A particle of mass m is fired from the origin of the co-ordinate axes making angle `45^(@)` with the horizontal . At an instant , its position vector is ` vec(r ) = 3hat(i) + 4hat(j)` and velocity is `vec(v) = 4hat(i) - 3hat(j)` . The angular momentum of the particle w.r.t the origin at the instant is

Options

A. 7 m `hat(k)`
B. ` - 7 m hat(k)`
C. `25 m hat(k)`
D. `- 25 mhat(k)`

Answer

To find the angular momentum of a particle with respect to the origin, we can follow these steps:

### Step 1: Identify the given quantities
The position vector \( \vec{r} \) and velocity vector \( \vec{v} \) of the particle are given as:
- \( \vec{r} = 3\hat{i} + 4\hat{j} \)
- \( \vec{v} = 4\hat{i} - 3\hat{j} \)

The mass of the particle is \( m \).

### Step 2: Calculate the momentum vector
The momentum \( \vec{p} \) of the particle is given by the product of its mass and velocity:
\[ \vec{p} = m \vec{v} = m(4\hat{i} - 3\hat{j}) = 4m\hat{i} - 3m\hat{j} \]

### Step 3: Use the formula for angular momentum
The angular momentum \( \vec{L} \) of a particle with respect to a point (the origin in this case) is given by the cross product of the position vector and the momentum vector:
\[ \vec{L} = \vec{r} \times \vec{p} \]

### Step 4: Calculate the cross product
Substituting the values of \( \vec{r} \) and \( \vec{p} \):
\[ \vec{L} = (3\hat{i} + 4\hat{j}) \times (4m\hat{i} - 3m\hat{j}) \]

Using the distributive property of the cross product:
\[ \vec{L} = 3\hat{i} \times (4m\hat{i}) + 3\hat{i} \times (-3m\hat{j}) + 4\hat{j} \times (4m\hat{i}) + 4\hat{j} \times (-3m\hat{j}) \]

### Step 5: Evaluate each term
1. \( \hat{i} \times \hat{i} = 0 \) (first term)
2. \( \hat{i} \times \hat{j} = \hat{k} \) (second term)
\[ 3\hat{i} \times (-3m\hat{j}) = -9m\hat{k} \]
3. \( \hat{j} \times \hat{i} = -\hat{k} \) (third term)
\[ 4\hat{j} \times (4m\hat{i}) = 16m(-\hat{k}) = -16m\hat{k} \]
4. \( \hat{j} \times \hat{j} = 0 \) (fourth term)

### Step 6: Combine the results
Now, combining the non-zero terms:
\[ \vec{L} = -9m\hat{k} - 16m\hat{k} = -25m\hat{k} \]

### Final Result
Thus, the angular momentum of the particle with respect to the origin at that instant is:
\[ \vec{L} = -25m\hat{k} \]

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