A boy of mass 50 kg jumps with horizontal velocity of 1 m/s on to a stationary cart. If mass of the cart will be 5 kg[Assume friction between the wheels the cart and road is zero ]

Options

A. ` 10/11` m/s
B. ` 11/10` m/s
C. `0*65 ` m/s
D. `0*8` m/s

Answer

To solve the problem step by step, we will use the principle of conservation of momentum.

### Step 1: Identify the given data
- Mass of the boy (m1) = 50 kg
- Velocity of the boy (V1) = 1 m/s
- Mass of the cart (m2) = 5 kg
- Initial velocity of the cart (V2) = 0 m/s (since it is stationary)

### Step 2: Write the formula for conservation of momentum
According to the law of conservation of momentum, the total momentum before the event must equal the total momentum after the event.

The momentum before the boy jumps onto the cart is:
\[ P_{\text{initial}} = m_1 \cdot V_1 + m_2 \cdot V_2 \]
Since the cart is stationary, \( V_2 = 0 \):
\[ P_{\text{initial}} = m_1 \cdot V_1 + 0 = m_1 \cdot V_1 \]

The momentum after the boy jumps onto the cart is:
\[ P_{\text{final}} = (m_1 + m_2) \cdot V \]
where \( V \) is the final velocity of the cart and the boy together.

### Step 3: Set the initial momentum equal to the final momentum
Using the conservation of momentum:
\[ m_1 \cdot V_1 = (m_1 + m_2) \cdot V \]

### Step 4: Substitute the known values into the equation
Substituting the values we have:
\[ 50 \cdot 1 = (50 + 5) \cdot V \]
\[ 50 = 55 \cdot V \]

### Step 5: Solve for V
To find \( V \), divide both sides of the equation by 55:
\[ V = \frac{50}{55} \]
\[ V = \frac{10}{11} \text{ m/s} \]

### Step 6: Conclusion
The final velocity of the cart after the boy jumps onto it is:
\[ V = \frac{10}{11} \text{ m/s} \]