User:Jeff/Physics answers

(Note to algorithmist.com visitors: I just need to use this page as a sandbox over the weekend... I'll take it down on Monday.)

Part A
The force of gravity that one object exerts on another is equal to G*m1*m2/r^2, where G is a constant, m1 and m2 are the masses of the two objects, and r is the distance between them. On Earth, the acceleration due to gravity equals 9.81 m/s^2. If the mass of Earth is 5.97*10^24 kg, and the radius of the Earth is 6.38*10^6 m, what is the value of G? Be sure to include SI units.

G * (5.97*10^24 kg)/(6.38*10^6 m)^2 * mass = 9.81 m/s^2 * mass G = (9.81 m/s^2) * (6.38*10^6 m)^2 / (5.97*10^24 kg) = 6.69*10^-11 m^3/(kg s^2) = 6.69*10^-11 Nm^2/kg^2

Part B
At what altitude above sea level does the acceleration of gravity become 3% less than the traditional value of 9.81 m/s^2?

9.81 * (6.38*10^6)^2 / r^2 = 0.97 * 9.81 = 9.52 r = sqrt(9.81*(6.38*10^6)^2/9.52) = 6.48*10^6 m h = r - 6.38*10^6 = 9.79*10^4 m

Problem 2
At time t=0, a 2.0 kg Jeff awakens to find himself at (0, 0) with an instantaneous velocity of 5.0 m/s in the –î direction. Exactly three forces are acting on Jeff: Fallie = 20 N ∠90°, Fgoogle = 10 N ∠210°, and Fbuttsex. At time t=3.0 s, Jeff's position is (13.5 cos 50° - 15, 13.5 sin 50°).

Part A
If the problem were changed such that Jeff was initially at rest, what would his position have been at time t=3.0 s? Where would he be if V(0) = 5.0 m/s ∠315°?

If you change V_x(0) from -5.0 m/s to 0 m/s, Jeff gets an additional velocity of 5.0 m/s in the î direction for 3.0 seconds. Therefore, his new x-position is 15 greater than it would have been otherwise. Since V_y(t) doesn't change, neither does P_y(t), so the answer is (13.5 cos 50°, 13.5 sin 50°).

P_x(t=3) = P_x(0) + V0_x*t + 0.5at^2 = 0 - 5t + 0.5at^2 = 13.5 cos 50° - 15 P_x_new(t=3) = P_x(3) - V0_x*t + V0_x_new*t = (13.5 cos 50° - 15) - (-5 * 3) + 0 = 13.5 cos 50°

If V(0) = 5.0 m/s ∠315°, then V_x(0) = 5 cos 315° = 5*sqrt(2)/2, and V_y(0) = 5 sin 315° = -5*sqrt(2)/2. Multiplying the components by time, and adding them to the V(0)=0 case, tells us that Jeff's new position would be (13.5 cos 50° + 15*sqrt(2)/2, 13.5 sin 50° - 15*sqrt(2)/2).

Part B
Find Jeff's acceleration vector and net force vector.

Since P0_y = 0 and V0_y = 0, P_y(t) = 0.5at^2. P_y(3) = 9a/2 = 13.5 a = 3.0 m/s^2 ∠50° f = ma = 6.0 N ∠50°

Part C
Find Fallie,x, Fallie,y, Fgoogle,x, Fgoogle,y, Fnet,x, Fnet,y, and Fbuttsex.

F_allie,x = 20 cos 90° = 0 N F_allie,y = 20 sin 90° = 20 N F_google,x = 10 cos 210° = -5*sqrt(3) = -8.66 N F_google,y = 10 sin 210° = -5 N F_net,x = 6 cos 50° = 3.86 N F_net,y = 6 sin 50° = 4.6 N F_buttsex,x = F_net,x - F_allie,x - F_google,x = 12.5 N F_buttsex,y = F_net,y - F_allie,y - F_google,y = -10.4 N F_buttsex = sqrt(12.5^2+(-10.4)^2) ∠arctan(-10.4/12.5) = 16.3 N ∠320°

Part D
What was Jeff's position at time t=-1.0 s? What was his instantaneous velocity?

P_x = P0_x + V0_x*t + 0.5at^2 = 0 + -5*-1 + 0.5*3*cos(50°)*(-1)^2 = 5.96 m P_y = P0_y + V0_y*t + 0.5at^2 = 0 + 0*-1 + 0.5*3*sin(50°)*(-1)^2 = 1.15 m V_x = V0_x + at = -5 + 3*cos(50°)*(-1) = -6.93 m/s V_y = V0_y + at = 0 + 3*sin(50°)*(-1) = -2.3 m/s V = sqrt((-6.93)^2+(-2.3)^2) ∠arctan(-2.3/-6.93) = 7.3 N ∠198°

Part E
Let Jeff's instantaneous velocity vector V(t) = Vx(t) î + Vy(t) ĵ. Which of the following is true?
 * Vx(tflip) = 0 for some tflip &gt; 0. The function Vx(t) is increasing.
 * Vy(tflip) = 0 for some tflip &gt; 0. The function Vy(t) is increasing.
 * Vx(tflip) = 0 for some tflip &gt; 0. The function Vx(t) is decreasing.
 * Vy(tflip) = 0 for some tflip &gt; 0. The function Vy(t) is decreasing.

The first statement is correct. Both the x- and the y-components of acceleration are positive, so Vx(t) and Vy(t) are both increasing functions. Since Vy(0) = 0, Vy(t) > 0 for all t > 0. But Vx(0) = -5 m/s is negative, so this component of velocity must equal zero for some positive value tflip.

Part F
Find tflip used in part E. What is Jeff's location and velocity at time t=tflip?

V_x = V0_x + at = -5 + 3*cos(50°)*t_flip = 0 t_flip = 5 / (3*cos(50°)) = 2.59 s P_x = P0_x + V0_x*t + 0.5at^2 = 0 + -5*2.59 + 0.5*3*cos(50°)*2.59^2 = -6.48 m P_y = P0_y + V0_y*t + 0.5at^2 = 0 + 0*2.59 + 0.5*3*sin(50°)*2.59^2 = 7.73 m V_x = V0_x + at = -5 + 3*cos(50°)*2.59 = 0 m/s V_y = V0_y + at = 0 + 3*sin(50°)*2.59 = 5.96 m/s V = 5.96 m/s ∠90°

Part G
If Fallie disappears at time t=10 s, Fbuttsex goes away at t=20 s, and Fgoogle vanishes at t=30 s, what is Jeff's velocity at t=50 s?

V_x = V0_x + F_allie,x*10/2 + F_buttsex,x*20/2 + F_google,x*30/2 = -5+0+125+130 = 250 m/s V_y = V0_y + F_allie,y*10/2 + F_buttsex,y*20/2 + F_google,y*30/2 = 0+100-104-75 = -79 m/s V = sqrt(250^2+(-79)^2) ∠arctan(-79/250) = 262 m/s ∠342°

Problem 3
An airplane is flying at a speed of 200 m/s. The airplane releases a package at an altitude of 10 km at time t=0. What is the x-displacement of the package when it hits the ground, if the airplane's velocity is parallel to the x-axis at t=0? If the velocity is parallel to 30°? -45°?

For theta = 0°: V0_x = 200 cos 0° = 200 m/s V0_y = 200 sin 0° = 0 P_y(t) = P0_y + V0_y*t + 0.5at^2 = 10000 - 4.9*t^2 = 0 t = sqrt(10000/4.9) = 45.2 s P_x = V0_x * t = 200*45.2 = 9035 m For theta = 30°: V0_x = 200 cos 30° = 100*sqrt(3) = 173 m/s V0_y = 200 sin 30° = 50 m/s P_y(t) = P0_y + V0_y*t + 0.5at^2 = 10000 + 50t - 4.9*t^2 = 0 t = (-50-sqrt(50^2+4*4.9*10000))/(2*-4.9) = 50.6 s P_x = V0_x * t = 173*50.6 = 8758 m For theta = -45°: V0_x = 200 cos -45° = 100*sqrt(2) = 141 m/s V0_y = 200 sin -45° = -100*sqrt(2) = -141 m/s P_y(t) = P0_y + V0_y*t + 0.5at^2 = 10000 - 141t - 4.9*t^2 = 0 t = (141-sqrt((-141)^2+4*4.9*10000))/(2*-4.9) = 33 s P_x = V0_x * t = 141*33 = 4666 m