OAT Test Prep: Physics Part 1
by: Dale Paynter
Hey everyone, today I will be giving an overview of the Physics section for the OAT. This will be the first article in a multi-part series covering Physics. At the end of this article, we have free, exclusive optometrystudents.com practice problems!
The Physics section covers material from both semesters of general physics, so ideally you will have completed these courses before you take your OAT.
Here is a list of most of the common topics covered in the Physics section:
- Mass and weight (not the same thing!)
- Kinematics (motion, velocity, acceleration, know how to work with vectors)
- Gravity (gravity is a type of force)
- Energy (potential energy and kinetic energy)
- Momentum (including angular momentum)
- Springs (period, angular frequency)
- Pressure (remember that pressure acts across an area)
- Gas law (PV=nRT) This really belongs in the General Chemistry section, not Physics but I list it here anyway because it is useful for understanding pressure
- Electromagnetism: Electrical Force, Electrical Potential, , Magnetism and its relationship to electric fields (and vice versa)
- Circuits (DC circuits only!): voltage, current, resistance, capacitors, transformers
- Wave Properties (mainly interference patterns)
- Optics (lenses and mirrors)
- Nuclear Physics (You only need to know the very basics of this, i.e. no real knowledge of quantum mechanics is necessary)
You also need to fully grasp the metric system! For example, you must know how to interconvert between grams and kilograms, centimeters and meters, joules and kilojoules, etc. You also should know the units for many of the concepts listed above; for example, force is in newtons (one newton is equal to one kilogram meter per second squared).
I’m sure you realize by now that there is much to know for the Physics section of the OAT, so let’s get straight to the practice problems! Many of the actual OAT Physics problems will be slightly easier than these, so as long as you can finish these at a reasonable pace, then you are probably in good shape.
A baseball player hits a baseball a certain distance. If, immediately after being struck, the ball was traveling at a velocity of 40 m/s at an angle of 30 degrees to the horizontal, then how far from the player does the ball land? Assume that it lands at the same height at which it was hit, and ignore air resistance. Use -10 meters per second squared for gravitational acceleration.
This is a classic projectile motion problem, and although it is more common to be asked about a cannonball, the same physics will apply to any airborne objects. The way that I’d solve this problem is by first breaking up the ball’s velocity into its horizontal and vertical components. To do this, I see that the total velocity (you can think of this as the sum of the horizontal and vertical components) is equal to 40 m/s as given in the problem. The angle to the horizontal is 30 degrees, so I now have enough information to draw my right triangle. I’ll first draw a horizontal line to represent the horizontal component of the velocity, then I’ll draw a second line forming a 30 degree angle with the horizontal line to represent the total velocity of 40 m/s, and I’ll connect the other ends of these two lines with a third, vertical line, representing the vertical component of the velocity (the intersection of the horizontal and vertical lines should obviously be at a right angle). Now, using basic trigonometry, I see that the sine of 30 degrees (which is 0.5) is equal to the vertical component of the velocity divided by the total velocity (40 m/s) and this will give me a vertical velocity component of 20 m/s. As soon as I know the vertical velocity component, I can immediately find the time it takes the object to travel, because the problem states that the ball lands at the same height at which it was hit. Each second the ball is in the air, gravity reduces the vertical (and only the vertical) component of the velocity by 10 m/s, so I can see that it will take 2 seconds for the ball to reach its apex (highest point) and, due to the symmetry of this function, it will take exactly 2 more seconds (or 4 seconds total) for the ball to return to the original height (i.e. it will take a total of 4 seconds for the ball to land). So I know the travel time is 4 seconds, so all I need now is to find the horizontal component of the velocity and multiply this by the time (4 seconds) to find the horizontal distance traveled. To find the horizontal component of the velocity, I will take the cosine of 30 degrees (sq. root of 3 over 2) and set this equal to the horizontal velocity component divided by 40 m/s. So 20 times the sq. root of 3 m/s is the horizontal velocity component, and I simply multiply this by the time of 4 seconds to give me a total distance of 80 sq. root of 3 meters.
An 18 V battery is attached to a circuit containing a 2 ohm and a 4 ohm resistor applied in series, and a 3 ohm resistor applied in parallel to the first two resistors. What is the total current of the circuit, and what is the current across the 2 ohm resistor?
The first thing I would do here is to draw a circuit diagram. I have an 18 V battery, and I’ll draw two branches of the circuit. On one branch I have a 2 ohm and 4 ohm resistor right next to one another (in series) and on the second branch I’ll draw a 3 ohm resistor. Next, I need to figure out the net resistance of the entire circuit. When I have resistors in series, they are simply added together to find net resistance, so a 2 ohm and a 4 ohm resistor in series is the same as a single 6 ohm resistor. So we can now pretend that there is a 6 ohm resistor (in place of the 2 and 4 ohm resistors) in parallel to the 3 ohm resistor. When figuring out net resistance of resistors in parallel, however, we use a specific equation: The inverse of the total resistance is equal to the sum of the inverse resistance of each resistor. What this means is that the inverse of the total resistance is equal to 1/6 plus 1/3 which means that the total resistance of the circuit will be 2 ohms (the inverse of 1/2). Now I can use the simple equation V=IR in order to find the total current of the circuit. The total current of the circuit will be 9 amps. To find the total current across the 2 ohm resistor, I will look at that branch of the circuit. The total voltage drop across that branch of the circuit must be 18 V (it’s easier to see if you draw it out) and the total resistance across that branch is 6 ohms (a 2 ohm and 4 ohm resistor in series act like a six ohm resistor) so I just use V=IR to come up with a current of 3 amps across that branch of the circuit. The current must be the same at all points along a single branch of a circuit, so the current across the 2 ohm resistor is 3 amps.
***Before I continue with the remaining problems, I’d like to mention something. Most problems on the Physics section can be solved by correctly applying the appropriate equation. Besides simply memorizing loads of formulas, however, you will need to be able to master a special problem type found on the Physics section of the OAT: I’ll refer to them as Conceptual problems. A Conceptual problem describes in words (not equations), an event that is taking place, and then you will be asked to answer what physical principles apply. The final two practice problems is are examples of this.
There is a 9 N force being applied horizontally to the right to a 30 kg crate on a flat surface. If the force of friction is equal to 9 N horizontal to the left, the gravitational and normal forces are of equal and opposite magnitude, and no additional forces are acting on the crate, then which of the following statements must be false?
A: The object is stationary.
B: The object has a non-zero acceleration.
C: The object is moving.
D: The potential energy of the object is not changing.
One way of solving must be false multiple choice problems is by eliminating every answer that can be true. Can choice A be true? Of course, the crate can simply be motionless in this problem, because no net force is acting on it (we know there is no net force because each force is exactly countered by an equal and opposite force). And if the crate can just be sitting there, then the potential energy of it certainly doesn’t have to change, right? So we can immediately eliminate answers A and D. But what about C? Can an object with no net force acting on it be moving? Of course it can! Remember, an object in motion remains in motion unless acted on by an outside force and since there is no net force, the crate could certainly be sliding at a constant velocity! Since we have eliminated choices A, C, and D, we know that either I suck at writing OAT problems or the answer is B. I certainly hope that I don’t suck at writing these articles, and yes choice B is the answer. Since the object has no net force acting on it, it cannot be accelerating, because of the equation F=ma and F=0 and m=30 kg, therefore a must equal 0.
For two satellites circularly orbiting the earth at the same distance from the planet, which of the following statements is true if no forces other than the earth’s gravitation act on the satellites?
A: Both satellites must be the same mass.
B: Both satellites must have the same orbital period.
C: Both satellites must have the same potential energy.
D: Both satellites must have identical velocities.
First, let’s think about satellites for a second. What holds satellites in an orbit? I’m sure you all know that gravitational acceleration is what causes and maintains orbits. It turns out that there is an important point about gravity that allows us to immediately eliminate two answer choices. Gravity has the same acceleration on an object regardless of the object’s mass. I’m sure you all know that, ignoring air resistance, a feather would fall at the same velocity as a bowling ball, and although it might be slightly less intuitive, the orbital dynamics of a satellite really have nothing at all to do with the mass of the satellite, so choice A is clearly false. And once we see that choice A is false, choice C must be false, since the equation for gravitational potential energy involves the mass of the object (and we know that the satellites don’t have to be the same mass). So that leaves us with choices B and D. This is a little difficult to explain, so please consult a physics book if necessary, but it works out that the period of any circularly orbiting satellite (the period is the time it takes to complete one orbit) is dependent upon just two things: the mass of the body being orbited (in this case the earth) and the distance at which the satellite is orbiting. They are both circularly orbiting the earth, so they must both have the same orbital period because they are both orbiting at the same distance. Choice B is, therefore, the answer. They must also be orbiting at the same speed, because their orbital path length and period are both the same, so why isn’t choice D also an answer? Well, because speed and velocity are not precisely the same thing. Velocity always includes a directional component, and the satellites do not have to be orbiting the earth in the same ‘direction’ (i.e. one could be orbiting in an east-west direction, and the other could be orbiting in a west-east direction, or a north-south direction, or a northeast-southwest direction, etc.).