Physics Se 2018 MEXT Monbukagakusho

Based on the problem description, here are the step-by-step solutions for the three parts.

**Given Data:**

* Mass (\(m\)): \(4.0 \text{ kg}\)

* Incline Angle (\(\theta\)): \(30^\circ\)

* Coefficient of Static Friction (\(\mu_s\)): \(0.20\)

* Coefficient of Kinetic Friction (\(\mu_k\)): \(0.050\)

* Acceleration due to gravity (\(g\)): \(9.8 \text{ m/s}^2\)

* Constant: \(\sqrt{3} \approx 1.73\)

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### (1) Maximal friction force between the object and the inclined plane

The maximal friction force is the limiting static friction, given by \(f_{max} = \mu_s N\).

First, we find the Normal force (\(N\)), which balances the component of gravity perpendicular to the plane:

\[ N = mg \cos(30^\circ) \]

\[ N = (4.0)(9.8)\left(\frac{\sqrt{3}}{2}\right) = (4.0)(9.8)(0.865) = 33.908 \text{ N} \]

Now, calculate the maximal friction force:

\[ f_{max} = \mu_s N = (0.20)(33.908) = 6.7816 \text{ N} \]

Rounding to two significant figures as requested:

**Answer (1): \(6.8 \text{ N}\)**

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### (2) Minimal force to pull up the object along the inclined plane

To pull the object up the incline at a constant speed (minimal force), the applied force (\(F\)) must overcome both the component of gravity pulling the object down the slope and the maximal static friction acting against the motion.

\[ F = mg \sin(30^\circ) + f_{max} \]

\[ mg \sin(30^\circ) = (4.0)(9.8)(0.5) = 19.6 \text{ N} \]

Using the three-significant-figure value for friction from the previous step (\(6.78 \text{ N}\)):

\[ F = 19.6 + 6.78 = 26.38 \text{ N} \]

Rounding to two significant figures:

**Answer (2): \(26 \text{ N}\)**

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### (3) Acceleration of the object after cutting the string

Once the string is cut, the only forces acting along the plane are the gravitational component down the slope and the kinetic friction acting up the slope.

Net force (\(F_{net}\)):

\[ F_{net} = mg \sin(30^\circ) - f_k \]

Where kinetic friction \(f_k = \mu_k N = \mu_k mg \cos(30^\circ)\).

Using Newton's Second Law (\(a = \frac{F_{net}}{m}\)):

\[ a = \frac{mg \sin(30^\circ) - \mu_k mg \cos(30^\circ)}{m} \]

\[ a = g(\sin(30^\circ) - \mu_k \cos(30^\circ)) \]

\[ a = 9.8(0.5 - (0.050)(0.865)) \]

\[ a = 9.8(0.5 - 0.04325) = 9.8(0.45675) = 4.47615 \text{ m/s}^2 \]

Rounding to two significant figures:

**Answer (3): \(4.5 \text{ m/s}^2\)**

 

The speeds of the objects after the collision are:

* **Speed of object A (\(v_A\)):** \(3.0 \text{ m/s}\)

* **Speed of object B (\(v_B\)):** \(5.2 \text{ m/s}\)

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### Physical Analysis

This is a 2D collision problem. Since the plane is smooth and there are no external horizontal forces, **linear momentum is conserved** in both the \(x\) (horizontal) and \(y\) (vertical) directions.

**Knowns:**

* Masses: \(m_A = m_B = m\) (identical)

* Initial state: \(u_B = 6.0 \text{ m/s}\), \(u_A = 0 \text{ m/s}\)

* Final angles: \(\theta_B = 30^\circ\), \(\theta_A = -60^\circ\) (below the axis)

### 1. Conservation of Momentum in the y-direction

The initial vertical momentum is zero.

\[ \sum p_{y, \text{initial}} = \sum p_{y, \text{final}} \]

\[ 0 = m v_B \sin(30^\circ) - m v_A \sin(60^\circ) \]

Dividing by \(m\) and substituting trig values (\(\sin 30^\circ = 0.5\), \(\sin 60^\circ = \frac{\sqrt{3}}{2}\)):

\[ v_B(0.5) = v_A\left(\frac{\sqrt{3}}{2}\right) \]

\[ v_B = \sqrt{3} v_A \quad \text{--- (Eq. 1)} \]

### 2. Conservation of Momentum in the x-direction

\[ \sum p_{x, \text{initial}} = \sum p_{x, \text{final}} \]

\[ m(6.0) = m v_B \cos(30^\circ) + m v_A \cos(60^\circ) \]

Dividing by \(m\) and substituting trig values (\(\cos 30^\circ = \frac{\sqrt{3}}{2}\), \(\cos 60^\circ = 0.5\)):

\[ 6.0 = v_B\left(\frac{\sqrt{3}}{2}\right) + v_A(0.5) \quad \text{--- (Eq. 2)} \]

### 3. Solving for Speeds

Substitute \(v_B = \sqrt{3} v_A\) from Eq. 1 into Eq. 2:

\[ 6.0 = (\sqrt{3} v_A)\frac{\sqrt{3}}{2} + 0.5 v_A \]

\[ 6.0 = \frac{3}{2} v_A + \frac{1}{2} v_A \]

\[ 6.0 = 2 v_A \implies v_A = 3.0 \text{ m/s} \]

Now find \(v_B\) using Eq. 1 and the provided value \(\sqrt{3} = 1.73\):

\[ v_B = \sqrt{3} \times 3.0 \]

\[ v_B = 1.73 \times 3.0 = 5.19 \text{ m/s} \]

### Final Results (Two Significant Figures)

* **\(v_A = 3.0 \text{ m/s}\)**

* **\(v_B = 5.2 \text{ m/s}\)**

*(Note: Because the scattering angles add up to \(90^\circ\) (\(30^\circ + 60^\circ\)), this confirms the collision is perfectly elastic, as expected for identical masses where one is initially at rest.)*

Answer: (1)

Step 1: Calculate the equivalent resistance ($R_p$) of the three parallel resistors (4.0 Ω, 5.0 Ω, 6.0 Ω).

$1/R_p = 1/4.0 + 1/5.0 + 1/6.0 = 0.25 + 0.2 + 0.1667 = 0.6167 \text{ S}$.

$R_p = 1 / 0.6167 \approx 1.621 \text{ Ω}$.

Step 2: Add the series resistor (1.0 Ω) to $R_p$.

$R_{total} = 1.0 + 1.621 = 2.621 \text{ Ω}$.

Rounding to two significant figures: 2.6 Ω.

Answer: Answer: 2.6 Ω.

(2)

Step 1: Calculate the equivalent capacitance ($C_p$) of the three parallel capacitors (4.0 F, 5.0 F, 6.0 F).

$C_p = 4.0 + 5.0 + 6.0 = 15.0 \text{ F}$.

Step 2: Calculate the total capacitance ($C_{total}$) with the series capacitor (1.0 F).

$1/C_{total} = 1/1.0 + 1/15.0 = 1 + 0.0667 = 1.0667 \text{ F}^{-1}$.

$C_{total} = 1 / 1.0667 \approx 0.9375 \text{ F}$.

Rounding to two significant figures: 0.94 F.

Answer: Answer: 0.94 F.

(3)

Step 1: Apply the Wheatstone bridge balance condition: $R_1/R_2 = R_3/R_4$.

Given $R_1 = 4.0 \text{ Ω}$, $R_2 = 5.0 \text{ Ω}$, $R_3 = 6.0 \text{ Ω}$, and $R_4 = R$.

$4.0 / 5.0 = 6.0 / R$.

Step 2: Solve for $R$.

$0.8 = 6.0 / R \implies R = 6.0 / 0.8 = 7.5 \text{ Ω}$.

Rounding to two significant figures: 7.5 Ω.

Answer: Answer: 7.5 Ω. 

Answer: (1) 110 Hz, (2) 3.1 m, (3) 97 Hz. 

Explanation: Explanation: The Doppler effect for sound is given by the formula $f_{obs} = f_s \left( \frac{v \pm v_{obs}}{v \mp v_s} \right)$, where $v = 340 \text{ m/s}$ is the speed of sound. We define the direction from the observer to the source as positive.

(1) The observer is moving toward the stationary source ($v_s = 0$). The observer's velocity $v_{obs} = 30 \text{ m/s}$ is in the direction that increases frequency.

Step 1: $f_{obs} = 100 \times \left( \frac{340 + 30}{340} \right) = 100 \times \frac{370}{340} \approx 108.82 \text{ Hz}$.

Step 2: Rounding to two significant figures, we get $110 \text{ Hz}$.

(2) The source is moving toward the stationary observer ($v_{obs} = 0$). The source velocity $v_s = 30 \text{ m/s}$ is in the direction that decreases the wavelength.

Step 1: The wavelength observed is $\lambda' = \frac{v - v_s}{f_s} = \frac{340 - 30}{100} = \frac{310}{100} = 3.10 \text{ m}$.

Step 2: Rounding to two significant figures, we get $3.1 \text{ m}$.

(3) Both are moving in the same direction. The source is behind the observer. Let the direction of motion be positive. The source velocity $v_s = 30 \text{ m/s}$ and observer velocity $v_{obs} = 40 \text{ m/s}$. Since the source is chasing the observer, the source is moving toward the observer ($+v_s$) and the observer is moving away from the source ($-v_{obs}$).

Step 1: $f_{obs} = 100 \times \left( \frac{340 - 40}{340 - 30} \right) = 100 \times \frac{300}{310} \approx 96.77 \text{ Hz}$.

Step 2: Rounding to two significant figures, we get $97 \text{ Hz}$.