Midyear Review Solutions

The solutions are posted below - I have updated some of the previous answers with a bit more details, so be sure to check here for the latest answers.

Please also post a comment to this blog post on what topics/problem types you personally want the most help with during class tomorrow.

These are the problems I have created for tomorrow - I suggest you try them and come in with questions: Review problems for Monday

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) change in velocity is -25 m/s (or 25 m/s directed upwards)
(c) 50 m/s^2 directed upwards

6.
(a) 4 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 directed upwards

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. The key thing to think about is that WALL-E always hangs from EVE's arms. The FBD of WALL-E always has EVE's force directed upwards, and his weight directed downwards. The difference is the direction of acceleration, which must always be directed towards the center of the circle.

For (a), the acceleration is upwards, so F – mg = mv^2/R. F = 669 N.

For (b), the acceleration is downwards, so mg – F = mv^2/R. F = 81 N.

5.
(a) T1 = 60 N, T = 100 N, minimum value of static friction coefficient = .27. Notice that this is the solution if you say that static friction is directed DOWN the inline.

If you instead make the static friction force go UP the incline in your FBD, you get a negative value for mu, which means that this is NOT possible in this situation.

(b) a = 1.31 0.670 m/s^2, T = 84.9 91.3 N. Notice that it says to assume the incline is smooth, which means no kinetic friction.

c) If the cord is cut, the right block will fall at g, or 9.8 m/s^2. The left block will slide at g sin(60) = 8.49 m/s^2. The answers must be different because the two blocks are no longer linked by the cord.

Midyear Review #4 – Circular Motion, Gravity, SHM

1. Be sure to use the static friction coefficient, as static friction is what keeps the car from slipping off the road. Kinetic friction occurs when two surfaces slide past each other.

Max speed = (mu*g*r)^(1/2) = 23.7 m/s

2.
(a) Mass is the same, 70 kg
(b) 700 N (16) = 11,200 N
(c)160 m/s^2


3. Remember to draw a FBD of the satellite, with the gravity force directed towards the Earth's center, and positive also directed towards the center. You should get that v = (GM/r)^(1/2).

4.use your answer to #3 to justify your answer here. Also, notice that the mass does NOT change your answer. The ratio is equal to (1/5)^(1/2)

5.
a) Since the mass is not moving when the spring is at maximum compression, the KE = 0.

b) U = 1/2*k*delta_x^2, so the potential energy = 1/9*maximum U = 8 J.

c) Since the total energy is constant (energy is conserved!), the KE = 64 J.

d) T = 2*pi(m/k)^(1/2), but since you aren't given the mass in this problem, you can't solve it.

MARLON: You're slacking!

I know, I know. So let's suppose the mass was 1 kg. This would mean the period would be 0.363 seconds.

e) If the pendulum has the same frequency as the mass on the spring, it must also have the same period. The period of oscillation for a pendulum is given by T = 2*pi(l/g)^(1/2) so for a period of 0.363 seconds, this means the length would have to be 3.27 cm. That's a small pendulum, so it will swing back and forth fairly quickly.

MARLON: That pendulum is NOT slacking!

I couldn't agree with you more.

6.
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