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AP Physics 1 Review - Vectors and Kinematics

Introduction - Vector Properties and 1D Kinematics

Recall that vectors are any quantity with both a magnitude and direction.

Magnitude represents how long the vector is, and is often denoted analytically in an equation by an absolute value symbol, |x|. Indicating both magnitude and direction is usually done by one of two ways:

1) By angle - Anything between 0º and 360º or using compass notation like 45º S of E, which is the same as 315º or -45º

2) By components, such as $${4\hat i+3\hat j}$$. The magnitude of vectors written in component notation can be calculated using the pythagorean theorem. So, the magnitude of this vector is 5.

These vectors can also be graphically represented as a field of vectors, where the length of each arrow represents the magnitude, and its direction is implied based on which direction the vector is pointing.

Vector vs Scalar Quantities

Certain wording indicates whether the quantities we are looking for should be a vector or scalar. "Displacement" is a vector quantity whereas "Distance" is a scalar. "Speed" is a scalar but "Velocity" is a vector. The key difference is that scalars don't have direction.

A good way to remember which is a vector and which is a scalar is to think of a car. A car always shows scalar quantities because it's not designed to handle vectors. On your speedometer, it shows the speed that you are going. On your odometer, it shows the distance you have traveled. Both are scalar quantities and don't have a direction associated with them.

What is interesting is that when we take into account the direction, strange things start happening. For example, a car that goes around a 1-mile loop and ends up at the beginning may have traveled a distance of 1 mile, but its displacement is 0 because displacement represents the shortest distance from one point to another.

Similarly, a ball or top spinning in place has a speed of 0, but it actually has an angular velocity that is positive or negative depending on your frame of reference and whether it is spinning counterclockwise (generally positive) or clockwise (generally negative).

Want more practice? Buy an eBook of 50 multiple choice questions on Amazon with similar topic ratios as the exam to prepare for test day by clicking on the book cover above.

Practice Question Set 1 - Graph-based

1.1) Given the above velocity vs time graph, assuming a particle starts at a displacement of 0, what is the total displacement of the particle in 4 seconds?

1. 3 m
2. 6 m
3. 4 m
4. 8 m

Discussion - Remember that displacement in a velocity vs time graph is represented by the area under the curve. In this case, the area under the curve for the first 2 seconds is 4m. Then we add the area under the curve until 4 seconds. Since the area between 2 and 3 seconds is 0, we can ignore it, and the area between 3 and 4 seconds is 2m. We add the two areas together to get 6m, or B as the answer.

1.2) Calculate the average velocity of the particle in the first 3 seconds.

1. 1.5 m/s
2. 1 m/s
3. 1.33 m/s
4. 2 m/s

Discussion - Recall that the average velocity in a velocity vs time graph, similar to the average displacement in a displacement vs time graph, is calculated by taking the area under the curve over the interval of interest, and then dividing by the interval of interest. In this case, we are interested in the first 3 seconds! Reading the question carefully is half the battle! The area under the curve for the first 3 seconds is 4m. Then dividing by the interval of interest (3 seconds) gives us $$\frac{4}{3} \frac{m}{s}$$ or 1.33 m/s.

1.3) Which of the following statements can describe the movement of the object in the above graph? Select all that apply. Assume no air resistance.

1. The object is an iron pellet traveling straight through air, is temporarily stopped by a strong magnet, and then instantly accelerated to its previous speed in the same direction
2. The object is a flying squirrel that flies towards a tree, slows down to a stop, and then jumps towards another tree in the same direction.
3. The object is a ball traveling straight through the air, briefly hits a wall and bounces off at the same speed.
4. The object is a ball in a wind tunnel with the wind traveling at a constant velocity. The ball briefly hits a very thin membrane of sticky goo, absorbs the goo in a perfectly elastic collision, and continues on.

Discussion - You will almost certainly see this type of question on the exam, where you are asked to relate a graph or a stimulus to a conceptual understanding of what is happening in the physical world. Let's start with (A) to see if it makes sense. The iron pellet is "temporarily stopped" (i.e. velocity reduced to 0) by a "strong" magnet, meaning the rate at which the velocity decreased to 0 can be instantaneous. And then it is "instantly" accelerated to its previous speed in the same direction. This answer makes sense and can feasibly describe the behavior of the graph. This is a multi-answer question, though. So we have to find all answers that apply. Next, let's take a look at (B). There is some wording here that tells us it's not exactly what we are looking for. The flying squirrel "slows down to a stop" meaning the deceleration is not instantaneous like we see in the graph. So this eliminates choice (B). (C) seems like it could be a likely candidate since bouncing is pretty instantaneous and it bounces off the wall at the same speed. BUT! And that's a big but. The direction is opposite so the velocity would actually be negative if the situation was as described in choice (C). So we can eliminate that too. Now let's look at choice (D). This is perhaps the trickiest choice because it uses concepts covered in another unit. But since this is review, you should know that in an elastic collision, there is no loss of momentum or kinetic energy. At the same time, it's implied that the goo has negligible mass because it's "very thin" and the ball is stopped instantly because the goo is "sticky". So, choices (A) and (D) are the best answer in this case.

1.4) Free response question (FRQ) - Using the graph above, assume that the object starts at displacement = 0 and draw its displacement over 4 seconds.

Solution: Similar to how we calculated displacement earlier with the area under the curve, we can graph the same object's displacement, by starting with the point t=1s, then calculating the area at t=2s, t=3s and t=4s. Mapping this out should give the following graph:

Notice that the slope of this graph corresponds to the value of the velocity graph. Key points to notice are labeled on the graph above and compare them to the graph of the velocity function shown earlier. Also note that velocity and acceleration have an analogous relationship to position and velocity. So the same question can be asked a different way with acceleration instead of velocity and you should be able to create a velocity over time graph.

Want more practice? Buy an eBook of 50 multiple choice questions on Amazon with similar topic ratios as the exam to prepare for test day by clicking on the book cover above.

2D Kinematics

The kinematic equations all rely on two things: the system being one-dimensional and the acceleration being constant. As long as we can have those two assumptions, the kinematic equations hold water and are usable.

For 2 dimensions, we need to do twice the work. We must divide the 2-dimensional problems into 2 one-dimensional problems. To do so, we resort to our old friend, vector components. By setting an x and y coordinate system, we can solve for any 2-dimensional problem. We just make a couple minor variations to the equations:

$$x = x_0 + \vec{v}_{0,x} \Delta t + \frac{1}{2} \vec{a}_x \left( \Delta t \right)^2$$

$$\left(\vec v_x\right)^2=\left(\vec v_{0,x}\right)^2+2\vec a_x \Delta x$$

$$\vec v_x = \vec v_{0,x}+ \vec a_x \Delta t$$

$$x=x_{0}+\frac 12\left(\vec v_x+\vec v_{0,x}\right)\Delta t$$

As you can see here, we have designated each vector as its component in the x direction. This allows us to define the kinematic equations in one direction, the x direction. The same can be applied to the y components, just replace each subscript x with y. This might be a little confusing, so although this is review, let's go through what each equation means in plain English.

The first equation allows us to find the position or displacement of an object after a time $$\Delta t$$ by taking its initial position $$x_0$$, adding its initial velocity times $$\Delta t$$ and then adding any acceleration / deceleration that occurs (watch the +/- signs) and multiplying it by $$\frac{1}{2} \Delta t ^ 2$$.

The second equation allows us to find the velocity of an object independent of time after that object has traveled a distance $$\Delta x$$ as long as we know the initial velocity, acceleration, and the distance the object has traveled. Note that the left hand side of the second equation is squared so in order to find the final velocity we must take the square root of the right hand side.

The third equation allows us to find the velocity after a certain time $$\Delta t$$ has passed as long as we know the initial velocity and acceleration over that time.

The fourth equation allows us to find the position of an object given an initial and final velocity. Notice that what we are doing in the fourth equation is taking the average velocity (which we learned how to calculate graphically earlier) and multiplying it by the time. Makes sense because we are assuming acceleration is constant in kinematics problems.

Note that it's not always the left hand side variable we are solving for, and we may be asked to solve for $$\Delta t$$ or any other variable on the right hand side equation. The secret to using these equations properly is recognizing which variables we are given and which variable we are solving for, then using the appropriate equation that utilizes all of the given variables except the one we are solving for.

Want more practice? Buy an eBook of 50 multiple choice questions on Amazon with similar topic ratios as the exam to prepare for test day by clicking on the book cover above.

Practice FRQ for 2D Kinematics

A cannonball is launched off of a 50m cliff at 40º with an unknown initial velocity. If the cannonball winds up hitting the target 450m away, what was the initial velocity of the cannonball?

Key concepts to understand: There are a few things we must understand conceptually in 2D kinematics problems that help us solve this. We are asked to find the initial velocity, but at first it doesn't seem like we're given enough information. All we are given is the height of the cannon from the ground (50m), the angle, and how far away the ball went. Below are the key things you must understand when solving a problem like this:

1. When the projectile is first launched, it has an initial velocity with some x and y component. The magnitude of this velocity is $$\sqrt{v_x ^2 + v_y ^ 2}$$.

2. At the apex or peak of the projectile's path, it only has the original x-component of its velocity, which stays constant throughout its flight (we are neglecting air resistance). Its y-component is 0 because the acceleration due to gravity has taken away its initial vertical momentum.

3. Once the projectile is again level with its initial launch position, the y-component of its velocity has the same magnitude as when it was first launched, except it's in the opposite direction because the projectile is falling downwards, so the sign is flipped.

4. Eventually the projectile will land on the ground. If the height at which it lands on the ground is lower than the height at which it was launched, then the magnitude of the final velocity at which it landed should be greater than the magnitude of the initial velocity because the x-velocity is constant and the y-velocity has been increased due to the acceleration from gravity.

It's important to understand the above concepts because they apply to all 2D kinematics problems and often times you will be asked about the concepts behind the kinematics problems rather than finding the mathematical solution.

Solution: Step 1 - Determine what we know

We are given the x distance, so we know $$x=450\text{m}$$ and $$x_0=0\text{m}$$. We also know that $$y=50\text{m}$$ and $$y_0=0\text{m}$$. The question doesn’t give us anything else to work with, though. We can assume that the horizontal acceleration is 0, and that the vertical acceleration is only due to gravity. That gives us $$\vec a_x$$ and $$\vec a_y$$.

What we want to know is the initial velocity. To use this in our kinematic equations, we have to split it up into its components. Once we do that, we get:

$$\vec v_{0,x}=\vec v_0\cos \theta$$

$$\vec v_{0,y}=\vec v_0\sin \theta$$

We know that the angle is 40º so $$\theta$$ is a known variable. So now we are working with the variables $$x\text{, }x_0\text{, }\vec v_0\text{, and }\vec a$$. We are still a variable short from picking a kinematic equation. But that’s it. There are no more assumptions that we can make. So what do we do now? We can use the time variable to help us out because the time will be the same for both situations, and therefore we can set up a system of equations.

Step 2 - Solve for the unknowns

With the chosen variables, we pick our equation. Here, we choose the 1st equation for both the horizontal and vertical direction since we know both the x and y variables. Setting up the system gives us the following two equations.

$$x=x_{0}+\vec v_{0,x}\Delta t+\frac 12 \vec a_x\left( \Delta t\right)^2$$

$$y=y_{0}+\vec v_{0,y}\Delta t+\frac 12 \vec a_y\left( \Delta t\right)^2$$

By inputting the knowns, as well as our chosen definitions for initial velocity, we get:

$$450\text{m}=0\text{m}+\left(\vec v_0\cos40^\circ\right)\Delta t$$

$$0\text{m}=50\text{m}+\left(\vec v_0\sin40^\circ\right)\Delta t - \frac 12 9.81\left(\Delta t\right)^2$$

We can use the first equation in the system to substitute for time. Doing so and performing a bit of algebra, we get:

$$0=50+450\tan40^\circ-\frac{99300}{\left(\vec v_0\right)^2\cos^240^{\begin{smallmatrix}\circ\end{smallmatrix}}}$$

Solving this for the initial velocity, we get $$\vec v_0=68.7$$ m / s.

Conclusion

Kinematics problems are one of the most common types of problems that can be solved purely algebraically in Physics 1. The hardest part is setting up the equations and recognizing which variables are knowns and unknowns. When they are in 1 dimension the kinematics equations are relatively easy to solve. In two dimensions, separate the kinematics equations into their horizontal and vertical components and solve for the in one dimension. Two dimensional kinematics equations sometimes require setting up systems of unknowns when asked to solve them mathematically.

Conceptually, it's important to recognize a few key moments in the path of a projectile, specifically the initial condition, the final condition, and when the projectile approaches its peak. Unless told otherwise (which is very rare), it is safe to make the following assumptions:

• No air resistance, so the horizontal acceleration is 0 and the horizontal velocity stays constant

• Gravity always acts downwards at a constant -9.8 m/s2

• The vertical velocity is 0 when the object is at its peak height