Physics 4A Solutions to Chapter 4 Homework

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1 Physics 4A Solutions to Chapter 4 Homework Chapter 4 Questions: 4, 1, 1 Exercises & Problems: 5, 11, 3, 7, 8, 58, 67, 77, 87, 11 Answers to Questions: Q 4-4 (a) all tie (b) 1 and tie (the rocket is shot upward), then 3 and 4 tie (it is shot into the ground!) Q 4-1 (a) c, b, a (b) a, b, c Q 4-1 (a) 9 and 7 (b) and 18 (c) 9 and 7 Answers to Problems: P 4-5 The aerage elocity of the entire trip is gien by Eq. 4-8: ag = Δr / Δt, where the total displacement Δ r =Δ r 1+Δ r +Δr 3 is the sum of three displacements (each result of a constant elocity during a gien time), and Δ t =Δ t t 1+Δ t +Δ 3 is the total amount of time for the trip. We use a coordinate system with +x for East and +y for North. (a) In unit-ector notation, the first displacement is gien by km 4. min = 6. ˆ i = (4. km)i. ˆ h 6 min/h Δr1 The second displacement has a magnitude of 4 north of east. Therefore, (6. )( ) =. km, and its direction is km h. min 6 min/h Δ r = (. km) cos(4. ) ˆi + (. km) sin(4. ) ˆj = (15.3 km) ˆi + (1.9 km) ˆ j. Similarly, the third displacement is km 5. min Δ r ˆ ˆ 3 = 6. i = ( 5. km) i. h 6 min/h Thus, the total displacement is

2 Δ r =Δ r +Δ r +Δ r = (4. km)i ˆ+ (15.3 km) ˆi + (1.9 km) ˆj (5. km) ˆi 1 3 = (5.3 km) ˆi + (1.9 km) ˆj. The time for the trip is Δ t = ( ) min = 11 min, which is equialent to 1.83 h. Equation 4-8 then yields ag (5.3 km) ˆi + (1.9 km) ˆj = = (.9 km/h) ˆi + (7.1 km/h) ˆj h The magnitude of ag is = (.9 km/h) + (7.1 km/h) = 7.59 km/h. ag (b) The angle is gien by θ = = 7.1 km/h = or.5 east of due north. 1 ag, y tan tan (north of east), ag, x.9 km/h The displacement of the train is depicted in the following figure: Note that the net displacement Δr is found by adding Δr 1, Δr and Δr 3 ectorially. P 4-11 In parts (b) and (c), we use Eq. 4-1 and Eq For part (d), we find the direction of the elocity computed in part (b), since that represents the asked-for tangent line. (a) Plugging into the gien expression, we obtain r [.(8) 5.()]i ˆ+ [6. 7.(16)] ˆj (6.ˆ = = = i 16ˆ j) m t.

3 (b) Taking the deriatie of the gien expression produces t t ˆ t 3 ( ) = (6. 5.) i 8. j ˆ where we hae written (t) to emphasize its dependence on time. This becomes, at t =. s, = (19.ˆi 4ˆj) m/s. (c) Differentiating the t () found aboe, with respect to t produces 1.tˆi 84.t ˆj, yields ˆ a =(4. i 336 ĵ) m/ s at t =. s. which (d) The angle of, measured from +x, is either 1 4 m/s tan = 85. or m/s where we settle on the first choice ( 85., which is equialent to 75 measured counterclockwise from the +x axis) since the signs of its components imply that it is in the fourth quadrant. P 4-3 (a) From Eq. 4- (with θ = ), the time of flight is h (45. m) t = = = 3.3 s. g 9.8 m/s (b) The horizontal distance traeled is gien by Eq. 4-1: Δ x = t = (5 m/s)(3.3 s) = 758 m. (c) And from Eq. 4-3, we find y = gt = = (9.8 m/s )(3.3 s) 9.7 m/s. P 4-7 We adopt the positie direction choices used in the textbook so that equations such as Eq. 4- are directly applicable. The coordinate origin is at ground leel directly below the release point. We write θ = 3. since the angle shown in the figure is measured clockwise from horizontal. We note that the initial speed of the decoy is the plane s speed at the moment of release: = 9 km/h, which we conert to SI units: (9)(1/36) = 8.6 m/s.

4 (a) We use Eq. 4-1 to sole for the time: 7 m Δ x = ( cos θ) t t = = 1. s. (8.6 m/s)cos ( 3. ) (b) And we use Eq. 4- to sole for the initial height y : 1 1 y y = ( sin θ ) t gt y = ( 4.3 m/s)(1. s) (9.8 m/s )(1. s) which yields y = 897 m. P 4-8 (a) Using the same coordinate system assumed in Eq. 4-, we sole for y = h: 1 h= y + sinθt gt which yields h = 51.8 m for y =, = 4. m/s, θ = 6., and t = 5.5 s. (b) The horizontal motion is steady, so x = x = cos θ, but the ertical component of elocity aries according to Eq Thus, the speed at impact is (c) We use Eq. 4-4 with y = and y = H: ( θ ) ( θ ) = cos + sin gt = 7.4 m/s. b g m. sinθ H = = g P 4-58 (a) The circumference is c = πr = π(.15 m) =.94 m. (b) With T = (6 s)/1 =.5 s, the speed is = c/t = (.94 m)/(.5 s) = 19 m/s. This is equialent to using Eq (c) The magnitude of the acceleration is a = /r = (19 m/s) /(.15 m) = m/s. (d) The period of reolution is (1 re/min) 1 = min, which becomes, in SI units, T =.5 s = 5 ms.

5 P 4-67 The stone moes in a circular path (top iew shown below left) initially, but undergoes projectile motion after the string breaks (side iew shown below right). (top iew) (side iew) Since a= / R, to calculate the centripetal acceleration of the stone, we need to know its speed during its circular motion (this is also its initial speed when it flies off). We use the kinematic equations of projectile motion (discussed in 4-6) to find that speed. Taking the +y direction to be upward and placing the origin at the point where the stone leaes its circular orbit, then the coordinates of the stone during its motion as a projectile are gien by x = t and y gt (since y = ). It hits the ground at x = 1 m and y =. m. = 1 Formally soling the y-component equation for the time, we obtain t = y/g, which we substitute into the first equation: b g b g g 98. m/s = x = 1 m = m/s. y. m Therefore, the magnitude of the centripetal acceleration is ( ) 15.7 m/s a = = = 16 m/s. R 1.5 m gx Note: The aboe equations can be combined to gie a =. The equation implies that the yr greater the centripetal acceleration, the greater the initial speed of the projectile, and the greater the distance traeled by the stone. This is precisely what we expect. P 4-77 This problem deals with relatie motion in two dimensions. Snowflakes falling ertically downward are seen to fall at an angle by a moing obserer. Relatie to the car the elocity of the snowflakes has a ertical component of = 8. m/s and a horizontal component of = 5 km/h = 13.9 m/s. The angle θ from the ertical is found from h

6 which yields θ = 6. h 13.9 m/s tanθ = = = m/s Note: The problem can also be soled by expressing the elocity relation in ector notation: rel = car + snow, as shown in the figure. P 4-87 This problem deals with the projectile motion of a baseball. Gien the information on the position of the ball at two instants, we are asked to analyze its trajectory. The trajectory of the baseball is shown in the figure below. According to the problem statement, at t 1 = 3.s, the ball reaches it maximum height y, max and at t = t1+.5s = 5.5s, it barely clears a fence at x = 97.5 m. Eq. -15 can be applied to the ertical (y axis) motion related to reaching the maximum height (when t 1 = 3. s and y = ): y max y = y t 1 gt. (a) With ground leel chosen so y =, this equation gies the result y 1 1 = gt = (9.8 m/s )(3.s) = 44.1 m max 1 P 4-11 Using Eq. -16, we obtain = gh, or h= ( )/g.

7 (a) Since = at the maximum height of an upward motion, with = 7. m/s, we hae h = (7. m/s) /(9.8 m/s ) =.5 m. (b) The relatie speed is r = c = 7. m/s 3. m/s = 4. m/s with respect to the floor. Using the aboe equation we obtain h = (4. m/s) / (9.8 m/s ) =.8 m. (c) The acceleration, or the rate of change of speed of the ball with respect to the ground is 9.8 m/s (downward). (d) Since the eleator cab moes at constant elocity, the rate of change of speed of the ball with respect to the cab floor is also 9.8 m/s (downward).