TBR Physics- Section 1 - Q64 and Q69

[SKaminski]

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Q64)
Preface: The question has two cannons aimed at each other, and asks if one of them fires, if it will ever hit the other one. They are 55.5m apart and the velocity is not given. Cannon A is 32 meters above cannon B.

Question: Using the setup shown in Figure 1, can a cannonball fired from Cannon A ever hit Cannon B?

A) Yes, if the speed of the ball upon exiting Cannon A is sufficiently large.
B) Yes, if the exit speed equals the speed the ball gains by falling 32m.
C) Yes, regardless of the exit speed.
D) No.

My answer: A.
Correct Answer: D.

Correction answers justification: Since the two cannons are aimed at each other, the cannonball would travel along a straight path and hit Cannon B, if there was no gravity. But there is gravity. Gravity will change the path of the ball, specifically causing the ball to drop below this path. No matter how fast the cannonball travels, gravity will always cause it to drop below this line and therefore miss Cannon B.

My objections: What about the cannon ball rolling? Even if it rolls and just barely TOUCHES the other cannon, isn't that considered contact, and thus a hit?

And even if its not, the picture (which i can't post) shows that its aimed at the UPPER portion of the cannon... that has a body below it. While we are not given height, if a cannonball is moving fast enough, it should hit the body of the cannon at a minimum right? What is the MCAT looking for in these type of questions?

Q69)

Background information: Above information, and that the length of the cannon's barrels are 2.5 meters long. The mass of the ball is 1.00kg.

Question: What force is required to give the cannonballs an exit speed of 25 m/s.

A. 125 Newtons
B. 250 Newtons
C. 2500 Newtons
D. It cannot be determined.

My math:
F=ma
a= change(V)/T
V=displacement/T

Here's what we know:
Displacement = 2.5m
Time = x
Vo = 0 m/s
Vf = 25 m/s
Other equations Vavg = 1/2(Vo + Vf)
Vavg = 12.5
m = 1.00kg

So, 2.5m / 12.5 m/s = .20 seconds (time it takes ball to travel to end of barrel)
V = 2.5 (displacement) / .20 (time) = 12.5.
A = 12.5 (velocity) / .20 (time) = 62.5.
F= ma, so F = (1)(62.5), so F = 62.5 newtons. This is obviously wrong.

The correct answer is A, which was exactly twice my answer. They used the formula:
Vf^2= Vo^2 + 2a(change)x

This gave them A = 125 m/s^2. WIth a mass of 1 kg, that's 125 newtons.

Why are the answers different?

 

[Yorick]

Acceleration is NOT average velocity/time. Acceleration is change in velocity over time or (vf-vi)/t. As you have, t=.2 and vf=25 and v0=0 so A=25/.2 =125, then F=ma=125 N.

 

[SKaminski]

Acceleration is NOT average velocity/time. Acceleration is change in velocity over time or (vf-vi)/t. As you have, t=.2 and vf=25 and v0=0 so A=25/.2 =125, then F=ma=125 N.

That makes a lot of sense! I completely see what I did wrong there. Thank you.

 

[MD Odyssey]

Let me speak a little bit to the first question about the cannonballs. You are completely right with all of your objections. The problem with the question is that in physics, particularly the first-year physics that the MCAT tests, real objects are nearly always replaced with point particles at their center of mass. For problems that ask about gravitation, distance traveled, work done, and things like that, this simplification works just fine. For some problems, like determining whether or not a cannonball from one cannon can hit the other, it's a terrible simplification. I remember reading this problem this summer and thought it was ridiculous. Don't fret too much - the MCAT does a much better job of being clear about the question than the Berkeley Review does. I say this with all respect for the authors of the Berkeley Review, which in my view was largely responsible for my performance on the exam.

Have no fear. If you set up the cannons as the picture indicates, point one directly at the other over such a short distance, and fire, the cannon is toast.

 

[ReggieD6868]

I saw the answer to the first question before I could read it, so I am only going to comment on the second question that you posted.

I think that one can use the Work-Energy Theorem to solve the second question.

Wnet = F*d = (1/2)mv^2,

so F = [(1/2)mv^2]/[d]

Plug in:

F = [(1/2)(1.0)(625)]/[2.5]

F=125N

I left out units except for the final answer so that it wouldn't get messy, but this method yielded the correct answer.

Did the book provide a solution similar to the one I mentioned above?

I hope I was able to help.