pcpkorea

Constant Force Friction?

Pre-College Physics Korea — Sun, 24 Aug 2025 22:41:11 +0900

[SUMMARY]

This blog page intuitively applies uniform acceleration, non-dimensionalization of speeds/time, velocity graphs, and momentum conservation to solve a Korean high school mechanics exam.

Link to the Original Exam Problem (Korean)

Main Drawing of the Problem

[Problem]

As shown in figure (a), object A moves on a horizontal plane with speed 4v. After passing through a friction zone, it continues with uniform motion. Object B moves toward A with speed v.

As shown in figure (b), after A and B collide, object A moves in the opposite direction to before the collision with speed v and then stops in the friction zone. Object B also moves in the opposite direction to before the collision with speed v.

In the friction zone, object A experiences a constant force opposite to its direction of motion. The time it takes for A to move through the friction zone in case (a) is twice that in case (b).

If the masses of A and B are m_A and m_B respectively, what is the ratio m_A/m_B?

[Think Along]

In this problem, there is a unique expression:
“An object receives a constant force of the same magnitude in the direction opposite to its motion while in the friction zone.”

This wording is actually a trick intended to mislead test takers. Instead of simply saying “a slope,” the problem deliberately uses the term “friction zone” to cause confusion. While an object moves uphill, it receives a constant force. Remember F=ma. If we know the initial velocity and the acceleration at every moment, we can predict the object’s motion at every moment. Whether it is a friction zone or a slope, F=ma makes no distinction.

Now, let’s look at the following figure.

The first friction zone can be expressed as follows:

Initial velocity = 4v
Time taken = 2T
Acceleration = -a

Meanwhile, the second friction zone is as follows:

Initial velocity = v
Time taken = T
Acceleration = -a

[Non-dimensionalization]
Now we need to decide on non-dimensionalization. In this problem, nothing is given in absolute quantities—whether mass, initial velocity, or time. Therefore, it is better to arbitrarily set some convenient values and proceed. The following shows the result after applying non-dimensionalization:

First friction zone: Initial velocity = 4 m/s, Time taken = 2 s, Acceleration = -1 m/s²
Second friction zone: Initial velocity = 1 m/s, Time taken = 1 s, Acceleration = -1 m/s²

Next, let’s look at the 1, 3, 5, 7 rule of uniformly accelerated motion.

[Regularity of uniformly accelerated motion]
Uniformly accelerated motion might seem very complicated moment by moment, but in fact, it has an amazing consistency. We may call this the 1, 3, 5, 7, 9 rule. Remember it well. In any uniformly accelerated motion, suppose an object starts from rest, and the distance traveled in the first time interval t is L. Then, in the next time interval t, the distance traveled is 3L, in the following t it is 5L, and in the next it is 7L. Of course, after that it continues with 9, 11, 13… and so on.

In the figure, the horizontal axis represents time and the vertical axis represents velocity. Since the motion is uniformly accelerated, the velocity increases linearly with time. In each diagram, the red lines are equally spaced, meaning the uniformly accelerated motion has been divided into equal time intervals of length t. The area of the blue shapes cut by the red lines represents the distance traveled.

In the left diagram, t is relatively large. The distance traveled in the first interval (area) is one triangle, and in the second interval the distance traveled is three triangles… showing that the 1, 3, 5, 7 rule is being followed well.

In the middle diagram, t has been made smaller. However, the rule of one triangle, three triangles, five triangles is still satisfied.

In the right diagram, the triangles are distorted. You may have noticed that the slope of the triangles in the figure represents the magnitude of the acceleration. Even though the acceleration has changed in this way, the distance traveled in the first interval is still one triangle, in the second interval three triangles, then five, then seven… showing that the 1, 3, 5, 7 rule is being maintained very well.

Now, let’s connect the 1, 3, 5, 7 rule of uniformly accelerated motion with this problem.

The figure is a time–velocity graph. Since the units of both the horizontal and vertical axes are 1, the acceleration is -1.

For object A on the first uphill section, its initial velocity is 4 and the time is 2 seconds. From the area of the “2 sec.” section in the graph, we can immediately see that the distance traveled is 6 m.

For object A on the second uphill section, by reading the area of the “1 sec.” section in the graph, we can immediately see that the distance traveled is 0.5 m.

However, the distance traveled is not actually useful for finding the answer.

What matters is that at the end of the “2 sec.” section, A’s velocity is 2 m/s. That is, right before colliding with B after climbing the hill, A’s velocity was 2v, and immediately after the collision it became -v. So now we know both A’s velocity just before and just after the collision.

Let’s apply the law of conservation of momentum: (Velocity to the Left = Negative Velocity)

Total momentum before collision = 2m_A - m_B
Total momentum after collision = -m_A + m_B

Calculating,

2m_A - m_B = -m_A + m_B
3m_A = 2m_B
m_A / m_B = 2/3

Therefore, the answer is choice 2.

선택지

[Examiner's Note]

[Source: May 2025 Korean National Physics I Exam, Grade 12, Problem 9]

2026-05-09-constant-force-friction-evaluated-by-chatgpt5.pdf

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Light Entering a Quarter-Pie Lens

Pre-College Physics Korea — Sun, 24 Aug 2025 08:02:43 +0900

[SUMMARY]

This blog page intuitively analyzes a Korean high school optics exam using reversibility of light, refraction angles, and total internal reflection.

Link to the Original Exam Problem (Korea)

Lens Shape

Experiment

[Question]
(a) As shown in the figure, quarter-circular media A and B, with the same radius, are placed on a horizontal plane, and monochromatic light is incident on either A or B.
(b) By varying the height of the point at which the monochromatic light enters A or B, it was examined whether the light underwent refraction or total internal reflection.
Based on the experimental results below, choose all the correct statements.

<Statements>

(a) In medium A, the wavelength of monochromatic light is shorter than in air.

(b) Medium A has a greater refractive index than medium B.

Experiment Result

[Think Along]
Before solving the problem, I would like to introduce the principle of reversibility of light, which can be useful in geometrical optics problems.

[Reversibility of Light]
There is a principle known as the reversibility of light. The idea is simple: if light passes through a certain path in a given optical system, then light can also travel along the same path in the reverse direction. In other words, as shown in the figure below, if monochromatic light travels from left to right along the blue path, then light entering from the opposite side along the red path will follow exactly the same path in the reverse direction.

Illustration of Reversability of Light

Now let us examine the truth or falsity of each condition one by one.

(a) In medium A, the wavelength of monochromatic light is shorter than in air.

First, (a) is true.
Looking at the red path in the figure, we see that as light enters lens A from air, it bends toward medium A. This is the typical behavior of a wave entering from a rarer medium into a denser medium. In addition, when entering a denser medium, the wavelength becomes shorter. The following figure illustrates why the direction of propagation bends when light enters a denser medium from a rarer one.

A car turning as it moves from a slippery road onto grass

This situation of light entering from a light medium (also called a rarer medium, where light travels faster) into a dense medium (a denser medium, where light travels slower) can be compared to a car entering from a smooth cement road into a grassy field. Suppose the car is moving roughly in the direction of (x, y = 1, 1) toward (x, y = 0, 0). As it crosses the boundary, the car’s right wheel touches the grass first, while the left wheel is still on the smooth road. During that brief moment of crossing the boundary, the friction on the two wheels is different, and as a result the car veers slightly to the right. The denser the grass, the more the car will turn.

Next, let us examine condition (ㄴ).

(b) Medium A has a greater refractive index than medium B.

To solve the problem, we need to determine which of the two lenses, A or B, is denser (i.e., has a greater refractive index).

When light enters a lens from air, the degree to which its direction bends is proportional to the refractive index. Therefore, lens A has a greater refractive index than lens B.

Thus, condition (b) is true.

Finally, let us look at condition (c).

(c) “Total internal reflection” corresponds to (Unknown).

Experiment Result

The condition “6 cm” means that the angle at which light enters from left to right is the same for both lens A and lens B. Therefore, what we must determine is this: when light enters the boundary between the lens and the air at the same angle, and total internal reflection occurred in lens B, can we also expect total internal reflection in lens A? This is the question to be answered.

Now look at the following figure.

The blue path shows light passing through the lens and exiting into the air. But which case looks more difficult for the light to break through the boundary—A or B? Lens A looks more difficult. (Here, if we express “difficult” in scientific terms: “Since A has a greater refractive index, light finds it harder to escape the boundary, making total internal reflection more likely.” However, since our goal is to solve physics problems quickly and intuitively, it is often more useful to develop the ability to judge such situations in terms of “easy” or “difficult” rather than strictly formal descriptions.)

Therefore, if at the same incident angle light in B could not penetrate the boundary and underwent total internal reflection, then in A it would certainly undergo total internal reflection as well. Hence, condition (c) is true.

Since (a), (b), and (c) are all true, the correct answer is option 5.

[Epilogue] Up to now, I’ve tackled three optics problems, and coincidentally, all of them worked out neatly with the car-on-grass analogy and the principle of the reversibility of light. I wonder what other kinds of problems might come along next.

[Source: May 2025 Korean National Physics I Exam, Grade 12, Problem 15]

2026-05-15-a-quarter-pie-lens-evaluated-by-chatgpt5.pdf

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2026-05-15-a-quarter-pie-lens-evaluated-by-gemini.pdf

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Exam Q/A Docs for 2024 (PDF)

Pre-College Physics Korea — Sun, 24 Aug 2025 00:53:43 +0900

November 2024 Final Enterance Exam - Screenshot

2024-11-entrance-exam-Answers.pdf

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2024-11-entrance-exam-Problems.pdf

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Exam Q/A Docs for 2025 (PDF)

Pre-College Physics Korea — Sun, 24 Aug 2025 00:50:49 +0900

May 2025 Screenshot

2025-05-Physics-Answers.pdf

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2025-05-Physics-Problems.pdf

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June 2025 Screenshot

2025-06-Physics-Answers.pdf

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2025-06-Physics-Problems.pdf

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2025-07-Physics-Answers.pdf

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2025-07-Physics-Problems.pdf

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A killer problem - easier tried than left unsolved

Pre-College Physics Korea — Sat, 23 Aug 2025 23:40:00 +0900

[SUMMARY]

This blog post intuitively uses action–reaction, non-dimensionalization of masses and speeds, energy partitioning, and height–energy relations to crack a Korean high-school mechanics exam.

Link to the Original Korean Problem

Main Diagram of the Problem

<Problem>

As shown in the figure, on a plane of height h1, objects A and B of masses 3m and 2m, respectively, compress spring P by a distance d from its natural length and are then released from rest.

Object A passes through frictional Interval 1 and on the horizontal plane compresses spring Q by at most d from its natural length.
Object B, immediately after separating from spring P, has speed v0. After passing through frictional Interval 2, it moves on a plane of height h2 with the same speed v0.

The mechanical energy lost by A and B when each passes once through Interval 1 and Interval 2, respectively, is equal to the kinetic energy of A immediately after separating from spring P. The spring constants of P and Q are the same.

Which of the following statements are correct?

(Assume the objects move in the same vertical plane, and neglect the mass and size of the springs, air resistance, and all friction except in the specified intervals.)

<Statements>

<Choices>
① a ② b ③ a,c ④ b,c ⑤ a,b,c

[Think Along]
This problem looks like a killer question, but it is not actually one. It can be solved step by step using the initial conditions.

The tricky part is correctly interpreting what happens at time t₀. We have two objects of different masses at rest, with a spring placed between them pushing them apart. How should we interpret this situation?

[Spirit] Although a spring is somewhat prickly and sharp, in this problem let us imagine it as an invisible spirit that cannot be seen or touched. That makes the thinking easier. This spirit pushes A and B apart from the middle. The spirit has no mass—only force. To an outside observer, it looks like A is pushing B and B is pushing A.

It’s like an astronaut in empty outer space pushing against a huge metal object. The metal object is obviously far heavier than the astronaut, so the one that actually moves back is the astronaut. If the astronaut had the exact same mass as the metal object, then both would recoil with the same speed.

Up to this point, it’s all in the realm of intuition. The real question is: how can we quantify this situation? That is the dividing line between being able to solve this problem or not.

An Astraunaut pushing against a spaceship [chatGPT]

[Action–Reaction Between Two Masses]
This problem can be solved either by applying conservation of momentum or by applying action–reaction. Since I always prefer F = ma, I will solve it using action–reaction.

Let’s apply action–reaction. This requires a bit of imagination. As long as the spring is simultaneously in contact with both A and B, A can push B, and B can push A. At every instant during which the spring touches both A and B, the force exerted by A on B and the force exerted by B on A are equal.

Because acceleration is inversely proportional to mass, the ratio of the accelerations caused by this action–reaction force is always 2:3. If the acceleration ratio is 2:3 at every moment, then the velocity ratio will also be 2:3 at every moment. Likewise, the displacement ratio will be 2:3, although displacement is not particularly important in solving this problem. The conclusion we can draw here is:

v_A : v_B = 2 : 3 (the velocities of A and B at the instant they are pushed apart by the spring).

[Non-dimensionalization]
Now it’s time to decide on a non-dimensionalization scheme. In this problem, nothing is given in absolute values—not the masses, nor the initial velocity, nor the height. Therefore, it is better in many ways to assign convenient arbitrary values at the start.

Let’s set the initial velocities of the two objects as follows:
v_A = 2 m/s
v_B = 3 m/s

Since the masses are given in the ratio 3:2, we might as well set them as follows. Instead of 3 kg and 2 kg, I will use 30 kg and 20 kg; I will explain later why I chose the larger values.
m_A = 30 kg
m_B = 20 kg

Now, it is time to calculate the initial kinetic energy:
E_A0 = 0.5 × 30 × 4 = 60 J
E_B0 = 0.5 × 20 × 9 = 90 J
E_0 = E_A0 + E_B0 = 150 J

Object A came down the slope, compressed the spring by distance d, and then stopped. At the moment it stopped, all of the kinetic energy that A had was stored in the spring. The problem states that all the springs involved are identical.

Now, let’s go back to the initial “big bang” moment. At that time, both objects A and B were at rest, and all the energy was stored in the spring. That energy was converted without any loss into the kinetic energy of A and B, totaling 150 J.

Therefore, when A comes down the slope and compresses the spring at the bottom by distance d, the energy stored in that spring must also be 150 J. Since it is the same type of spring and it is compressed by the same amount d, the energy stored must be the same.

E_S = 150J

[Clue 1] Immediately after B separated from A, its speed was v0 (= v_B = 3 m/s), and after passing through the frictional interval it moved at constant speed v0 (= v_B = 3 m/s) on the plane at height h2.
B initially had 90 J of kinetic energy, and after coming down the slope it still had 90 J of kinetic energy. In other words, the potential energy that B lost while descending the slope must have been completely dissipated by the frictional interval.

[Clue 2] The mechanical energy lost when passing through Intervals 1 and 2 is equal to the kinetic energy of A immediately after separating from P (= E_A0 = 60 J).

[Conclusion] Now it is time to gather all the pieces of information we have collected and determine which of the given statements are correct.

First, let’s look at (a).
Immediately after separating from P, the speed of A is 2 m/s, which is equal to 0.666 × (v_B = 3 m/s). This is correct.

Next, let’s look at (b).
We need to determine the spring constant.

We have already calculated the energies at each position, so we should use the relation between spring constant and energy. For a spring constant k, the energy stored in the spring is defined as:

E_S = 0.5 k d²

When A came down the slope, the energy stored in the spring was 150 J. Therefore,

2 E_S = k d²

300 / d² = k (the spring constant)

Now, let’s check whether 10 × m × v₀² equals 300.

10 × 10 × 9 = 900 ≠ 300

Therefore, (b) is not correct.

Earlier I skipped over explaining why I used 30 kg and 20 kg for m_A and m_B instead of 3 kg and 2 kg.

If we had used 3 kg and 2 kg, then in the formula above m would have been 1. In that case, it would have been difficult to tell from the numbers alone whether there was any influence of m or not. Normally, most proportional problems can be solved by non-dimensionalization, but in a problem like this, where you must check an equation, it can be helpful if the variables being normalized are relatively prime. Originally, the velocities contained 2 and 3, but for the masses I multiplied by 10 to add a factor of 5. This also seems to be a useful trick.

Finally, let’s look at (c).
We need to find both h1 and h2.
Ah, I almost forgot—we can set the gravitational acceleration g = 1. What matters here is only the ratio of h1 and h2, not their exact values. So whether we take g = 1 or g = 9.8, the result will be the same.

Let’s follow the motion of A. Initially, A had 60 J of kinetic energy. It was stated that after passing through the frictional interval, 60 J of energy was lost. Then, just before meeting the spring on the ground, all of A’s remaining energy was absorbed by the spring, and the amount stored was 150 J. Therefore, this 150 J must have come entirely from the potential energy at height h1. Thus:

m_A × g × h1 = 30 × h1 = 150 J
h1 = 5 m

Next, let’s follow the motion of B.
Initially, B had 90 J of kinetic energy. Later, even at height h2, it still had 90 J of kinetic energy. This means that during its descent, all of the potential energy gained from falling through the height difference (h1 – h2) was lost to friction. Thus:

m_B × g × (h1 – h2) = 60 J
20 × 1 × (5 – h2) = 60
5 – h2 = 3
h2 = 2 m

Therefore, h2 / h1 = 2/5, not 1/3. So (c) is also false.
Hence, the correct answer is choice 1.

[Source: June 2025 Korean National Physics I Exam, Grade 12, Problem 20]

2026-06-20-killer-problem-easier-tried-than-left-unsolved-evaluated-by-chatgpt5.pdf

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Travelling the Same Distance in Different Times

Pre-College Physics Korea — Sat, 23 Aug 2025 21:08:54 +0900

[SUMMARY]

This blog page intuitively applies uniform acceleration, time-triangle area analysis, non-dimensionalization, energy change comparison, and F=ma to solve a Korean high-school mechanics exam.

Main Drawing of the Problem

Experiment Result

<Problem> As shown in the figure, on a cart of mass M placed on a horizontal plane, one of the two weights A and B is placed on top of the cart, and the other is hung from a string connected to the cart. The table shows the time t it takes for the cart, initially at rest on the starting line, to move with uniform acceleration and pass the reference line, as well as the magnitude of the change in the mechanical energy of A. The masses of A and B are m_A and m_B, respectively. What is the value of M/m_B? (Assume the string has no mass, and ignore all friction and air resistance.)

<Choices>

① 1/9 ② 2/9 ③ 1/3 ④ 4/9 ⑤ 5/9

<Think Along> When a problem involves constant acceleration motion of multiple masses connected by a pulley, recall F = ma. Before solving the problem directly, let’s do a simple thought experiment. The key is to determine sensibly what constitutes F and what constitutes m, considering how the objects and the string are connected. For practice, let’s look at the following figure.

We will apply F = ma to the figure above.
But before that, let’s test our intuition. In the left figure, do you see a 97 kg washing machine placed on a 2 kg cart? A very strong but massless string passes over a pulley and is connected to a 1 kg milk pack. In the right figure, the positions of the milk pack and the washing machine are reversed.
In either case, it is clear that the cart will move from the starting line to the reference line. And doesn’t it also seem obvious that the cart will move much faster in the right-hand case? That’s excellent intuition. But that alone won’t fully solve the problem.

Let’s look at the next figure.

Do you feel how much simpler this new setup is compared to the earlier figure with pulleys, strings, and so on?

Now, focus on the horizontal motion of the object B (the milk pack or the washing machine) in the middle. Compare the earlier figure with pulleys and strings to this new figure, and you can sense how much simpler the new one is. Do you like it better now? Let’s apply F = ma here.

[Left figure]
F = 10 N
m = (97 + 2 + 1) = 100 kg
a = 0.1 m/s²

[Right figure]
F = 970 N
m = 100 kg
a = 9.7 m/s²

As a result, by adding physics to our intuition, we can quantitatively predict the motion of the given system. Another thing we discovered: when we swapped the washing machine and the milk pack in the setup, the acceleration of the uniform motion turned out to be proportional to the mass of the object hanging by the pulley. This fact will later be useful for solving problems.

And one more note: from a student’s perspective, redrawing the figures like this every time is not ideal. To solve exam problems within the time limit, you need to practice looking at the given figure and directly building F = ma.

Practice is over now. Let’s solve the problem.

Main Drawing of the Problem

Looking at the t column in the table, it says that the time ratio between hanging A and hanging B was 1:3, for moving the distance from the starting line to the reference line with uniform acceleration.

Since the same distance was traveled, the areas of the two triangles are equal.
Because the time ratio is 1:3, the ratio of the base lengths of the triangles is 1:3.
But note: as the horizontal axis becomes three times longer while the area must remain the same, the height also becomes 1/3.

For the faster motion, it took time t0, and the acceleration is 1/1 = 1.
For the slower motion, it took 3 t0, and the acceleration is 0.333/3 = 1/9.

Earlier we said that in this system the ratio of accelerations matches the ratio of the masses of the weights hung on the string. So the ratio of m_A to m_B must be 9:1 or 1:9. Looking at the table, it is clear that B is heavier, so m_A : m_B = 1:9.

At this point, let’s begin solving the problem step by step.
The value we need to calculate is M/m_B.

A bit of frustration begins to creep in—another ratio again…
But wait, this actually means there isn’t a single unique solution that satisfies the conditions, but rather several possible ones, and yet the ratio M/m_B will always be the same.

Oh? Is this actually convenient?
Yes. For this problem, we can simply set m_A = 1 kg and m_B = 9 kg and proceed.

Main Drawing of the Problem

Another thing: the problem does not mention the distance between the starting line and the reference line. Turning this around, it means that even if the distance between the starting line and the reference line changes, as long as the given conditions are satisfied, the mass ratio of B to the cart will remain constant. Since I like the number 1, I will set the distance between the starting line and the reference line to 1 m and proceed with the solution. Finally, let’s also take the gravitational acceleration to be 1 m/s² for convenience.

In classical mechanics problems—especially like this one, where the goal is to find a ratio of masses (a dimensionless quantity)—nondimensionalizing the main variables can simplify the solution process. Doing less nondimensionalization is harmless, though it can make the algebra heavier with more variables. Doing too much nondimensionalization, however, is risky—it can cause the answer to disappear. It is important to develop the skill of carefully judging which quantities can be nondimensionalized and which should not.

Now let’s consider the change in the mechanical energy of A in [Experiment-1: A on the cart, B bound to the string on the right].
A starts from rest, then undergoes uniformly accelerated motion, moving 1 m. Since it moves horizontally, there is no change in potential energy—only an increase in kinetic energy. Therefore, A’s final kinetic energy equals the change in its mechanical energy. Let’s calculate A’s final kinetic energy.

Final kinetic energy = Force on A × Distance moved
Force on A = Acceleration of A × 1 kg
Acceleration of A = Acceleration of the whole system (A, B, and the cart)

F = m_B × g = 9 = ma = (1 + 9 + M) × Acceleration of A

Final kinetic energy = 9 / (M + 10)

Change in mechanical energy of A (Experiment 1) = 9 / (M + 10)

[Experiment-2] is a little more complicated. Here, A acts as the hanging weight. The decrease in A’s potential energy is what accelerates A, B, and the cart. The decrease in A’s potential energy is distributed into the kinetic energies of A, B, and the cart according to their mass ratios. Therefore, the change in A’s mechanical energy is as follows:

Change in mechanical energy of A = Decrease in A’s potential energy – A’s final kinetic energy

Change in mechanical energy of A (Experiment 2) = 1 × 1 × 1 – 1/(M+10) = 1 – 1/(M+10) = (M+9)/(M+10)

(Kinetic energy is proportional to the square of velocity for the same mass. Earlier we said that the final velocity in Experiment 1 was 3 times faster than in Experiment 2.)

Now it is time to apply the last condition of the problem: the ratio of A’s change in mechanical energy between the two experiments is 9:11.

9:11 = 9/(M+10) : (M+9)/(M+10)
9:11 = 9 : (M+9)
99 = 9M + 81
9M = 18
M = 2
M/m_B = 2/9

Therefore, the answer is choice ②.

[Source: June 2025 Korean National Physics I Exam, Grade 12, Problem 16]

2026-06-16-same-distance-in-different-time-evaluated-by-chatgpt5.pdf

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Understanding Wave Motion at 0.75 Hz

Pre-College Physics Korea — Sat, 23 Aug 2025 20:13:01 +0900

[SUMMARY]

This blog page intuitively uses imagined particle motion, wave animation, periodic shifts, and phase tracking to solve a Korean high-school wave exam, with a Python program provided to animate wave motion.

Link to Original Exam Problem(Korean)

<Problem>

The figure shows the displacement of a wave with frequency 0.75 Hz propagating in the +x direction, as a function of position x. The solid line and the dotted line represent the wave at time t = 0 and t = t₀, respectively. At t = 0, the displacement of the wave first becomes 2 m at time t = t₀ after t = 0. Which of the <Statements> are correct?

<Statements>

a. The propagation speed of the wave is 3 m/s.

b. t₀ = 1 second.
c. At t = 2 seconds, the displacement of the wave at x = 3 m is 2 m.

<Choices>

① a ② c ③ a,b ④ b,c ⑤ a,b,c

[Think Along]
This problem can be solved more easily than expected if you can imagine the motion of the wave. However, imagining the motion of a wave is harder than it seems. It takes practice, and for practice, it can help to think of a rubber duck in the bathtub.

A rubber duck

A rubber duck is floating in the bathtub, and waves are passing through. This is a magical rubber duck—it moves up and down with the waves but never slips sideways. Using the up-and-down motion of the duck, we can define the period of the wave. In the problem, the frequency is given as 0.75 Hz, so the duck will move up and down 0.75 times per second. That means 1/4 of a cycle in 0.333 seconds, 2/4 in 0.666 seconds, 3/4 in 1 second, and 1 full cycle in 1.333 seconds. In other words, it takes 1.333 seconds for the duck’s vibration to complete one full cycle.

Now, let’s try to imagine the motion of the wave.

Animated illustration of the wave motion

Main Drawing of the Problem

Looking at the black-and-white figure above, it may seem as if the wave is moving to the left, but if you carefully observe the motion of the green graph and the changes at each moment t, it is clear that the wave is moving to the right. Now that you are familiar with the motion of the wave, let’s solve the problem.

Condition (a)

a. The propagation speed of the wave is 3 m/s.

In the animation below, you can see that it travels 1 meter in 0.333 seconds. Therefore, in 1 second it will travel 3 meters, so the propagation speed is indeed 3 m/s.

Animated illustration of the wave motion

Next, condition (b)

b. t₀ = 1 second.

This is also correct. During 1 second, the wave slides to the right by 3/4 of a wavelength. Since the wavelength is 4 m, sliding 3 m to the right matches the orange dotted graph.

Finally, condition (c)

c. At t = 2 seconds, the displacement of the wave at x = 3 m is 2 m.

To think of it simply, from t = 0 to t = 1 the graph shifted 1 m to the left, so from t = 0 to t = 2, the graph would shift 2 m to the left. In reality, it goes 3 steps to the right and then 3 steps further to the right, but because of the wave’s periodicity, that ends up being the same as going 1 step to the left and then another step to the left.

Animated illustration of the wave motion

So, if you look closely at the graph in the animation at t = 0.666, the displacement at x = 3 is 2. Therefore, (c) is correct.

Hence, the answer is 5.

[Epilogue]

Wave problems require the ability to animate the motion of the wave in your head using the given conditions. You could try to memorize formulas for each type, but since there are so many cases, even a slight change in the conditions can lead to a wrong answer and cause you to lose interest. It seems there is no other way but practice.

[Python Program for the Wave Animation]

# Wave shift animation with pause segments and optional GIF saving or live display.
# Smooth loop across one full period (T = 1/f). No duplicated first/last frame to prevent flicker.
# Pauses at t = 1/3 T and 2/3 T, then continuous to T, and loops back to 0 seamlessly.

import numpy as np
import matplotlib.pyplot as plt
from matplotlib.animation import FuncAnimation, PillowWriter

# ===== User options =====
OUTPUT_MODE = "gif"                 # "gif" or "show"
GIF_PATH = "wave_slide_to_right.gif"     # path to save GIF when OUTPUT_MODE == "gif"
FPS = 30                            # frames per second
PAUSE_FRAMES = 15                   # pause duration at t=1/3 T and t=2/3 T (frames)
END_REPEAT_DELAY_MS = 0             # used only for live display (show mode)

# ===== Wave parameters =====
A = 2.0               # amplitude [m]
f = 0.75              # frequency [Hz]
lam = 4.0             # wavelength [m]  (so in 1 s, shift = 0.75*λ to +x)
omega = 2*np.pi*f     # angular frequency
k = 2*np.pi/lam       # wave number
phi = 0.0             # initial phase at t=0
T = 1.0 / f           # period = 4/3 s

# Spatial axis (0~8 m like the figure)
x = np.linspace(0, 8, 1000)

# Reference curves for t=0 (solid light) and t=T (same shape as t=0)
y_t0 = A * np.sin(k*x - omega*0 + phi)
y_tT = A * np.sin(k*x - omega*1 + phi)

# ===== Build timeline with pauses at t=1/3 T and t=2/3 T =====
# Important for smooth loop: DO NOT include the final frame at t=T, and do not pause there.
segments = [(0.0, T/4.0), (T/4.0, 2.0*T/4.0), (2.0*T/4.0, 3.0*T/4.0), (3.0*T/4.0, 4.0*T/4.0)]
frames_per_segment = int(FPS * (T/3.0))  # equal frames per segment across one period

t_values = []
for (t_start, t_end) in segments:
    # frames inside the segment, excluding endpoint to avoid duplicate at boundaries
    seg_ts = np.linspace(t_start, t_end, frames_per_segment, endpoint=False)
    t_values.extend(seg_ts.tolist())
    # add a pause at internal breakpoints only (not at the very end t=T)
    if t_end <= T:  # internal boundary
        t_values.extend([t_end] * PAUSE_FRAMES)

# ===== Set up the figure =====
fig, ax = plt.subplots(figsize=(7, 3.2), dpi=150)
ax.set_xlim(0, 8)
ax.set_ylim(-2.4, 2.4)
ax.set_xlabel("x (m)")
ax.set_ylabel("Amplitude (m)")

# Static references
ax.plot(x, y_t0, linewidth=1.0, alpha=0.4, label="t=0")
ax.plot(x, y_tT, linestyle="--", linewidth=1.0, alpha=0.8, label="t=1")

# Animated line
line, = ax.plot(x, y_t0, linewidth=2.0, label="Sliding Wave")
# show time modulo T to make the loop feel natural
title = ax.set_title("Sliding Wave: t = 0.000 s")
ax.legend(loc="upper right")


def update(frame_index):
    t = t_values[frame_index]
    y = A * np.sin(k*x - omega*t + phi)
    line.set_data(x, y)
    t_mod = (t % T)
    title.set_text(f"Sliding Wave: t = {t_mod:0.3f} s")
    return line, title

ani = FuncAnimation(
    fig,
    update,
    frames=len(t_values),
    interval=1000/FPS,
    blit=False,
    repeat=True,
    repeat_delay=END_REPEAT_DELAY_MS  # used for live display only
)

# ===== Output selection =====
if OUTPUT_MODE.lower() == "gif":
    # Save GIF via PillowWriter (GIF will loop; we avoided duplicate end/start frames)
    writer = PillowWriter(fps=FPS)
    ani.save(GIF_PATH, writer=writer)
    print(f"Saved: {GIF_PATH}")
elif OUTPUT_MODE.lower() == "show":
    plt.show()
else:
    raise ValueError("Unknown OUTPUT_MODE. Use 'gif' or 'show'.")

[Source: June 2025 Korean National Physics I Exam, Grade 12, Problem 10]

2026-06-103-0.75hz-wave-propagation-evaluated-by-chatgpt5.pdf

0.04MB

Estimating the critical angles using pencil and ruler

Pre-College Physics Korea — Sat, 23 Aug 2025 17:01:22 +0900

[SUMMARY]

This blog page intuitively uses light refraction, car-analogy, triangle symmetry, reversibility of rays, and simple geometry to solve a Korean high-school optics exam.

Link to the Original Exam Problem (Korean)

Main Drawing of the Problem

<Problem>
As shown in the figure, monochromatic light X enters medium B from medium A with an angle of incidence θ, and then undergoes total internal reflection at the boundary between medium C and medium A. Which of the following statements is correct?

<Options>
a. The speed of X in A is greater than in B.
b. The refractive index of B is greater than that of C.
c. The critical angle between C and A is greater than θ.

<Choices>
① a ② b ③ a, c ④ b, c ⑤ a, b, c

[Think Along]
Let’s first look at the big picture. In the center, there are two triangles back-to-back, and they happen to form the shape of a square cut in half. Each triangle has angles of 90°, 45°, and 45°. With this overall structure in mind, it’s time to move on to solving the problem.

First, (a).

a. The speed of X in A is greater than in B.

From the way the light bends at the A/B boundary, (a) is true. It’s as if the light gets “slowed down” or “hindered” as it enters B from A. You can compare the situation—light entering from a rare (low-density; light travels faster) medium into a dense (high-density; light travels slower) medium—to the following picture: a car is coming roughly from the (x, y) = (1, 1) direction toward (x, y) = (0, 0). As it crosses from a smooth concrete surface into a grassy field, the right wheel hits the grass first while the left wheel is still on the smooth surface. For a brief moment while crossing the boundary, the friction on the left and right wheels is different, so the car veers slightly to the right. The denser the grass, the more it will turn.

Enough small talk—back to the problem.

Let’s look at condition (b).

b. The refractive index of B is greater than that of C.

We’re asked to compare which of the two triangles is “denser” (i.e., optically tougher). To check this condition, we need a bit of imagination. Look at the following figure.

At first, the path of X passed through A and B, then entered the boundary between C and A at an incident angle of 45°. But it couldn’t break through the boundary—instead, it bounced back by total internal reflection. To break through the boundary, the incident angle would have to be sharper with respect to the surface. Apparently, it wasn’t sharp enough yet. The light just skimmed along the boundary without entering medium A. But with just a slightly sharper angle, it could pierce through… and in the end, the blue ray managed it. By making the angle just a bit sharper, it was able to slip through the boundary, just barely.

Now, I need to talk about the symmetry of optical systems. The physical laws of light remain valid even if the direction is reversed. In other words, if light travels along a certain path in an optical system, then it can also travel along the exact same path in the opposite direction. This principle is extremely useful when solving optics problems, and in geometrical optics it is called the principle of reversibility of light rays.

The following figure shows the previous diagram with the principle of reversibility applied.

On the path where the blue ray enters from A into C, its incident angle is such that it grazes along the slanted side of the right-hand triangle. Looking at the figure, you can see that the blue ray entering from A into C bends more than 45° relative to the boundary. In other words, for all angles of entry from A into C, the light bends more than 45°. Meanwhile, in the case of the black path X, the entry from A into B bent at 45° relative to the boundary. Therefore, since A→C bends more than A→B, we can conclude that the refractive index satisfies C > B, meaning condition (b) is false.

b. The refractive index of B is greater than that of C.

Finally, let’s examine condition (c).

c. The critical angle between C and A is greater than θ.

This one can be checked almost instantly just by looking carefully at the last diagram.

In the figure, the critical angle is clearly less than 45°, while θ is greater than 45°. Therefore, condition (c) is false.

In conclusion, the correct answer is choice 1.

[Reference] Definition of the Critical Angle

if( Angle of incidence θ < critical angle) --> Light bypasses the boundary

[Source: June 2025 Korean National Physics I Exam, Grade 12, Problem 13]

Estimating-critical-angles-evaluated-by-chatgpt5.pdf

0.04MB

Just for the sake of a problem kind of problem

Pre-College Physics Korea — Sat, 23 Aug 2025 11:56:20 +0900

[SUMMARY]

This blog page intuitively uses inverse-problem reasoning, energy flow tables, non-dimensionalization, and backward inference to solve a Korean high-school mechanics exam.

Link to Original Exam Problem (Korean)

Main Drawing of the Problem

<Problem>

As shown in the figure, object A is released from rest at a height of 4h. It passes through friction region I and collides with object B, which was initially at rest on a plane at height h. After the collision, A again passes through region I and comes to rest at point p, while B passes through friction region II and comes to rest at point q. The height of q above the horizontal plane is twice that of p.
Immediately after the collision, the kinetic energies of A and B are E and 3E, respectively, and no mechanical energy is lost in the collision. The mechanical energy lost when A passes through region I once is the same as the mechanical energy lost when B passes through region II, denoted as E0. The mass of B is three times that of A.

What is E/E0? (Assume that the objects move within the same vertical plane, and neglect the sizes of the objects, air resistance, and any friction other than in the specified regions.)

<Choices>

① 1/3 ② 11/21 ③ 4/7 ④ 7/9 ⑤ 21/23

[ On "Killer" Question – Not directly related to the problem. Can skip ]

This problem looks difficult at first glance. Somehow, it feels very different from the ones before. It makes me angry.

Since I’m angry, let me pause to talk a bit about my life. Fortunately, I had diligent parents who made sure I received a good education, and I was able to join a good company, living comfortably among capable and kind colleagues. Of course, as with all things in life, work life also inevitably brings conflicts between people from time to time.

Through years of working in an organization, one piece of wisdom I have learned is this: when someone makes me angry, the best first step is to talk to that person. Once you talk, most of the time the reasons why that person acted harshly toward you become clear. And once you know the reason, your options expand. Whether you choose to retaliate, or to find common ground for cooperation, knowing the cause makes it much more likely that your response will be strategically effective compared to when you did not know.

The worst choice, on the other hand, is cutting off communication completely. That almost always drives things toward a breakdown, and then either you or the other person ends up bearing the responsibility for the fallout — roughly a fifty-fifty chance. Anyway, that was a long digression.

Why such a story all of a sudden? Earlier, I said that the killer problem itself made me angry. When angry, it’s better to listen to what your anger has to say. So I decided to reflect more deeply on killer problems.

The biggest peculiarity of this type of problem is that you cannot simply follow the motion of the objects step by step from the initial conditions. Instead, you have to combine observations from later outcomes and work backward to reconstruct the earlier events. It’s similar to how a detective or investigator connects small clues from a crime scene to identify the culprit.

If solving an ordinary problem from initial conditions is like a police officer chasing a thief down the street, then solving a killer problem is more like a homicide detective (or Detective Conan) identifying an unknown suspect at a crime scene. That’s why it’s so difficult. But now, I feel like I’m ready to give it a try.

With a bit more research, I found that this style of problem-solving approach is called an “inverse problem.” For more details about inverse problems, please refer to the relevant Wikipedia page: https://en.wikipedia.org/wiki/Inverse_problem

[Think Along]

Now, let’s calm down and carefully read through the problem.

Main Drawing of the Problem

Steps	State Description	Total Energy of A	Total Energy of B
1	initially	4mgh	3mgh
2	Energy Change	-E0
3	After collision	mgh + 0.5mv2	3mgh + 1.5mv2
4	Energy Change	-E0	-E0
6	Stopped at the peak	mgx	*6mgx*

This problem talks only about the increase, decrease, and exchange of energy. The table above shows the flow of events from the perspective of energy.

[Step 1] At the beginning, objects A and B were at rest. Therefore, all the energy they had was potential energy.

[Step 2] B was stationary at height h, while A approached, losing an amount of energy equal to E0 along the way.

[Step 3] Then A collided with B, and it is stated that energy was conserved during the collision. After the collision, the kinetic energy ratio of A to B was 1:3. Since the mass ratio of A to B is 1:3, the only way for the kinetic energy ratio to be 1:3 is if the two have the same velocity (vA = vB = v).

[Step 4] After that, A and B each lost an amount of energy equal to E0.

[Step 5] A and B then rose to their maximum heights and came to a temporary stop, though the problem does not specify whether they stopped at the same time. What we do know is that each stopped at some moment at their respective maximum points, and from this situation we can only infer the ratio of their total mechanical energies: 1 to 6.

At this point, the problem still isn’t very easy to solve. Should I get angry again? Just kidding.

Now it’s time once more to let the imagination take flight.

[Replacing variables with constants?] In this problem, m, g, and h can each be replaced with the constant 1. If you look carefully, the total amount of energy given as the initial condition is directly proportional to m, g, and h. Therefore, if the result value is x when m, g, h = 2, 2, 2, then the result will be twice as large as x when m, g, h = 1, 1, 1. Revising the table with this simplification, it becomes as follows.

Steps	Stete Description	Total Energy of A	Total Energy of B
1	initially	4	3
2	Energy Change	-E0
3	After collision	*1 + E*	*3 + 3E*
4	Energy Change	*-E0*	*-E0*
6	Stopped at the peak	x	6x

Now we’ve got equations that are actually workable.

7 - E0 = 4 + 4E --> If you subtract E0 from the initial total energy, it equals the total energy of A and B immediately after the collision.

1 + E - E0 = x --> If you subtract E0 from A’s energy right after the collision, it equals A’s potential energy when it comes to rest at the highest point.

3 + 3E - E0 = 6x --> If you subtract E0 from B’s energy right after the collision, it equals B’s potential energy when it comes to rest at the highest point.

6 + 6E - 6E0 = 6x
3 + 3E - E0 = 6x
3 + 3E = 5E0
4 + 4E = (20/3) E0

7 - E0 = (20/3) E0
E0 = 21/23
E = 12/23
E/E0 = 12/23 × 23/21 = 12/21 = 4/7

Therefore, the answer is choice 3.

[Translation]

20. [Examiner’s Intention] Applying the Law of Conservation of Reversed Energy

Let the gravitational potential energies of A at the plane of height hh and at point pp be UU and UpU_p, respectively.
When applying the law of conservation of reversed energy,

From (1), (2), and (3), we obtain:

[Source: July 2025 Korean National Physics I Exam, Grade 12, Problem 20]

Mechanics Problem List

Pre-College Physics Korea — Fri, 22 Aug 2025 23:11:05 +0900

1. https://pcpkorea.tistory.com/13

The 1, 3, 5, 7 Rule in Uniform Acceleration

[Problem]As shown in the figure, when car A passes reference line P with speed v₀, car B, which had been at rest at reference line Q, starts moving. Both A and B undergo uniformly accelerated motion parallel to the straight road and pass reference line R

pcpkorea.tistory.com

2. https://pcpkorea.tistory.com/14

Taming the Pulley: How to Apply F=ma to Connected Masses

[Problem]As shown in figure (a), three objects A, B, and C with masses m, 2m, and M, respectively, are connected by strings. When object B is placed at point p and released, A, B, and C move with uniform acceleration.As shown in figure (b), at the instant

pcpkorea.tistory.com

3. https://pcpkorea.tistory.com/20

Just for the sake of a problem kind of problem

pcpkorea.tistory.com

4. https://pcpkorea.tistory.com/23

Travelling the Same Distance in Different Times

As shown in the figure, on a cart of mass M placed on a horizontal plane, one of the two weights A and B is placed on top of the cart, and the other is hung from a string connected to the cart. The table shows the time t it takes for the cart, initially at

pcpkorea.tistory.com

5. https://pcpkorea.tistory.com/24

A killer problem - easier tried than left unsolved

As shown in the figure, on a plane of height h1, objects A and B of masses 3m and 2m, respectively, compress spring P by a distance d from its natural length and are then released from rest.Object A passes through frictional Interval 1 and on the horizonta

pcpkorea.tistory.com

6. https://pcpkorea.tistory.com/28

Constant Force Friction?

[Problem]As shown in figure (a), object A moves on a horizontal plane with speed 4v. After passing through a friction zone, it continues with uniform motion. Object B moves toward A with speed v.As shown in figure (b), after A and B collide, object A moves

pcpkorea.tistory.com

7. To be continued...