Questions I Missed In Physics

Front 1

If you are asked to figure out from answer choices what could be a pair of component vectors for a vector, what do you do?

Back 1

Find the resultant vector of the component vectors and compare to the original vector. If it doesn't match the original vector, it's not the right vector.

Front 2

If you multiply 2 vectors and get a vector, what will be the direction of the vector and the magnitude?

Back 2

The vector product will point in a direction perpendicular to both Q and R, and have a magnitude proportional to the sine of the angle between Q and R.

Front 3

If an object moves up and comes down back to the same position, what is it's vertical velocity?

Back 3

The vertical velocity is zero. Vertical velocity is vertical displacement over time. The vertical displacement is 0.

Front 4

Consider a particle moving along the circumference of half a circle. The particle moves along the circumference at a constant speed of 1 m/s. What is the average acceleration of the particle as it moves from one side of the circle to another?

Back 4

The answer is 0.4/pi m/s^2. Acceleration is the change in velocity divided by time. The initial velocity is 1 m/s^2 up; final velocity is 1 m/s^2 down. The change in velocity is therefore 2 m/s^2 (the magnitude). The time is found from speed - the distance is 5 pi, so you find the seconds by multiplying distance with the speed. Then you divide the velocity by the time to get acceleration.

Front 5

Can an object accelerate and have a constant speed at the same time?

Back 5

Yes because a change in direction will result in a change in velocity and may or may not result in a change in speed. Think of a particle moving in a circle at a constant speed. It has centripetal acceleration.

Front 6

What is a requirement for acceleration in uniform linear motion equations?

Back 6

It is required that the acceleration is constant.

Front 7

When using a equation that only involves velocity, distance, and time and constant acceleration, how should the final and initial velocity values be used?

Back 7

First solve for acceleration. Do vf^2 = vi^2 + 2a(x). Once you get a, then use the equation a = vf-vi/t or vf = vi + at. Then solve for t. Do not use average velocity unless the question explicitly asks for it or gives it in the equation. Do no use vf-vi = xf-xi/t either.

Front 8

A particle moving at 5 m/s reverses its direction in 1 s to move at 5 m/s in the opposite direction. If its acceleration is constant, what is its speed at 0.5 s?

Back 8

The answer is 0 m/s. First find the acceleration for the trip. Use the equation a = vf-vi/tf-ti. a = (-5 m/s - 5 m/s)/1s = -10 m/s^2. Then find vf = vi + at. This time plug t = 0.5s, since we are looking for the speed (velocity) at 0.5 second. Speed and velocity are interchangeable as long as the distance is the same (the displacement is not different from what the distance would be).

Front 9

How do you solve for distance questions in uniform linear motion questions?

Back 9

Find the displacement in both directions and add them together. So, if you were given a problem where vi = 10m/s and vf = -10m/s, then you would solve for delta x in the direction of vi = 10m/s and vf = 0m/s, then you would solve for delta x in the direction of vi = 0m/s and vf = -10m/s. You would find the acceleration, which is the same in both directions because acceleration is always constant for these problems. You would also find the time in both directions, and you would find the time by setting the constant acceleration equal to the different vi and vf. Then you would add up the distances in both directions, with the magnitude of the direction that is negative.

Front 10

If you were looking at a graph of displacement versus time, what kind of slope would represent zero acceleration?

Back 10

Velocity is the slope of these graphs. A change in velocity would represent acceleration. You would pick a graph that has a constant, straight slope, in either the up or down or horizontal direction. A constant slope would mean zero acceleration, a changing slope would mean that there is acceleration.

Front 11

If you are given a graph of displacement versus time, what graph would represent positive velocity?

Back 11

Any graph with a positive slope. The slope does not have to be a straight, constant line. The line could curve up, like a parabola. As long as the slope (rise/run) of the line is positive, it would be a positive slope.

Front 12

If you are looking a velocity versus time graph, what kind of graph would have a positive velocity?

Back 12

As long as the velocity is greater than zero, it is positive. It can be a constant horizontal line, it can be a straight slope up, and it can curve up like a parabola, all would be a positive velocity.

Front 13

What does the total area between the curve and the zero time axis on any velocity versus time graph represent?

Back 13

The total area would assume that the area beneath the x-axis is positive, so the total area would be distance, not displacement.

Front 14

A projectile has R for the range, v for the initial velocity, h for the maximum height, theta for the angle from the horizontal from which hte projectile is launched, and t is the time required for the entire flight. What equation would represent the maximum height h reached by the projectile?

Back 14

1/8gt^2. Remember that t is the time for the entire flight of the projectile. If we examine only the second half of the flight, the projectile drops from the maximum height, and starts from an intial velocity of zero. This would occur in half the time or 1/2t seconds. Using the equation x = initial velocity*t +1/2at^2, we would get x = 1/2a(1/2t)^2. x = 1/8at^2.

Front 15

A projectile has R for the range, v for the initial velocity, h for the maximum height, theta for the angle from the horizontal from which hte projectile is launched, and t is the time required for the entire flight. As the angle theta increases from 0 degrees to 90 degrees, what happens to the range R?

Back 15

It increases to a maximum at 45 degrees, then decreases.

Front 16

In order to maximize the range of a projectile, what should the angle theta be?

Back 16

45 degrees

Front 17

Which two angles theta will result in the same range R of a projectile?
A) 45 degrees and 30 degrees
B) 30 degrees and 60 degrees
C) 45 degrees and 60 degrees
D) 45 degrees and 90 degrees

Back 17

30 degrees and 60 degrees. Angles that are equidistant from 45 degrees will result in the same range where there is no air resistance. 45 degrees is the maximum angle for a projectile range where there is no air resistance.

Front 18

Air resistance would decrease all of the following except:
A) R B) h C) initial velocity D) t

Back 18

Initial velocity is a given and is not affected by air resistance. Air resistance slows a projectile resulting in a smaller maximum height and shorter range. Time is trickier since air resistance decreases the trip upwards, but increases the trip downward. Since air resistance has a greater affect on faster moving bodies, the trip upward is decreased by more than the trip downward is increased.

Front 19

How are time t and angle theta related for projectiles?

Back 19

As theta increases, the vertical velocity increases. uy = u*sintheta. uy is the initial velocity. The time of the flight is directly proportional to the vertical velocity. The equation for velocity reveals that the magnitude of velocity is reduced by an amount “gt” after a time interval of “t” during upward motion. The projectile is decelerated in this part of motion (velocity and acceleration are in opposite direction). The reduction in the magnitude of velocity with time means that it becomes zero corresponding to a particular value of “t”. The vertical elevation corresponding to the position, when projectile stops, is maximum height that projectile attains. For this situation ( vy = 0), the time of flight “t” is obtained as : vy=uy−gt
⇒0=uy−gt
⇒t=uy/g
uy is the initial velocity and vy is the final velocity.

Front 20

A bullet fired from a high powered rifle at target 20 meters away. The bullet will begin to fall when?

Back 20

The moment it reaches terminal velocity. The barrel is the only thing applying a force upward to counteract gravity. The projectile is projected with certain force at certain angle to vertical direction. The force that initiates motion is a contact force. Once the motion of the ball is initiated, the role of contact force is over. It does not subsequently affect or change the velocity of the ball as the contact is lost.

Front 21

An object is dropped from a height h and strikes the ground with a velocity v. If the object is dropped from a height 2h, which of the following represents its velocity when it stikes the ground?
A) v B) 1.4v C) 2v D) 4v

Back 21

If we use the equation vf^2 = vi^2 +2ax, and use x = h and a = g, then we would find that v = sqrt(2gh). If we double h, then we would increase the velocity by sqrt(2), because then v = 2*sqrt(gh). So the answer is 1.4v.

Front 22

When solving for the maximum height of projectiles, do you need to know the horizontal velocity?

Back 22

No, the horizontal velocity is irrelevant. All you need to know is the vertical velocity.

Front 23

An object strikes the ground with a speed of 25 m/s. If it was originally thrown with a horizontal velocity of 15 m/s, and a vertical velocity of zero, from how high was it dropped?

Back 23

We have to find the vertical component of the speed. We cannot just use the speed of 25m/s as the speed to solve the problem. So, to find the vertical component of speed, we have to use the Pythagorean theorem. The horizontal component of velocity is 15 m/s and the hypotenuse is the speed and it's 25m/s. So 15^2 + x^2 = 25^2. So the vertical component of velocity is 20m/s. Then we use that component to solve for the height. We don't know the time, so we use vf^2 = vi^2 + 2ax. x = 20 m.

Front 24

When solving for total time of a trip for projectiles, what do you have to make sure to do with the time?

Back 24

Make sure to multiply the time by 2, because the time up is equal to the time down. So the total time of the trip is the double of the time found at the maximum height.

Front 25

A projectile is launched at an angle of 30 degrees to the horizontal from a 15 m platform. Its initial velocity is 20 m/s. How far does it travel?

Back 25

First find out how long the projectile is in the air, and then multiply the air time by the horizontal velocity to get the total distance. Air time is dependent entirely on vertical velocity. The vertical velocity is 20 m/s times sin30, or 10 m/s. From vf = vo + at, we fnd that the time for the projectile to reach the maximum height is 1 second. This is the time to reach the maximum height after being thrown from 15 m from the ground. The maximum height from the 15 m is y = vot + 1/2at^2. y = 10m/s(1s) + 1/2(-10m/s^2)(1s)^2 = 5m. So the total height, the true maximum height is 15m of the platform + 5 m from the launch. The time required to fall from the total height of 20 m is 2 seconds. We find this from y = vot + 1/2at^2. vot is equal to 0 because vy = 0 at the maximum height. so 20m = 1/2(10m/s^2)(t^2). So t = 2s. This is the total time required to fall from 20m. So the total flight time is 3 seconds (1 second to reach the maximum height from 15 m platform, and 2 seconds to fall from 20 m). The we multiply 3 seconds by the horizontal velocity, which is vicos30, which is 20m/s*cos30 which is about 51m.

Front 26

A projectile reaches its maximum height in approximately 1.8 seconds. It has a horizontal velocity of 24 m/s. At what speed is it launched?

Back 26

First we must find the vertical component of the velocity from the time of flight. It is vf = vi +at. vi is 18m/s. Then to find speed, we need to do vector addition. We would do the Pythagoream theorum with the horizontal and vertical velocity components. So (24m/s)^2 + (18m/s)^2 = (speed)^2. Then you solve for speed. The answer is 30 m/s.

Front 27

A projectile launched over ground level reaches its maximum height in 10 seconds. Approximately what was the range of the projectile if it was launched with a speed of 200 m/s?

Back 27

The answer is 3400m. To solve for range problems, you have to find the horizontal velocity and multiply it by the total flight time. Since we were said that it takes 10 seconds to reach the maximum height, the total flight time must be 20 seconds. To find the vertical component of velocity, we would use the equation vf = vo + at. We would set vf = 0 and t = 10 seconds. The initial vertical component of velocity would be 100 m/s. Then we could solve for the horizontal component of the inital velocity in a couple of ways. We could use the Pythagoream theorum and say (100m/s)^2 + (x)^2 = (200m/s)^2. Then we could solve for x. Or we could say vy = vsintheta. v = the initial speed which is 200 m/s. vy = 100m/s. So sintheta = 1/2, so theta is equal to 30 degrees. Then we know that horizontal component of velocity = vcostheta. So 200m/s*cos(30) = about 3400m/s.

Front 28

Two objects are dropped from a height of 100m. If one object is heavier than the other object, how will the force and acceleration of the objects compare to each other? (ignoring air resistance)

Back 28

The heavier object will experience more force and will accelerate at the same rate as the lighter object. Acceleration is 10m/s^2 for both heavy and light objects on earth. Acceleration does not change, so the acceleration will stay at the same rate for both objects. The heavier object has more inertia, so requires more force to keep its motion at the same rate.

Front 29

Two balls with exactly the same size and shape are launched with the same initial velocity from the surface of a perfectly flat plane. When air resistance is considered, how will the ball with the greater mass's flight time and maximum height compare to the ball with the smaller mass?

Back 29

The ball with the greater mass will have a longer flight time and a greater maximum height. Air resistance acts against motion. The air resistance is based on size and shape and will be the same for both balls, so the ball with the greater inertia (mv = p) will resist the change in its motion the most. So say if you throw a heavy ball versus a feather, the heavy ball is going to higher and stay in the air longer than the feather.

Front 30

All of the following increases the force of air resistance on a projectile except:
A) greater mass of the projectile
B) greater surface area of the projectile
C) greater velocity of projectile
D) greater density of air

Back 30

The answer is A - mass does not increase the force of air resistance. Air resistance increases with velocity, surface area, and air density.

Front 31

Which of the following would most likely experience the greatest force of air resistance if dropped from an airplane?
A) a feather
B) an elephant
C) a bullet
D) a toaster

Back 31

The answer is B. The surface area, not the mass, changes the air resistance. The elephant has the greater surface area.

Front 32

How many significant figures are in 0.008010?

Back 32

4. Just count the total (8,0,1,0). This is 4.

Front 33

The following values of a sample of gas were measured in the lab: P=1.5atm, V = 10.3 L, T = 298.0K. If we were to calculate nR from the equation PV = nRT, how many significant figures would be in the answer?

Back 33

2. So the answer could be 0.XX, where the Xs are numbers.

Front 34

Which of the following has a mass of approximately 1 amu?
A) one proton B) one electron
C) one atom of 12C D) one mole of 12C

Back 34

An amu is precisely defined as 1/12 the mass of an atom of 12C, which is approximately (not exactly) the mass of 1 proton

Front 35

How much does 3-mole sample of Na weigh?
A) 23 amu B) 69 amu
C) 23 grams D) 69 grams

Back 35

3 mole sample of Na weighs 69 grams, not amu.

Front 36

The charge on one mole of electrons is given by Faraday's constant (F = 96,500 C/mol). What is the total charge of all the electrons in 2 grams of He?

Back 36

The answer is 96,500 C/mol. First, we find out how many moles of He we are working with. We are working with 1/2 moles of He. But the conversion is that Faraday's constant is for the moles of electrons. We only found the moles of He. There are 2 moles of electrons with He, based on the periodic table. So 1/2 mol of He * 2 mol of electrons yields 1 mol of electrons. So 1 mol of electrons yields 96,500 C/mol.

Front 37

Which of the following represents the charge on one mole of electrons?
A) 1 e B) 6.02 x 10^23 e
C) 1 C D) 6.02 x 10^23 C

Back 37

The answer is B. The charge on one electron is e, so one mole is Avogadro's number multiplied by e.

Front 38

69% of naturally occurring copper is copper-63. If only one other isotope is present in natural copper, what is it?
A) Copper-61 B) Copper-62
C) Copper-64 D) Copper-65

Back 38

The answer is D. The atomic weight of copper is 63.5. Therefore, the other isotope of copper must be heavier than the predominant copper-63. Since more than 2/3 of the copper is copper-63, the atomic weight should be closer to 63 than to the other isotope, so that rules out copper-64. If copper-64 were the other isotope, the atomic weight would have to be less than 63.5. This leaves D.

Front 39

A 0.5kg ball is dropped from 2 m and achieves a speed of 10 m/s immediately before it hits the ground. Upon rebound, it has a speed of 8 m/s in the opposite direction. What is the total change in momentum from the maximum downward speed to the initial rebound speed?


(a) 1
(b) 9
(c) 10
(d) 18

Back 39

The initial momentum is (0.5 kg)(10 m/s) = 5 kg m/s, while the final momentum (0.5 kg)(8 m/s) = 4 kg m/s. The two are added since they are in opposite directions, yielding a change in momentum of 9 kg m/s.

Front 40

A solid dodecahedron has a weight of 100 N with a cross-sectional length of 50 cm from each pair of the closest opposite vertices. It is immersed in water and its weight is now 80 N. What is the density of the dodecahedron? Assume ρwater = 1.00 g/mL and g = 10 m/s2.


(a) 0.2 g/mL
(b) 2 g/mL
(c) 5 g/mL
(d) 50 g/mL

Back 40

The fact that the solid is a dodecahedron, and the fact that the radius is given, is essentially useless information. You are not required to know how to calculate the volume of a dodecahedron without an equation, so ignore this part. You are told the weight is 100 N, and that g = 10 m/s2, hence the mass must be 10 kg. Underwater, the apparent mass must be 8 kg. This means the water has displaced 2 kg worth of mass. Since the density of water is 1.00 g/mL, it is also 1.00 kg/L. Thus 2 L of water have been displaced for 2 kg of the mass of the solid. Hence the volume of the solid must be 2 L in order to displace 2 L of water when immersed.

Now we know the mass of the solid is 10 kg, and that its volume is 2 L. 10/2 = 5 kg/L; 5 g/mL.

Front 41

The total volume of a ship is 106 L. The ballast tanks have a volume of 105 L; these tanks can take in and expel water to raise and lower the boat. The mass of the empty ship and empty ballast tanks is 105 kg. If 50,000 kg of cargo are taken on to the boat, and the ballast tanks are filled to half their capacity, how much more cargo can the ship take before it sinks?


(a) 200,000 kg
(b) 750,000 kg
(c) 800,000 kg
(d) 850,000 kg

Back 41

If the volume is 1,000,000 L of which is 100,000 L of ballast tanks, when the tanks are half-full, the mass of the boat will be 50,000 kg of water plus 100,000 kg of boat. Another 50,000 kg of cargo will bring the mass to 200,000 kg. Since the volume of the ship is 1,000,000 L, the ship can have a total mass of 1,000,000 kg before it will sink. Since 200,000 are already occupied, 800,000 kg more cargo can be placed on the boat. 1 Kg is equal to 1 L and 1 mL is equal to 1 g, which is also equal to 1 cm^3.

Front 42

A satellite is in a perfectly circular orbit around earth. Which of the following describes the direction of the satellite's acceleration?

(a) A vector with its tail tangent to the sun and its head toward earth.
(b) A vector with its tail at the center of the satellite and its head toward earth.
(c) A vector with its tail at the center of the satellite and its head in the direction of the satellite's orbit.
(d) A vector with its tail tangent to the satellite and its head toward earth.

Back 42

Recall that centripetal acceleration governs the motion of an object in a circular path. The vector for this acceleration begins from the center of mass of the object in orbit and points toward the center of the object's circular path, in this case, the planet.
The answer is b.

Front 43

Two bodies, one charged and one uncharged, come into contact. Which of the following options is (are) true regarding this event?

I. If either of the bodies are poor conductors, they will both retain their charge.
II. If both bodies are good conductors, they will both retain their charge.
III. If both bodies are good conductors, protons will move from the charged body to the uncharged body.

(a) I only
(b) II only
(c) I and III only
(d) II and III only

Back 43

II is incorrect since good conductors will tend to release or gain charges easily. III is incorrect since static charges are governed by the movement of electrons, not protons.
A is the answer.