A big olive (m=0.50 kg) lies at the origin of an xy coor dinate system, and a big Brazil nut (M = 1.5 kg) lies at the point (1.0,2.0) m. At t=0, a force `vecF_o = (2.0hati + 3.0hatj)N` begins to act on the olive, and a force `vecF_n = (-3.0hati – 2.0hatj)N` begins to act on the nut. In unit-vector notation, what is the displacement of the center of mass of the olive-nut system at t = 4.0 s, with respect to its position at 1= 0?

Answer

To find the displacement of the center of mass of the olive-nut system at \( t = 4.0 \, s \), we will follow these steps:

### Step 1: Define Given Quantities
- Mass of the olive, \( m_0 = 0.50 \, kg \)
- Mass of the Brazil nut, \( M = 1.5 \, kg \)
- Initial position of the olive: \( \vec{r}_0 = (0, 0) \, m \)
- Initial position of the nut: \( \vec{r}_n = (1.0, 2.0) \, m \)
- Force on the olive: \( \vec{F}_0 = (2.0 \hat{i} + 3.0 \hat{j}) \, N \)
- Force on the nut: \( \vec{F}_n = (-3.0 \hat{i} - 2.0 \hat{j}) \, N \)

### Step 2: Calculate Acceleration of Each Object
Using Newton's second law, \( \vec{F} = m \vec{a} \), we can find the acceleration of each object.

1. **Acceleration of the Olive**:
\[ \vec{a}_0 = \frac{\vec{F}_0}{m_0} = \frac{(2.0 \hat{i} + 3.0 \hat{j})}{0.50} = (4.0 \hat{i} + 6.0 \hat{j}) \, m/s^2 \]

2. **Acceleration of the Nut**:
\[ \vec{a}_n = \frac{\vec{F}_n}{M} = \frac{(-3.0 \hat{i} - 2.0 \hat{j})}{1.5} = (-2.0 \hat{i} - \frac{4}{3} \hat{j}) \, m/s^2 \]

### Step 3: Calculate Displacement of Each Object
The displacement \( \Delta \vec{r} \) can be calculated using the formula:
\[ \Delta \vec{r} = \frac{1}{2} \vec{a} t^2 \]

1. **Displacement of the Olive**:
\[ \Delta \vec{r}_0 = \frac{1}{2} (4.0 \hat{i} + 6.0 \hat{j}) (4.0)^2 = \frac{1}{2} (4.0 \hat{i} + 6.0 \hat{j}) (16) = (32.0 \hat{i} + 48.0 \hat{j}) \, m \]

2. **Displacement of the Nut**:
\[ \Delta \vec{r}_n = \frac{1}{2} (-2.0 \hat{i} - \frac{4}{3} \hat{j}) (4.0)^2 = \frac{1}{2} (-2.0 \hat{i} - \frac{4}{3} \hat{j}) (16) = (-16.0 \hat{i} - \frac{32}{3} \hat{j}) \, m \]

### Step 4: Find New Positions of Each Object
1. **New Position of the Olive**:
\[ \vec{r}_0' = \vec{r}_0 + \Delta \vec{r}_0 = (0, 0) + (32.0 \hat{i} + 48.0 \hat{j}) = (32.0, 48.0) \, m \]

2. **New Position of the Nut**:
\[ \vec{r}_n' = \vec{r}_n + \Delta \vec{r}_n = (1.0, 2.0) + (-16.0 \hat{i} - \frac{32}{3} \hat{j}) = (-15.0, 2.0 - \frac{32}{3}) = (-15.0, -\frac{26}{3}) \, m \]

### Step 5: Calculate the Center of Mass
The center of mass \( \vec{R}_{cm} \) is given by:
\[ \vec{R}_{cm} = \frac{m_0 \vec{r}_0' + M \vec{r}_n'}{m_0 + M} \]

Calculating the x-coordinate:
\[ R_{cm,x} = \frac{0.5 \times 32.0 + 1.5 \times (-15.0)}{0.5 + 1.5} = \frac{16.0 - 22.5}{2.0} = \frac{-6.5}{2.0} = -3.25 \, m \]

Calculating the y-coordinate:
\[ R_{cm,y} = \frac{0.5 \times 48.0 + 1.5 \times \left(-\frac{26}{3}\right)}{0.5 + 1.5} = \frac{24.0 - 13.0}{2.0} = \frac{11.0}{2.0} = 5.5 \, m \]

### Final Result
Thus, the coordinates of the center of mass at \( t = 4.0 \, s \) are:
\[ \vec{R}_{cm} = (-3.25 \hat{i} + 5.5 \hat{j}) \, m \]