Midyear Review Answers

I'll be updating this post with answers throughout the night and day, so check them when you can.
Midyear Review #1 - 1D kinematics
1.
(b) slope is zero, zero acceleration
(c) zero
(d)&(e) 200 meters
2.
(b) 8 m/s^2
(c) area under v vs. t graph, distance = 100 m.
(d) Y = vo*t + .5at^2 = 100 m.
(e) 16 m/s
(f) graph should be a parabola with its vertex at t = 0, y = 0.

3.
(a) 4 seconds
(b) horizontal line from t = 0 to t = 3, then a diagonal line going down to zero.
(c) 60 meters
(d) 100 meters

4. 24 m/s

5.
(a) +15 m/s, -10 m/s
(b) 25 m/s
(c) 50 m/s^2

6.
(a) 40 m/s^2
(b)East
(c) velocity directed south, acceleration directed East


Midyear Review #2 - 2D Kinematics and Projectile Motion
1.
(a) 1.563 seconds
(b) 7.813 meters
(c) 13.53 meters

2.
(a) 4.0 seconds
(b) 5 m/s
(c) (We can assume that the net force is mostly vertical since the horizontal velocity (5 m/s) is so much less than the vertical velocity (40 m/s). acceleration = .769 m/s^2, displacement = 1040 meters (that's a deep target!)

3.
(a) 412.3 km at 14 degrees North of East.
(b) 42.7 km/h
(c) 51.7 km/h


Midyear Review #3 - Newton's Laws
1. 4 N
2. T1 (diagonal cord) = 200 N, T2 (horizontal cord) = 173 N
3.
(a) a = F/6m
(b) Block 1 (3m), Block 2 (m), Block 3 (2m)
Fnet1 = F/2
Fnet2 = F/6
Fnet3 = F/3
(c) T1 (between 3m and m) = F/2, T2 (between m and 2m) = 2/3F
4.
(a) T1 = 60 N, T2 = 100 N, minimum value of static friction coefficient = .27
(b) T1 = 59 N, T2 = 98 N, a = .17 m/s^2
5.
(a) 22.5 m
(b) 2.5 m
(c) Snoopy is going fast enough in the horizontal direction that the normal force of the plane acting on him holds him in a circular path, rather than a parabolic one.
6. 24 m/s


Midyear Review #4 - Work and Energy
1. 23.8 J
2. 0.8 m
3. 27 meters
4. (Note that the height of the building should be 80 meters, not 30 as written on the sheet.)
(a) 400 J
(b) 5 N
5.
(a) 1400 J
(b) 60 W
6.
(a) 0 J
(b) 8 J
(c) 64 J
7.
(a) 230 J
(b) 10.0 m/s



Midyear Review #5 - Gravity, Momentum, Torque
1.
(a) Mass is the same, 70 kg
(b) 700 N (16) = 11,200 N
(c) 160 m/s^2
2. (GM/r)^(1/2)
3. (1/5)^(1/2)
4. 70 kg
5.
(a) (3/10)v0.
(b) -21/20mv^2
6.
(a) 1.183 m/s
(b) 782 m/s
7.
(a)(.4 kg)(2 m/s) + 0 = (.4 kg)(1.2 m/s) cos 30 + (.4 kg)Vbx
0 = (.4 kg)(1.2 m/s) sin30 + (.4 kg)Vby
(b) Vbx = 0.96 m/s, Vby = -.6 m/s, so Vb = (.96^2 + .6^2)^(1/2) = 1.13 m/s
tan(x) = .6/.96 so x = 32 degrees.
Final velocity of ball b is 1.13 m/s at 32 below the horizontal.
(Ignore the part that says the collision is elastic - if you calculate change in KE, it is NOT zero!)
8.
speed, max D, min A
kinetic energy max D, min A
potential energy, max A, min D
total energy, constant
angular momentum, constant
net force on the satellite, max D, min A
acceleration, max D, min A