AP1 EXAM SOLUTION GUIDE

AP EXAM QUESTIONS 1-5 2 Dim Motion

1. CIf both balls are thrown from the same height horizontally, they will both strike the ground at the sametime (because all of the vertical variables are the same). Also note that vertical acceleration (9.8 m/s2)does not depend on mass.⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯2. EIn projectile motion (which is free-fall), the acceleration is always 9.8 m/s2 down, so the accelerationstays the same during the entire flight.⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯3. BThe vertical component of the ball’s velocity obeys the equation y y v gt v0 = − + . This equation is a linewith a slope of –g and an intercept of v0y. The negative slope means the graph can only be Graph 2.⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯4. CThe horizontal component of the ball’s velocity is constant and not zero, and Graph 3 is the graph of aconstant but non-zero function.⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯5. CThe speed of the ball is never zero, but decreases on the way up and increases on the way down. Since itdecreases on the way up, speed is the least at the peak of the trajectory.⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

AP EXAM QUESTIONS 6-10 Momentum

AP EXAM QUESTIONS 11-12 2 DIM MOTION/VECTORS

Q 13 Energy

Q 14 Energy

Q 15 Energy

Q 16.

Q 17

Q 18

19-21

DYNAMICS

Questions 22-27 did not supply answer guides

22 D

23 B

24 A

25 A

26 C

27 B

VARIOUS MULTIPLE CORRECT

28 AC

29 AB

30 AD

FREE RESPONSE 1 ENERGY WORK POWER

FREE RESPONSE 2

Problem 2 (13 Points) (Teacher Made)

(a) 4 points

Showing (average velocity) = (change in position)/(change in time) anywhere in (a)

(This point is not given for merely saying distance/time unless point is also given.)

Choosing two positions between t = 0 and t = 0.70 s (inclusive) for part (a-i), AND choosing two positions between t = 1.30 s and t = 2.00 s (inclusive) for part (a-ii)

Pairing appropriate positions with appropriate times in a vavg = x/t equation in EITHER (a-i) or (a-ii)

Example: vavg = (1.20 m – 1.06 m) / (2.00 s – 1.30 s) for part (a-ii) (units & sig-figs not necessary)

Stating vavg = 1.0 m/s for (a-i) AND vavg = 0.2 m/s for (a-ii)

(b) 2 points

Acceleration starts at either t = 0.7 s or t = 0.8 s, and acceleration ends at either t = 1.2 s or t = 1.3 s.

Stating that, between these times, the distance between markings is changing

(c) 3 points

A horizontal line at the value found in (a-i) from t = 0 to the start of the interval stated in (b)

A horizontal line at the value found in (a-ii) from the end of the interval stated in (b) to t = 2.00 s

A negative-slope line that connects the two horizontal segments (only awarded if and awarded)

(d) 3 points

Showing (acceleration) = (change in velocity)/(change in time) OR acceleration is the slope of v vs. t.

(This point is not awarded for merely saying that acceleration = velocity/time unless point was given)

Showing substitutions consistent with the answers to (a-i), (a-ii), and (b) OR stating that the slope of the middle, diagonal section of the graph should be taken.

Acceleration is between 1.33 m/s2 and 2.00 m/s2 (only awarded if and awarded)

The acceleration is directed to the left.

FREE RESPONSE 3 16 PTS 1 DIM KINEMATICS

Problem 3: Consider the following eight velocity vs. time graphs. Positive velocity is “forward” velocity.

Graph A Graph B Graph C Graph D

Graph E Graph F Graph G Graph H

(a) Rank the objects based on their acceleration from “most forward” to “most backward”. (If any two graphs show the same acceleration, put an equal sign between those letters.)

Most “forward” acceleration Most “backward” acceleration

Briefly explain how you determined the accelerations. H> B=C > D=F > A=G >FCAN BE EXPLAINED WITH ANGLE OF SLOPE, POSITIVE VS NEGATIVE SLOPE, EQUATIONS, DIRECTION OF SLOPE3 PTS CORRECT RANKING 2 PTS FOR EXPLANATION.

(b) Rank the objects based on their displacement over the 4-second interval. F> A=C> E=H> B=G > D

3 PTS CORRECT RANKING

Most “forward” displacement Most “backward” displacement

(c) Rank the objects based on their distance traveled over the 4-second interval. Remember that distance is a scalar that can never be less than zero.

Longest distance

Shortest distance

Briefly explain how you determined the displacements and distances. Make sure you explain the difference between distance and displacement. F=D> A=B=C=E=G=H CAN BE EXPLAINED WITH DEFINTIONS OF AREA, NEGATIVE VS POSITIVE OR MATMATICAL PROOF3 PTS CORRECT RANKING 2 PTS FOR EXPLANATION

(d) Positive net work is done on an object if it gains kinetic energy. Negative net work is done on an object if it loses kinetic energy. 3 PTS 1 PT FOR EACH ANSWER

(i) For which two objects was positive net work done during the 4-second interval? C G

(ii) For which two objects was negative net work done during the 4-second interval? AB

(iii)For which four objects was zero net work done during the interval? DEFH

Problem 1(a) H > B=C > D=F > A=G > E

Ranking 3 pts, reasoning 2 pts

B and C are equal

D and F are equal

A and G are equal

H is the highest

E is the lowest

H, B, and C greater than D and F

H, B, and C greater than A, G, and E

D and F greater than A, G, and E

The entire ranking is correct

Acceleration is the slope

Acceleration is the slope of a v vs. t graph

Positive slope means positive (forward) acceleration

Negative slope means negative (backward) acceleration

Zero slope is zero acceleration

(b) F > A=C > E=H > B=G > DRANKING 3 PTS

A and C are equal

E and H are equal

B and G are equal

F is the highest

D is the lowest

F, A, and C greater than E and H

F, A, and C greater than B, G, and D

E and H greater than B, G, and D

(D Graphs A, E, and G Graph A

Graph E

Graph G

Acceleration points down the plane

Down the plane is backwards

These graphs have backward acceleration

The entire ranking is correct

(c) F=D > A=B=C=E=G=HRANKING 3 PTS, REASONING 2 PTS

F and D are equal

A, B, C, and G are equal

E and H are equal

F and D the first two ranked

The entire ranking is correct

Displacement is the area

Displacement is the area of a v vs. t graph

Positive area means positive (forward) displacement

Negative area means negative (backward) displacement

Zero area is zero displacement

Distance is found by taking all areas a positive

(