the 35 kilogram girl is standing on a 20 kilogram wagon and jumps off. Giving the wagon a kicks that sends it off at 3.8 metres per second. How fast is the girl moving just after the jump off

Options

A. `1.2 ms^-1`
B. `3 ms^-1`
C. `4 ms^-1`
D. `2.2 ms^-1`

Explanation

To solve the problem, we will use the principle of conservation of momentum. Here are the steps to find the speed of the girl just after she jumps off the wagon:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the girl (Mg) = 35 kg
- Mass of the wagon (Mw) = 20 kg
- Velocity of the wagon after the jump (Vw) = 3.8 m/s

2. **Understand the Conservation of Momentum:**
- According to the law of conservation of momentum, the total momentum before the jump must equal the total momentum after the jump.
- Initial momentum (before the jump) = 0 (since both the girl and the wagon are at rest).
- Final momentum (after the jump) = Momentum of the girl + Momentum of the wagon.

3. **Set Up the Equation:**
- Let the velocity of the girl after jumping off be Vg.
- The momentum of the girl after jumping off = Mg * Vg.
- The momentum of the wagon after the jump = Mw * Vw.
- Since the initial momentum is zero, we can write:
\[ Mg \cdot Vg + Mw \cdot Vw = 0 \]

4. **Rearranging the Equation:**
- This can be rearranged to find Vg:
\[ Mg \cdot Vg = -Mw \cdot Vw \]
- Thus, we have:
\[ Vg = -\frac{Mw \cdot Vw}{Mg} \]

5. **Substituting the Values:**
- Substitute the values of Mw, Vw, and Mg into the equation:
\[ Vg = -\frac{20 \, \text{kg} \cdot 3.8 \, \text{m/s}}{35 \, \text{kg}} \]

6. **Calculating Vg:**
- Calculate the value:
\[ Vg = -\frac{76}{35} \approx -2.17 \, \text{m/s} \]
- Since we are interested in the speed (magnitude), we take the absolute value:
\[ Vg \approx 2.17 \, \text{m/s} \]

7. **Final Answer:**
- The speed of the girl just after she jumps off is approximately **2.17 m/s**.