Ga Physics 2019 MEXT Monbukagakusho English

Based on the velocity-time graph provided in Fig. 1-1, the distance traveled by the car is equal to the **area under the curve**. The shape formed is a **trapezoid**.

### **Method 1: Using the Trapezoid Area Formula**

The area $A$ of a trapezoid is given by:

$$A = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$$

1. **Bottom base ($b_1$):** The total time from start to stop is $80 - 0 = 80 \text{ s}$.

2. **Top base ($b_2$):** The time during which the car maintains constant velocity is $40 - 20 = 20 \text{ s}$.

3. **Height ($h$):** The maximum velocity reached is $60 \text{ m/s}$.

$$\text{Distance} = \frac{1}{2} \times (80 + 20) \times 60$$

$$\text{Distance} = \frac{1}{2} \times 100 \times 60$$

$$\text{Distance} = 50 \times 60 = 3000 \text{ m}$$

---

### **Method 2: Summing Individual Parts**

You can also split the shape into two triangles and one rectangle:

* **Part 1 (0 to 20s):** Area of triangle = $\frac{1}{2} \times 20 \times 60 = 600 \text{ m}$

* **Part 2 (20 to 40s):** Area of rectangle = $20 \times 60 = 1200 \text{ m}$

* **Part 3 (40 to 80s):** Area of triangle = $\frac{1}{2} \times 40 \times 60 = 1200 \text{ m}$

* **Total Distance:** $600 + 1200 + 1200 = 3000 \text{ m}$

**Answer:**

The total distance is **3000 m**, which corresponds to option **(c)**.

 Here are the step-by-step solutions for the two problems shown in the image.

---

### **Question (2): Bouncing Ball Height**

**Problem Statement:** A ball falls from height $h$ with coefficient of restitution $e$. Find the maximum height between the $n$th and $(n+1)$th impact.

**Step-by-step Solution:**

1. **Velocity before 1st impact ($v_0$):** Using conservation of energy, $mgh = \frac{1}{2}mv_0^2 \implies v_0 = \sqrt{2gh}$.

2. **Velocity after 1st impact ($v_1$):** By definition of the coefficient of restitution, $v_1 = e \cdot v_0$.

3. **Velocity after $n$th impact ($v_n$):** Each bounce reduces the velocity by a factor of $e$. Therefore, after $n$ impacts, the upward velocity is $v_n = e^n \cdot v_0 = e^n \sqrt{2gh}$.

4. **Height after $n$th impact ($h_n$):** This is the maximum height between the $n$th and $(n+1)$th impact.

    Using $v^2 = 2gh_n$:

    $$h_n = \frac{v_n^2}{2g} = \frac{(e^n \sqrt{2gh})^2}{2g} = \frac{e^{2n} \cdot 2gh}{2g} = h e^{2n}$$

**Answer:** The correct option is **(d) $he^{2n}$**.

---

### **Question (3): Magnetic Field at Point P**

**Problem Statement:** Two long straight wires with current $I$ in opposite directions are separated by $2d$. Find the magnitude of magnetic field $H$ at point $P$.

**Step-by-step Solution:**

1. **Distance to Point P ($r$):** Point $P$ forms a right triangle with each wire. The horizontal distance is $d$ and the vertical distance is $d$.

    $$r = \sqrt{d^2 + d^2} = \sqrt{2}d$$

2. **Magnetic Field Magnitude from one wire ($H_{wire}$):** The formula for the magnetic field $H$ (in A/m) at distance $r$ is $H = \frac{I}{2\pi r}$.

    $$H_1 = H_2 = \frac{I}{2\pi(\sqrt{2}d)}$$

3. **Direction and Vector Addition:**

    * For the **left wire** (current coming out $\odot$): By the right-hand rule, the field at $P$ is perpendicular to the radius line, pointing "up and right" at a $45^\circ$ angle to the vertical.

    * For the **right wire** (current going in $\otimes$): The field at $P$ is perpendicular to its radius line, also pointing "up and left" at a $45^\circ$ angle to the vertical.

4. **Resultant Field ($H_{total}$):** The horizontal components cancel out, and the vertical components add up.

    $$H_{total} = 2 \cdot H_1 \cdot \cos(45^\circ) = 2 \cdot \frac{I}{2\pi\sqrt{2}d} \cdot \frac{1}{\sqrt{2}}$$

    $$H_{total} = \frac{2I}{2\pi (\sqrt{2} \cdot \sqrt{2}) d} = \frac{2I}{2\pi(2)d} = \frac{I}{2\pi d}$$

**Answer:** The correct option is **(a) $\frac{I}{2\pi d}$**.

For the first question (4):

**Given:**  

- Wave speed \(v = 2\ \text{m/s}\) in the \(+x\) direction.  

- From the graph, amplitude \(A = 3\ \text{m}\).  

- At \(t=0\), the wave passes through \(x=0,\ y=0\), and the first maximum is at \(x=1\ \text{m}\).

**Step 1: Determine the wavelength**  

From the graph, one full wavelength is the distance between two successive maxima, e.g. \(x=1\ \text{m}\) to \(x=5\ \text{m}\).  

\[

\lambda = 5-1 = 4\ \text{m}

\]

**Step 2: Determine the wave number**  

\[

k=\frac{2\pi}{\lambda}=\frac{2\pi}{4}=\frac{\pi}{2}\ \text{rad/m}

\]

**Step 3: Determine the angular frequency**  

\[

\omega = kv = \left(\frac{\pi}{2}\right)(2)=\pi\ \text{rad/s}

\]

**Step 4: Write the wave equation**  

For a wave traveling in the \(+x\) direction:

\[

y(x,t)=A\sin(kx-\omega t+\phi)

\]

Substitute \(A=3,\ k=\frac{\pi}{2},\ \omega=\pi\):

\[

y(x,t)=3\sin\left(\frac{\pi}{2}x-\pi t+\phi\right)

\]

\[

y(x,t)=3\sin\left[\frac{\pi}{2}(x-2t)+\phi\right]

\]

**Step 5: Determine the initial phase**  

At \(t=0,\ x=0\), the graph gives \(y=0\).  

\[

0=3\sin(\phi)\Rightarrow \phi=0

\]

Thus:

\[

y(x,t)=3\sin\left[\frac{\pi}{2}(x-2t)\right]

\]

**Answer:**  

\[

\boxed{\text{(b) } y=3\sin\left[\frac{\pi}{2}(x-2t)\right]}

\]

For the second question (5), just the answer:  

\[

\boxed{\text{(c) }333\ \text{m/s}}

\]

**Summary:**  

For question (4), the wavelength was found from the graph to be \(4\ \text{m}\), giving \(k=\pi/2\ \text{rad/m}\). Using \(v=2\ \text{m/s}\), the angular frequency was \(\omega=\pi\ \text{rad/s}\). The wave equation was then formed as \(y=3\sin\left[\frac{\pi}{2}(x-2t)\right]\), matching option (b).

For the first question (4):

**Given:**  

- Wave speed \(v = 2\ \text{m/s}\) in the \(+x\) direction.  

- From the graph, amplitude \(A = 3\ \text{m}\).  

- At \(t=0\), the wave passes through \(x=0,\ y=0\), and the first maximum is at \(x=1\ \text{m}\).

**Step 1: Determine the wavelength**  

From the graph, one full wavelength is the distance between two successive maxima, e.g. \(x=1\ \text{m}\) to \(x=5\ \text{m}\).  

\[

\lambda = 5-1 = 4\ \text{m}

\]

**Step 2: Determine the wave number**  

\[

k=\frac{2\pi}{\lambda}=\frac{2\pi}{4}=\frac{\pi}{2}\ \text{rad/m}

\]

**Step 3: Determine the angular frequency**  

\[

\omega = kv = \left(\frac{\pi}{2}\right)(2)=\pi\ \text{rad/s}

\]

**Step 4: Write the wave equation**  

For a wave traveling in the \(+x\) direction:

\[

y(x,t)=A\sin(kx-\omega t+\phi)

\]

Substitute \(A=3,\ k=\frac{\pi}{2},\ \omega=\pi\):

\[

y(x,t)=3\sin\left(\frac{\pi}{2}x-\pi t+\phi\right)

\]

\[

y(x,t)=3\sin\left[\frac{\pi}{2}(x-2t)+\phi\right]

\]

**Step 5: Determine the initial phase**  

At \(t=0,\ x=0\), the graph gives \(y=0\).  

\[

0=3\sin(\phi)\Rightarrow \phi=0

\]

Thus:

\[

y(x,t)=3\sin\left[\frac{\pi}{2}(x-2t)\right]

\]

**Answer:**  

\[

\boxed{\text{(b) } y=3\sin\left[\frac{\pi}{2}(x-2t)\right]}

\]

For the second question (5), just the answer:  

\[

\boxed{\text{(c) }333\ \text{m/s}}

\]

**Summary:**  

For question (4), the wavelength was found from the graph to be \(4\ \text{m}\), giving \(k=\pi/2\ \text{rad/m}\). Using \(v=2\ \text{m/s}\), the angular frequency was \(\omega=\pi\ \text{rad/s}\). The wave equation was then formed as \(y=3\sin\left[\frac{\pi}{2}(x-2t)\right]\), matching option (b).

For the first question, the object leaves the horizontal plane with initial horizontal speed \(v\) and no horizontal acceleration (assuming no air resistance).  

The horizontal motion is uniform, so:

\[

x = v t

\]

Thus the \(x\)-coordinate at time \(t>0\) is \(v t\).

**Answer for (1):** \(\boxed{(d)\ vt}\)

**Answers for the remaining questions:**

- (2) \(\boxed{(f)\ h-\frac12 gt^2}\)

- (3) \(\boxed{(c)\ \frac{\sqrt{2gh}}{2\tan\theta}}\)

- (4) \(\boxed{(e)\ \frac{2v^2\tan\theta}{g}}\)

**Summary:**  

The first part uses the fact that horizontal velocity remains constant, so \(x=vt\). The remaining parts follow from projectile motion and intersection with the slope.

To find the wavelength of the light ray in the film:

1. When light enters a medium, its frequency \(f\) does not change; it is determined by the source.

2. The speed of light in a medium with refractive index \(n\) is  

   \[

   v=\frac{c}{n}

   \]

   where \(c\) is the speed of light in air/vacuum.

3. Using the wave relation \(v=f\lambda_n\) and \(c=f\lambda\), we have  

   \[

   f=\frac{c}{\lambda}

   \]

4. Therefore, the wavelength in the film is  

   \[

   \lambda_n=\frac{v}{f}=\frac{c/n}{c/\lambda}=\frac{\lambda}{n}

   \]

**Answer:**  

\[

\boxed{\frac{\lambda}{n}}

\]  

So the correct choice is **(b)**.

**Summary:**  

The wavelength in the film is the vacuum wavelength divided by the refractive index, giving \(\lambda/n\).

Answers for the remaining questions:  

(2) **(d)** \(\sin\theta=n\sin\phi\)  

(3) **(a)** \(QB+BC=\left(m+\frac12\right)\frac{\lambda}{n}\)  

(4) **(b)** \(d=\frac{(2m+1)\lambda}{4n\cos\phi}\)

For the first question, we solve step by step.

**Given:**  

- Object leaves the horizontal plane at \(x=0,\ y=h\) at \(t=0\) with horizontal speed \(v\).  

- Gravity \(g\) acts downward.  

- No air resistance, so horizontal acceleration is zero.

**Step 1: Horizontal motion**  

Since there is no horizontal force, the horizontal velocity remains constant:

\[

v_x = v

\]

Thus, the horizontal position at time \(t>0\) is:

\[

x(t)=v t

\]

**Step 2: Compare with options**  

The expression \(x(t)=vt\) matches option (d).

**Answer for (1):**  

\[

\boxed{x = vt}

\]

Correct choice: **(d)**

**Summary:**  

The object moves horizontally with constant speed \(v\), so its \(x\)-coordinate at time \(t\) is simply \(x=vt\).

---

**Answers for the remaining questions:**

(2) \(y = h - \frac{1}{2}gt^2\) → **(f)**  

(3) \(v_c = \frac{\sqrt{2gh}}{2\tan\theta}\) → **(c)**  

(4) \(x = \frac{2v^2}{g}\tan\theta\) → **(e)**