A particle of mass 15 kg has an initial velocity `vecv_(i) = hati - 2 hatjm//s` . It collides with another body and the impact time is 0.1s, resulting in a velocity `vecc_f = 6 hati + 4hatj + 5 hatk m//s` after impact. The average force of impact on the particle is :

Options

A. `15[5hati + 6hatj+ 5hatk]`
B. `15[5hati + 6hatj - 5hatk]`
C. `150[5hati - 6hatj+ 5hatk]`
D. `150[5hati + 6hatj+ 5hatk]`

Explanation

To find the average force of impact on the particle, we can use the impulse-momentum theorem, which states that the impulse (force multiplied by the time duration of the impact) is equal to the change in momentum of the particle.

### Step-by-step Solution:

1. **Identify Given Values:**
- Mass of the particle, \( m = 15 \, \text{kg} \)
- Initial velocity, \( \vec{v_i} = \hat{i} - 2 \hat{j} \, \text{m/s} \)
- Final velocity, \( \vec{v_f} = 6 \hat{i} + 4 \hat{j} + 5 \hat{k} \, \text{m/s} \)
- Impact time, \( \Delta t = 0.1 \, \text{s} \)

2. **Calculate Initial Momentum:**
\[ \vec{p_i} = m \vec{v_i} = 15 \, \text{kg} \cdot (\hat{i} - 2 \hat{j}) = 15 \hat{i} - 30 \hat{j} \, \text{kg m/s} \]

3. **Calculate Final Momentum:**
\[ \vec{p_f} = m \vec{v_f} = 15 \, \text{kg} \cdot (6 \hat{i} + 4 \hat{j} + 5 \hat{k}) = 90 \hat{i} + 60 \hat{j} + 75 \hat{k} \, \text{kg m/s} \]

4. **Calculate Change in Momentum:**
\[ \Delta \vec{p} = \vec{p_f} - \vec{p_i} = (90 \hat{i} + 60 \hat{j} + 75 \hat{k}) - (15 \hat{i} - 30 \hat{j}) \]
\[ = (90 - 15) \hat{i} + (60 + 30) \hat{j} + 75 \hat{k} = 75 \hat{i} + 90 \hat{j} + 75 \hat{k} \, \text{kg m/s} \]

5. **Calculate Average Force:**
Using the impulse-momentum theorem:
\[ \vec{F} \Delta t = \Delta \vec{p} \]
Rearranging gives:
\[ \vec{F} = \frac{\Delta \vec{p}}{\Delta t} = \frac{75 \hat{i} + 90 \hat{j} + 75 \hat{k}}{0.1} \]
\[ = 750 \hat{i} + 900 \hat{j} + 750 \hat{k} \, \text{N} \]

### Final Answer:
The average force of impact on the particle is:
\[ \vec{F} = 750 \hat{i} + 900 \hat{j} + 750 \hat{k} \, \text{N} \]