A body of mass 5 kg moving with a uniform speed `3 sqrt 2 ms^(-1)` in X – Y plane along the line y = x + 4. The angular momentum of the particle about the origin will be ______ `kg m^2s^(–1)`.

Answer

To find the angular momentum of the body about the origin, we will follow these steps:

### Step 1: Identify the parameters
Given:
- Mass of the body, \( m = 5 \, \text{kg} \)
- Speed of the body, \( v = 3\sqrt{2} \, \text{m/s} \)
- The line of motion is given by \( y = x + 4 \).

### Step 2: Determine the position vector \( \mathbf{r} \)
The line \( y = x + 4 \) intersects the y-axis at \( (0, 4) \). Therefore, when the body is at this point, the position vector \( \mathbf{r} \) from the origin to the point can be represented as:
\[ \mathbf{r} = (0, 4) \, \text{m} \]

### Step 3: Determine the velocity vector \( \mathbf{v} \)
Since the body is moving along the line \( y = x + 4 \), we need to find the components of the velocity vector. The slope of the line is \( 1 \), indicating that the angle \( \theta \) with the x-axis is \( 45^\circ \).

The components of the velocity can be calculated as:
\[ v_x = v \cos(45^\circ) = 3\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 3 \, \text{m/s} \]
\[ v_y = v \sin(45^\circ) = 3\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 3 \, \text{m/s} \]
Thus, the velocity vector is:
\[ \mathbf{v} = (3, 3) \, \text{m/s} \]

### Step 4: Calculate the angular momentum \( \mathbf{L} \)
The angular momentum \( \mathbf{L} \) about the origin is given by the formula:
\[ \mathbf{L} = \mathbf{r} \times \mathbf{p} \]
where \( \mathbf{p} = m \mathbf{v} \) is the linear momentum.

First, calculate the linear momentum:
\[ \mathbf{p} = m \mathbf{v} = 5 \cdot (3, 3) = (15, 15) \, \text{kg m/s} \]

Now, using the cross product:
\[ \mathbf{L} = \mathbf{r} \times \mathbf{p} = (0, 4) \times (15, 15) \]
The magnitude of the angular momentum can be calculated as:
\[ L = m r v \sin(\alpha) \]
where \( \alpha \) is the angle between \( \mathbf{r} \) and \( \mathbf{v} \). The angle \( \alpha \) can be calculated as:
\[ \alpha = 90^\circ - 45^\circ = 45^\circ \]
Thus,
\[ L = m r v \sin(45^\circ) \]
Substituting the values:
\[ L = 5 \cdot 4 \cdot (3\sqrt{2}) \cdot \frac{1}{\sqrt{2}} = 5 \cdot 4 \cdot 3 = 60 \, \text{kg m}^2/\text{s} \]

### Final Answer
The angular momentum of the particle about the origin is \( \mathbf{L} = 60 \, \text{kg m}^2/\text{s} \).

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