Physics 101 Fall 2007: Pledged Problems 2 1-D and 2-D Kinematics

Transcription

1 Physics 11 Fall 27: Pledged Problems 2 1-D and 2-D Kinematics Time allowed: 2 hours in one sitting Due: Monday, September 17 at 5PM in the box marked Phys11/12 in the Physics Lounge You may use your own textbook, your notes and a non-programmed calculator. For the purposes of this problem set, you may also use the online solutions to the corresponding suggested problems. You should consult no other help. Please follow the standard format: Write legibly on one side of 8-1/2 white or lightly tinted paper. Staple all sheets (including this one) together in the upper left corner Make one vertical fold. On the outside (staple side up) on successive lines PRINT your last name in CAPITAL letters. PRINT your first name. Print the phrase Pledged Problems 2 and the due date. Print the times at which you started and finished the problems. Write and sign the Pledge, with the understanding you may consult the resources described above. 1. During the opening night football game in Rice stadium, Fred catches a football while standing directly on the goal line (x = ). He immediately starts running forward with an acceleration of 6 ft/s 2. At the moment the catch is made, Tommy Guns is 2 yards away and heading directly toward Fred with a steady speed of 15 ft/s. Use the coordinate system defined in the figure where x = corresponds to the point where Fred catches the ball. (a) Sketch a figure showing the positions of Fred and Tommy at.5 s intervals. Use dots to represent the positions of Fred and Tommy.

2 (b) Sketch the x vs. t graphs for Fred and Tommy on a common graph Pledged problem #2.1 Fred Tom 15 x(yd) t(s) (c) Sketch the v x vs. t graphs for Fred and Tommy on a common graph. 1 Velocity vs. time 5 v(yd/s) t(s)

3 (d) At what time (after the catch has been made) does the tackle occur. According to the graph (x vs t), t 2.6 s. Analytically, x fred = 1 2 a fred t 2 (1) x fred = 1 2 (2m/s2 )t 2 (2) x tom = x + v t (3) x tom = 2m 5m/s t (4) x fred = x tom (5) t 2 = 2m 5m/s t (6) t t 2 =. (7) t = 5 ± (8) t = 2.62s or t = 7.62s (9) t = 2.62 s (1) (e) Where does the tackle occur? To find where the tackle occurs, insert Eq. 1 into either Eq. 2 or Eq. 4. I will choose to insert Eq. 1 into Eq. 2 since Fred starts at x = m. x fred = 1 2 2m/s t2 1 (11) t 1 = 2.62s (12) x f = t 2 1 (13) x f = 6.86m (14)

4 2. A rocket is fired vertically upward. The rocket motor burns for t R seconds and gives the rocket a constant upward acceleration of a R during that time. Neglecting effects of wind and rotation of the earth, you can assume the rocket lands alongside the place from which it was fired. Leave all answers in terms of some or all of the following quantities: g, a R, and t R. (a) Draw a v vs. t curve for the rocket. Be sure to label your graph. v(max) v. vs. t. curve for acclerating rocket Acceleration = m/s 2 Velocity (Speed) Acceleration > -v(max) t_r TIME (b) By inspection of the v vs. t graph for the flight, argue that regardless of the value of a R or t R, the rocket always takes longer to go up than to come down. The area under the v t curve should equal zero since the rocket returns back to earth (net displacement is zero). By inspection of the v t curve, the positive area is equal to the negative area. Since the height of the positive area triangle is smaller than the height of the negative are triangle, the time (base of triangle) interval is larger going up (positive area) than coming down. (c) What was the maximum speed reached on the upward flight? The maximum speed reached on the upward flight is attained immediately at the end of the acceleration time (t R ). Since the rocket starts at rest: v max = a R t R (15) (d) How high did the rocket rise with respect to the earth? The height of the rocket at the end of the acceleration period is given by the position equation derived under constant acceleration. y f1 = y + v t a R t 2 R (16) y f1 = 1 2 a R t 2 R (17) After the rocket stops positively accelerating, we can find the maximum height the rocket rises with respect to the earth by considering it an object in free fall with an initial speed of v = a R t R at an initial height of y f1. y f = 1 2 a R t 2 R + (a R t R ) t 1 2 g t2 (18) We need to find the time it takes the rocket to reach its maximum height (v = ).

5 v f = (a R t R ) g t (19) = a R t R g t z (2) t z = a R t R g (21) where t z is the time interval (after t R ) that it takes the rocket to approach zero speed. h = 1 2 a R t 2 R + (a R t R ) t z 1 2 g t2 z (22) h = 1 2 a R t 2 R ( ) 1 + ar g (23) (e) How fast was the rocket moving when it hit the ground? The rocket is at rest a distance h above the earth so v 2 f = 2 g ( h) (24) v f = 2 g h (25) v f = 2 g [ ( 1 2 a R t 2 R 1 + a )] R g (26) v f = a R t R ( 1 + g a R ) (27)