A girl of mass `40kg` jumps with a horizontal velocity of `5m//s` onto a stationary cart with frictionless wheels. The mass of the cart is `3kg`. What is her velocity as the cart starts moving? Assume that there is no external unbalanced force working in the horizontal direction.

Options

A. 100/43 m/s
B. 200 m/s
C. 200/43 m/s
D. NONE

Explanation

To solve this problem, we will use the principle of conservation of momentum. The total momentum before the girl jumps onto the cart must equal the total momentum after she jumps onto it, as there are no external unbalanced forces acting on the system.

### Step-by-Step Solution:

1. **Identify the masses and initial velocities:**
- Mass of the girl, \( m_g = 40 \, \text{kg} \)
- Initial velocity of the girl, \( v_g = 5 \, \text{m/s} \)
- Mass of the cart, \( m_c = 3 \, \text{kg} \)
- Initial velocity of the cart, \( v_c = 0 \, \text{m/s} \)

2. **Calculate the initial momentum (Pi):**
The initial momentum of the system is the sum of the momentum of the girl and the cart.
\[ P_i = m_g \cdot v_g + m_c \cdot v_c \]
Since the cart is stationary, \( v_c = 0 \):
\[ P_i = (40 \, \text{kg} \cdot 5 \, \text{m/s}) + (3 \, \text{kg} \cdot 0 \, \text{m/s}) = 200 \, \text{kg m/s} \]

3. **Calculate the final momentum (Pf):**
After the girl jumps onto the cart, the total mass of the system is \( m_g + m_c \) and they move together with a common velocity \( V \).
\[ P_f = (m_g + m_c) \cdot V \]
\[ P_f = (40 \, \text{kg} + 3 \, \text{kg}) \cdot V = 43 \, \text{kg} \cdot V \]

4. **Apply the conservation of momentum:**
According to the law of conservation of momentum:
\[ P_i = P_f \]
\[ 200 \, \text{kg m/s} = 43 \, \text{kg} \cdot V \]

5. **Solve for the final velocity (V):**
Rearranging the equation to find \( V \):
\[ V = \frac{200 \, \text{kg m/s}}{43 \, \text{kg}} \approx 4.65 \, \text{m/s} \]

### Final Answer:
The velocity of the girl and the cart as they start moving together is approximately \( 4.65 \, \text{m/s} \).

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