EK Physics - Section 1 - 30 Minute Quiz

[SKaminski]

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30 Minute Quiz:
Q19)
Background information:
Students conduct an experiment to study projectile motion. A projectile is launched from a spring-loaded gun. The gun launches the projectile from a hill and with the same speed each time. The gun is aimed so that the initial velocity o fhte projectile has an angle (theta) from a hill and with the same speed each time.

Question:
Which of the following statements is true concerning the flights of the projectile in the experiment?
A) At its maximum height, the speed of the projectile was zero for every flight.
B) All projectiles reached maximum acceleration just before hitting the ground.
C) The speed of the projectile changed at a constant rate throughout the experiment.
D) The distance traveled through the air by the projectile was smallest when launched at 0' from horizontal.

My answer: C
Correct Answer: D.

Obviously D is true. It mentions DISTANCE not DISPLACEMENT. So i get that D is true.
Why isn't C true as well? Change in speed is nearly identical to change in velocity, except that speed doesn't have a direction. But the Upward/downward component of speed is changing at a constant rate... at 9.8 m/s^2. So the formula for its rate is going to be equal to

Speed = Vo * sin(theta) + (Vo * cos(theta) + (9.8m/s^2 *(change)t)

So, why is C wrong? Whats wrong with my thinking?

 

[MD Odyssey]

My thought process for this problem would be the following.

A) Clearly incorrect, since the horizontal speed is a constant, but non-zero for all angles except for 90 (i.e., launched straight up).

B) The maximum acceleration would depend upon when the force had the greatest magnitude. If there is a constant force on the particle, then I'm inclined to exclude this answer as well. If gravity is the only force acting upon the particle, then this answer is incorrect. Once you skip to the next two answers, it becomes obvious that the question is not assuming gravity is the only force at work. Since we know that the drag force is always in the opposite direction of motion and is proportional to the square of the velocity, we know that the point where the net force has the greatest magnitude is where the speed is a maximum. This occurs at the moment the particle is launched, so this answer must also be false.

C) Since the acceleration is a function of time, the speed of the projectile does not change uniformly with time.

D) This must be the correct answer.

In reality, this is an awfully worded questions since you have to sleuth out the answer. If I were only given answer choices A-C I would have picked C as well. Answer choices C and D are clearly both correct, but C is more accurate if you include air resistance. Don't worry. The MCAT will make it clear when you should neglect air resistance.

Of course...the MCAT will have questions that are far more obtuse than this one was.

 

[Yorick]

C is not true because the rate that speed changes, dv/dt, is equal to acceleration, the magnitude of which is constant for the whole experiment, g. However, when the object is initially launched and is traveling up, it is traveling in the opposite directoin of gravity (directed down) so a= -g = dv/dt. But after the object passes the maximum height and is traveling down in the same direction as gravity, a= g = dv/dt, so the speed did NOT change at a constant rate.

 

[MD Odyssey]

C is not true because the rate that speed changes, dv/dt, is equal to acceleration, the magnitude of which is constant for the whole experiment, g.

The magnitude of the acceleration is not equal to g because there are two forces at work, namely gravity and drag. On the way up, the gravitational force and the drag force are in the same direction. On the way down, the gravitational force is still down, but the drag force is now in the opposite direction. Since drag is typically proportional to the square of the velocity, the drag force is not a constant and thus the rate of change of the velocity is not a constant either.

However, when the object is initially launched and is traveling up, it is traveling in the opposite directoin of gravity (directed down) so a= -g = dv/dt. But after the object passes the maximum height and is traveling down in the same direction as gravity, a= g = dv/dt, so the speed did NOT change at a constant rate.

This is a common mistake. The sign of the acceleration has absolutely nothing to do with the direction you are traveling. The acceleration is determined by the net force on the particle - that's Newton's 2nd law. Your statement that a = g on the way down is false. You changed the signs of the force - your statement implies that gravity pulls on the object on the way up but pushes on it (i.e., is positive) on the way down, which is silly because gravity doesn't work that way.

If one ignores air resistance, as you have done, then the force on the way up is the same as the force on the way down - equal magnitude and the same sign. The vertical component of the velocity is the greatest at the very beginning. As it rises, it decreases because there is a net force on the particle. It continues to decrease, eventually passes zero, becomes negative, and eventually hits the ground. At every point along the way, it's acceleration is -g. If one neglects air resistance, then answer choice C would be correct.

In fact, if you ignore air resistance, then the slope of the velocity vs. time curve will be linear, with a slope of -g, an x-intercept of the time it took to reach maximum height, and a y-intercept of the initial velocity.

That's what makes this problem so poorly worded. Answer choice C requires an assumption not given in the problem that answer choice D does not.

 

[sazerac]

This is an extremely well-written question to separate the clueless (A&B), from the folks who sort of know what is going on (C), from the masters of the topic (D).

C is patently incorrect because the question is really testing if you know the difference between the velocity vector v and the magnitude of the velocity vector s=|v|. And you guys all seem to be missing this point.

SKaminski, speed is NOT Vo * sin(theta) + (Vo * cos(theta) + (9.8m/s^2 *(change)t). You gave the equation for the velocity vector. This question asked about speed, which is the magnitude of the velocity vector.

MDOdyessy, acceleration is not a function of time for the MCAT. The MCAT is written for constant acceleration scenarios only (anything else would require calculus), and coincidentally gravity is a constant. Also, you are expected to ignore air resistance in problems like these.

The velocity vector of the projectile changes at a constant rate throughout its flight. Every second it points 10 m/s^2 further down than it did the last second. However answer choice C didn't ask about velocity, it asked about speed.

The SPEED is some positive value that shrinks for a while (if the projectile is shot straight up it would shrink all the way to zero) but then the speed starts increasing again as the projectile falls back to earth. Since the speed is a positive number that shrinks and then it grows again, the change in speed is not a constant. The change in speed must be a negative number for a while and then it must be a positive number.

 

[MD Odyssey]

MDOdyessy, acceleration is not a function of time for the MCAT. The MCAT is written for constant acceleration scenarios only (anything else would require calculus), and coincidentally gravity is a constant. Also, you are expected to ignore air resistance in problems like these.

Quantitatively speaking, I'd agree with you - the MCAT won't be asking you to make calculations by incorporating air resistance, but I'm pretty sure a couple of the practice exams tested it qualitatively. The Berkeley Review absolutely included it in several places.

The velocity vector of the projectile changes at a constant rate throughout its flight. Every second it points 10 m/s^2 further down than it did the last second. However answer choice C didn't ask about velocity, it asked about speed.

Now that I read the question again, you're absolutely correct. That's the key - to pick up on the fact that it mentions speed and not velocity. I'm not sure why I presumed it was talking about velocity instead of speed - probably didn't read the question all that well. If you plot the magnitude of the y-component of the velocity as a function of time, it will have a negative slope initially, intersect the x-axis at whatever time it reaches the apex of its curve, and then have a positive slope as the particle speeds up again. So, the slope is not a constant and that's the reason that answer C was incorrect.

Good example of understanding the physics but misreading the question - I'm glad I don't have to take the exam again. I remember reading this question the first time and concluded that D had to be the right answer because it was true regardless whether or not one accounted for drag. I think I let that stick in my mind and ended up having to invent a complication that negated answer choice C. This is also a good example of getting the right answer for the wrong reasons - thanks for pointing out my error.

 

[SKaminski]

Hey guys,

Sorry to revive this older thread. I retook the 30 minute quiz again today (to review), and came up with a formula to illustrate that the speed changes at a consistent rate. Looking at my original comment, I believe i posted the formula there:

Speed = Vo * sin(theta) + (Vo * cos(theta) + (-9.8m/s^2 *(change)t)

So, it would be changing by a constant amount in the vertical component, while being at equilibrium in the horizontal component. That is still changing at a constant speed, however.

SKaminski, speed is NOT Vo * sin(theta) + (Vo * cos(theta) + (9.8m/s^2 *(change)t). You gave the equation for the velocity vector. This question asked about speed, which is the magnitude of the velocity vector.

If we were to isolate the vertical component, specifically, then wouldn't the magnitude of the vector equal the speed? In which case, it is changing at a constant rate during the entire flight?

Explaining this idea more, if we declare the upward direction to be "positive" and the downward direction to be "negative", then the vertical component IS changing at a constant rate. (And i believe that scalars CAN have positive and negative components. A unit of charge is considered a scalar, yet you can have positive and negative charges). So, if you have a constant change in the Y direction, and a non-changing speed in the X direction, then it would actually be changing at a constant rate!

 

[slz1900]

With respect to d, isn't the distance traveled smallest when it is launched at 90 degrees from the horizontal? Since it is being launched off of a hill, an angle of 0 degrees would still give it some displacement in the X direction, right?

 

[milski]

Hey guys,

Sorry to revive this older thread. I retook the 30 minute quiz again today (to review), and came up with a formula to illustrate that the speed changes at a consistent rate. Looking at my original comment, I believe i posted the formula there:

Speed = Vo * sin(theta) + (Vo * cos(theta) + (-9.8m/s^2 *(change)t)

So, it would be changing by a constant amount in the vertical component, while being at equilibrium in the horizontal component. That is still changing at a constant speed, however.



If we were to isolate the vertical component, specifically, then wouldn't the magnitude of the vector equal the speed? In which case, it is changing at a constant rate during the entire flight?

Explaining this idea more, if we declare the upward direction to be "positive" and the downward direction to be "negative", then the vertical component IS changing at a constant rate. (And i believe that scalars CAN have positive and negative components. A unit of charge is considered a scalar, yet you can have positive and negative charges). So, if you have a constant change in the Y direction, and a non-changing speed in the X direction, then it would actually be changing at a constant rate!

The magnitude of the vector is |v|=sqrt(vx^2+vy^2). You cannot really split the magnitude in vertical or horizontal component. Even if vx is constant and vy is changing at constant rate, sqrt(vx^2+vy^2) is not changing at a constant rate.

 

[SKation]

30 Minute Quiz:
Q19)
Background information:
Students conduct an experiment to study projectile motion. A projectile is launched from a spring-loaded gun. The gun launches the projectile from a hill and with the same speed each time. The gun is aimed so that the initial velocity o fhte projectile has an angle (theta) from a hill and with the same speed each time.

Question:
Which of the following statements is true concerning the flights of the projectile in the experiment?
A) At its maximum height, the speed of the projectile was zero for every flight.
B) All projectiles reached maximum acceleration just before hitting the ground.
C) The speed of the projectile changed at a constant rate throughout the experiment.
D) The distance traveled through the air by the projectile was smallest when launched at 0' from horizontal.

My answer: C
Correct Answer: D.

Obviously D is true. It mentions DISTANCE not DISPLACEMENT. So i get that D is true.
Why isn't C true as well? Change in speed is nearly identical to change in velocity, except that speed doesn't have a direction. But the Upward/downward component of speed is changing at a constant rate... at 9.8 m/s^2. So the formula for its rate is going to be equal to

Speed = Vo * sin(theta) + (Vo * cos(theta) + (9.8m/s^2 *(change)t)

So, why is C wrong? Whats wrong with my thinking?

Hey,

I am stuck in the same problem I dont understand why exactly D is right, How would we prove that distance is smallest at 0 so I can see it visually?

Thanks!