still equals 20 m/s. It takes 3.0 s. to move 60 m (see x position of object just before it hits the ground in Fig. 3 above) at a constant horizontal velocity of 20 m/s. At t = 3.0 s, the vertical velocity v fy(3.0s) = viy + ayt = 15m/s + (-10 m/s2)(3.0 s) = -15 m/s. For this symmetrical situation, vfy = -viy. The final velocity is 25 m/s at an angle of -370. Notice also for this case where yi = yf = 0, the time to go half the distance horizontally equals one-half the total time of flight. The velocity at h and f are shown in Fig. 3 above. The acceleration is a constant. It is vertically downward and we took it to be -10 m/s2. The acceleration vectors are shown at x = 20.0, 30.0, and 40.0 m in Fig. 3 above.

10. Divide this problem into two one-dimensional problems:

 X Y xo = 0 yo = 20 m vox = 25 m/s (cos 37o) voy = 25 m/s (sin 37o) = 25 m/s (0.80) = 20 m/s = 25 m/s (0.06) = 15 m/s ax = 0 ay = -10 m/s2 vx = vox + axt = vox = 20 m/s vy = voy + ayt = 15 m/s - 10 m/s2t (c) In general, vy2(y) = vo2 + 2ay(y - yo) For this case, = (15 m/s)2 -20 m/s2(y - 20m) At highest point with y = ymax, 0 = 225 m2/s2 -20 m/s2(ymax - 20m)                                                                (ymax - 20 m) = 11.25 m                                                                ymax = 31.25 m (d) In general, y(t) = yo + voyt + 1/2 ayt2 For this case, y(t) = 20 m + 15 m/s t - 5 m/s2t2 When the object returns to earth, 0 = 20 m + 15 m/s t - 5 m/s2t2  or t2 - 3 s t - 4 s2 t2 = 0  or factoring (t - 4 s)(t + 1 s) = 0  and  t = 4 s (e) x(t) = xo + voxt + 1/2 axt2 x(t) = 0 + 20 m/s t + 0 x(4s) = (20 m/s)(4 s) = 80 m vx = vox vy(t) = voy + ayt (f) vx(4s) = 20 m/s vy(4s) =15 m/s -10 m/s2(4s)           = -25 m/s (g) v(4s) = {(20 m/s)2 + (-25 m/s)2}1/2 = 32.0 m/s. tan Θ= -25/20 Θ= -51.3o

Velocity vectors are drawn on Fig. 4 above.

 11 (a)  ro = 0i + 20 m j = 20 m j (b)  vo = (20 i + 15 j) m/s (c)  a = -10 m/s2 j (d)  v(t) = 20 m/s i + (15 m/s - 10 m/s2 t)j (e)  v(t) = (20 m/s i +15 m/s j) - (10 m/s2 t)j        v(t) = vo +  at (f)  r(t) = (20 m/s i)t + {20 m +15 m/s t + 1/2 (-10m/s2)t2}j (g)  = 20 m j + {(20 i + 15 j) m/s}t + 1/2(-10 m/s2)j t2        = ro + vo t+ 1/2 at2 The figure above shows how to find r(1s), r(2s), r(3s) and r(4s) using r(t) = ro + vo t + 1/2 at2.  To find r(3s) start out at the origin and go up 20 m, add a vector 37o above the X-axis of vo x 3s = 25 m/s x 3 s = 75 m. Finally add a vector down = 1/2 (10 m/s2)(3s)2 = 45 m and you arrive at the head of r(3s).
 12 To catch up to the ball, the horizontal velocity vb of the boy must equal the initial horizontal velocity of the ball. vb = vox = vo cos Θ.  cos Θ= vb/vo = (20 m/3s)/20 m/s = 1/3. Θ = 70.5o.  voy = vo sin Θ= 20 m/s sin 70.5o = 18.9 m/s. For motion in the y-direction, vy(y) = voy2 + 2 ay(y - yo). For the highest point, vy = 0  and  0 = (18.9 m/s)2 - 20 m/s2(y - yo). The ball rises (y - yo) = (18.9)2m/20 = 17.8 m.

13.

 X Y xo = 0 yo = 0 vox = vo cosΘ voy = vo sin Θ ax = 0 ay = -10 m/s2 xmax = 180 m for which y = 0 x(t) = voxt y(t) = voyt + 1/2 ayt2 For x = 180 m = (vo cos Θ) t  (Equation 1) y = 0 = vo sin Θ t - 1/2 ayt2 t = 2vo sin Θ/ay       (Equation 2)

Substituting Eq. 2 into Eq. 1:
180 m = (vo cos Θ)(2vo sin Θ/ay) = (2vo2 sin Θ cos Θ/ayor
vo2 = 90 m ay/sin Θ cos Θ       (Equation 3)

When the ball clears the tree,
x = 30 m = (vo cos Θ)t  and  y = 15 m = vo sin Θ t - 1/2 ayt2       (Equation 4)

t = 30 m/(vo cos Θ)        (Equation 5)

Substituting Eq. 5 into Eq. (4):
15 m = (vo sin Θ)(30 m/vo cos Θ) - 1/2 ay(30 m/vo cos Θ)2   or
15 m = 30 m tan Θ - 1/2 ay(30 m/vo cos Θ)2              (Equation 6)

Substituting Eq. 3 into Eq. (6):
15 m = 30 m tan Θ - 1/2 ay (30 m/cos Θ)2(sin Θ cos Θ/90 m ay)
15 m = 30 m tan Θ - 5 m tan Θ= 25 tan Θ.  tan Θ= 15/25.  Θ = 310
vo2 = 90 m ay/sin Θ cos Θ .   Since sin 2Θ= 2 sin Θcos Θ,
vo2 = 180 m ay/sin 2Θ= 180 m(10 m/s2)/0.88.
vo = 45 m/s.

 14 When, and only when, a projectile starts at the ground and ends up at the ground, the time to reach the highest point equals the time to go from the highest point to the ground. Since they all go up the same vertical height, the time involved must be the same. In more detail, To go from the highest point take yo = ymax.  At that point, the vertical velocity is zero, so y = yo + 1/2 at2, where a is the acceleration of the object. When the object hits the ground, y = 0 = ymax + 1/2 at2.  Since ymax and a are the same for all objects, the time for the object to fall or to be in flight is the same. At the highest point, the vertical velocity of the object is zero.  0 = vy = voy+ ayt  and the time is the same for all three projectiles so voy = vo sin Θ = ay/t = constant. Thus vo = constant/sin Θ or vo is inversely proportional to sin Θ. The larger the angle, the larger the sine of the angle and the smaller the initial velocity vo. The initial velocity is greatest for the smallest angle Θ3. The horizontal distance x = (vo cos Θ)t.  As the angle decreases the cosine increases and vo increases, so the range is the greatest for the smallest angle Θ3.
 15 (a) and (b).  The velocity v and the acceleration a at positions A, B, and C for an object moving with uniform circular motion are shown in Fig. 6 above. The magnitude of the velocity is a constant and always tangent to the circle at the point where you wish to find it. The centripetal acceleration is constant in magnitude and always in toward the center of the circle. The centripetal acceleration a = v2/r = (2.0 m/s)2/0.4 m = 10 m/s2. (c)  While the magnitudes of the velocity and acceleration are constants, the directions are not. Thus neither v nor a is constant. (d)  A change in direction of velocity results in an acceleration just as much as a change in the magnitude of the velocity. You could use this device to see how many "g's" an object could withstand.
 16 (a) The period = distance moved in one complete rotation/velocity T = 2 πr/v = 2 π (0.4 m)/2.0 m/s = 0.4 π s (b) The frequency = 1/period = 2.5/ π s-1 (c) From (a),  v = 2 πr/T (d) The centripetal acceleration a = v2/r = (2 πr/T)2/r = 4 π2r/T2 = 4 π2rf2 = (d)
 17 By definition of an angle in radians, ΔΘ= Δs/r        (Equation 1) Dividing both sides of Eq. 1 by Δt,     ΔΘ/ Δt= (1/r) Δs/ Δt limΔt 0 ω= dΘ/dt = (1/r) ds/dt = 1/r (v) By definition, the velocity v = rate of change of position = ds/dt. ω = v/r. Curling the fingers of your right hand counterclockwise, your thumb points out of the page. Since the magnitude of the angular velocity = v/r = a constant and the direction of  ω  for counterclockwise motion is out of the page, uniform circular motion is a case of constant angular acceleration. Note: use "out of the page" rather than up because up is in the plane of the page.
 18 rx = r cos ωt ry = r sin ωt r(t) = r(cos ωt i + sin ωt j) v(t) = dr/dt = v(t) = ωr(-sin ωt i + cos ωt j) vx = -ωr sin ωt vy = ωr cos ωt v = (vx2 + vy2)1/2 = ωr(sin2ωt + cos2 ωt)1/2 = ωr a(t) = dv/dt a(t)= -ω2r(cos ωt i + sin ωt j) = -ω2r(t) ax = -ω2r cos ωt. ay = -ω2r sin ωt a = (ax2 + ay2)1/2 = ω2r(cos2 ωt + sin2 ωt)1/2 = ω2r Since ω = v/r, a = (v/r)2 r = v2/r
 19 Since v1 is perpendicular to r1 and v2 is perpendicular to r2, the two angles Θ are equal. Since the two triangles are isosceles (r1 = r2 = r and v1 = v2 = v), they are similar and Δv/v = Δr/r  or Δv = Δr(v/r)  and Δv/ Δt = ( Δr/ Δt) (v/r). Taking the limit of this equation,      Δv/ Δt = ( Δr/ Δt) (v/r)  or  limΔt 0                            dv/dt = (v)(v/r)  since dr/dt = v. limΔt 0 Since dv/dt = a, a = v2/r. As Δt goes to 0,  Θ goes to 0, and Δv is perpendicular to v. Since a = Δv/ Δt, a is perpendicular to v or parallel to r into the center of the circle. For this reason it is called a centripetal acceleration.

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 Website Designed By: Questions, Comments To: Date Created: Date Last Updated: Susan D. Kunk Phyllis J. Fleming October 8, 2002 November 2, 2002