AP Physics 1, Unit 1 Test Review

  1. You drop a small ball, and then a second small ball. When you drop the second ball, the distance between them is 3.0 cm. What statement below is correct?

Answer: C. Because acceleration is governed bythe “x = ½ at2” of our formula, as time increases, displacement increases exponentially. Therefore the distance will grow over time. The type the ball has no impact.

  1. The distance between the balls stays the same as they fall.
  2. The distance between the balls decreases as they fall.
  3. The distance between the balls increases as they fall.
  4. The distance depends on the type of ball used.
  1. You have two small metal balls. You drop the first ball and throw the other one in the downward direction. Choose the statements that are not correct.

Answer: C.

  1. The second ball will spend less time in flight.True due to its initial velocity.
  2. The first ball will have a slower final speed when it reaches the ground.True because the second ball is adding in an initial velocity.
  3. The second ball will have a larger acceleration.False because both are accelerated by the exact same g (-9.8 m/s2).
  4. Both balls will have the same acceleration.True becauseboth are accelerated by the exact same g (-9.8 m/s2).
  1. You throw a small ball upward and notice the time it takes to come back. If you then throw the same ball so that it takes twice as much time to come back, what is true about the motion of the ball the second time?

Answer: A. Acceleration due to gravity is -9.8 m/s2, and the acceleration formula is change in velocity over change in time.Let’s focus on the ball’s return trip only, so initial velocity and time are both 0.

a = vf –vi-9.8 m/s2=vf

tf – ti tf

So if the time doubles, and the only acceleration is due to gravity, the velocity must double as well.

  1. Its initial speed was twice the speed in the first experiment. Correct!
  2. It traveled an upward distance that is twice the distance of the original toss. In double the time an object will always travel 4 times the distance.
  3. It had twice as much acceleration on the way up as it did the first time.Only gravity accelerates it.
  4. The ball stopped at the highest point and had zero acceleration at that point.Gravity accelerates it throughout the path.
  1. While running, how should you throw a ball with respect to you so that you can catch it yourself?

Answer: C. Its horizontal velocity component is already the same as you when running, and that doesn’t change.

  1. Slightly forward
  2. Slightly backward
  3. Straight up
  4. It is impossible.
  1. The graph above shows the velocity versus time for an object moving in a straight line. At what time after t = 0 does the object again pass through its initial position?

Answer:B. The area under a velocity/time graph represents distance traveled. The area to the negative that the ball moves is represented by the downward facing triangle. That is only equaled by the area above between 1 and 2 seconds, which is when the object passes through the initial position. See the new diagram I included that shows where the areas match up, which is somewhere between 1 and 2s.

a. 1 s b.Between 1 and 2 s c. 2s d. Between 2 and 3 s

  1. A diver initially moving horizontally with speed v dives off the edge of a vertical cliff and lands in the water a distance d from the base of the cliff. How far from the base of the cliff would the diver have landed if the diver initially had been moving horizontally with speed 2v?

Answer: C. The diver begins with only horizontal velocity. Since the vertical descent time is identical (remember the ball I shot across the room hitting the floor at the same time as the dropped ball) the times in the two cases are identical. Since we know that: v = dt and v doubled so: 2v = 2dt
a.d b. c.2d d.4d e. can’t be determined without knowing the height of the cliff

7. / Which graph of v versus t best describes the motion of a particle whose velocity is constant and negative?
Answer: B. Graph 2 shows a zero slope line in the negative quadrant (which indicates negative velocity), so acceleration is zero (zero slope = no acceleration and constant velocity).
A) 1 B) 2 C) 3 D) 4 E) 5
  1. In which graph of v versus t does the particle end up closest to its starting point?
Answer: D. Using the area under the graph to determine displacement, graph 4 appears to have the least area between the line and the origin.
A) 1 B) 2 C) 3 D) 4 E) 5
  1. In which graph of v versus t does the particle end up farthest from its starting point?
Answer: A. Using the area under the graph to determine displacement, graph 1 appears to have the largest area between the line and the origin.
A) 1 B) 2 C) 3 D) 4 E) 5
  1. An eagle flies at constant velocity horizontally across the sky, carrying a turtle in its talons. The eagle releases the turtle while in flight. From the eagle’s perspective, the turtle falls vertically with speed v1. From an observer on the ground’s perspective, at a particular instant the turtle falls at an angle with speed v2. What is the speed of the eagle with respect to an observer on the ground?

Answer: D.Think of this as a reverse projectile problem. In my sketch below, the eagle is traveling to the right with velocity “V eagle”, the turtle’s apparent angular descent is represented by “V2”, and the vertical component of that (which is the same as his relative descent to the eagle) is represented by “V1”. The turtles horizontal vector follows that of the eagle. Using the Pythagorean theoremV12 + Ve2 = V22, so we solve for Ve.

V12 + Ve2 = V22

Ve2 = V22 -V12

Ve = square root of: V22 - V12

a.b.c.d.

  1. If an object is moving at uniform speed in a straight line, its instantaneous velocity halfway through any time interval is
C. Uniform speed by definition is constant. Since it is in a straight line, it is also the same vector, and therefore the same as the average velocity.
  1. greater than its average velocity.

  1. less than its average velocity.

  1. the same as its average velocity.

  1. impossible to determine from this information alone.

  1. The graphs represent the velocity of a particle along the x axis as a function of time. The particle begins at the origin. In which graph is the particle the farthest from the origin at t = 5 s?
Answer:E. Remember that the area under a velocity/time graph represents distance traveled. Each of the others has a negative and positive component which will cancel. Only E has constant positive velocity and the largest area between the line and the 0 velocity axis.

  1. An object is released from rest on a planet that has no atmosphere. The object falls freely for 3.0 meters in the first second. What is the magnitude of the acceleration due to gravity on the planet?
Answer: B. a = vf –vi where vi and ti are zeroa = vf = 3 m/s= 3.0 m/s2
tf – tivf = 3 m / 1 s tf1 s
a. 1.5 m s2 b. 3.0 m/s2 c. 6.0 m/s2 d. 10.0 m/s2 e. 12.0 m/s2


A. 24 right – 8 left = 16 rightB. 10 right – 30 left = 20 leftC. 4 left – 20 right = 16 right
D. 12 right – 20 right = 32 rightE. 4 left – 16 left = 20leftF. 8 right – 16 right = 24right
So listed from Greatest to Least: D. > F. > B. = E. > C. = A.
An object slides one meter down a frictionless ramp of constant slope as shown below. A student measures the time it takes for the object to travel various displacements using a stopwatch. Three consecutive trials are measured, and the data is recorded as shown below.

Displacement (m) / Avg. Time (s)
0 / 0
0.2 / 0.68
0.4 / 0.98
0.6 / 1.18
0.8 / 1.38
1.0 / 1.52

Determine the acceleration of the object.Answer: a = vf – vi= 1.0 m/1.52 s – 0 m/s = 0.422 m/s2
tf – ti 1.52 s – 0 s

The military wants to determine the speed with which a bullet leaves the barrel of a gun, but all of their high tech equipment is currently unavailable. Using only a meter stick, determine an experimental procedure that would allow this velocity to be determined. Be sure to address the measurements that would be taken, the calculations made, and the ways error would be minimized.

In freefall we know that x = xo + vxot + ½axt2, xo = 0, andvxot = 0

That leaves x = ½axt2, and means that the distance an object falls in 1 second is: x = ½(-9.8m/s2)(1 s)2

Or -4.9 meters. This is a projectile problem. The bullet drops 4.9 meters in one second, so place the rifle on a stand of 4.9 meters height, fire it and wherever it hits the ground, it will have traveled for 1 second. You can only measure distance, but that allows us to have both a distance to the point of impact and a time of 1 second, so we can now calculate muzzle velocity using v = d/t. To minimize error you could repeat the procedure, and even increase the height of the gun. For a 2 second time to impact, you’ll need to place the gun up 19.6 meters:

x = ½(-9.8m/s2)(2s)2= or 19.6 meters high

While looking out of your apartment window one day, you notice a flower pot falling. You know from buying your blinds that your windows are 1.2 meters tall, and you estimate that the flower pot falls by in 0.24 seconds. In order to assist the investigation of whose flower pot that fell, you, as a faithful physics student, must determine the height above your window from which the flower pot fell.

V = d/t at the instance the pot falls past your window: 1.2 meters / 0.24 seconds = 5 m/s. Assuming it is in free fall from an above window sill, it has fallen for approximately 1 second. To calculate how far up it was originally:

x = xo + vxot + ½axt2, xo = 0, andvxot = 0

That calls forx = ½vx+vx0 * t = ½ 5 m/s +0 m/s * 1 second = 2.5 meters up!

A director wishes to have an actress fall onto a moving truck from an overpass as a part of a stunt scene for a new movie. If the overpass is 8.6 m tall, and the truck moves at a constant 16 m/s, how far away should the truck start for the actress to land safely?

↑ +y

VERT

Vo = 0 m/s

V = ?m/s

Δy = 8.6 m

a = -9.8 m/s2

t = FIND

+x

HORZ

Vo = 16 m/s

V = 16 m/s

Δx = FIND

a = 0 m/s2

t = same as vertical, so 1.8 s

X = v * t, or 16 m/s * 1.8 s = 29 meters!

y = ½ at2

t = √y
½ a

t = √8.6 m = 1.8 s
(0.5) (-9.8m/s2)