A trolley having mass of 200kg moves with uniform speed of `36km h^(-1)` on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of `4ms^(-1)` relative to the trolley in a direction opposite to its motion and ultimately jumps out of the trolley. with how much velocity has the trolley moved from the time the child begins to run?

Options

A. `10.36 ms^(-1)`
B. `11.36 ms^(-1)`
C. `12.36ms^(-1)`
D. `14.40ms^(-1)`

Explanation

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

### Step 1: Convert the speed of the trolley from km/h to m/s
The initial speed of the trolley is given as 36 km/h. We need to convert this to m/s.

\[ \text{Speed in m/s} = \frac{36 \text{ km/h} \times 1000 \text{ m/km}}{3600 \text{ s/h}} = 10 \text{ m/s} \]

### Step 2: Identify the masses and velocities
- Mass of the trolley, \( m_t = 200 \text{ kg} \)
- Mass of the child, \( m_c = 20 \text{ kg} \)
- Initial velocity of the trolley, \( v_t = 10 \text{ m/s} \)
- Relative speed of the child with respect to the trolley, \( v_{c/t} = 4 \text{ m/s} \) (opposite to the direction of the trolley)

### Step 3: Determine the velocity of the child with respect to the ground
Since the child is running in the opposite direction to the trolley's motion, the velocity of the child with respect to the ground is:

\[ v_c = v_t - v_{c/t} = 10 \text{ m/s} - 4 \text{ m/s} = 6 \text{ m/s} \]

### Step 4: Apply the conservation of momentum
According to the conservation of momentum, the total momentum before the child starts running must equal the total momentum after the child starts running.

**Initial momentum:**
\[ P_{initial} = (m_t + m_c) \cdot v_t = (200 \text{ kg} + 20 \text{ kg}) \cdot 10 \text{ m/s} = 2200 \text{ kg m/s} \]

**Final momentum:**
Let \( v' \) be the final velocity of the trolley after the child runs. The final momentum will be:

\[ P_{final} = m_t \cdot v' + m_c \cdot (v' - 4 \text{ m/s}) \]

### Step 5: Set up the equation
Setting initial momentum equal to final momentum:

\[ 2200 = 200 \cdot v' + 20 \cdot (v' - 4) \]

Expanding this gives:

\[ 2200 = 200v' + 20v' - 80 \]
\[ 2200 + 80 = 220v' \]
\[ 2280 = 220v' \]

### Step 6: Solve for \( v' \)
\[ v' = \frac{2280}{220} = 10.36 \text{ m/s} \]

### Step 7: Calculate the distance moved by the trolley
Now, we need to find out how far the trolley has moved while the child runs from one end to the other (10 m).

The time taken by the child to run 10 m relative to the trolley is:

\[ t = \frac{10 \text{ m}}{4 \text{ m/s}} = 2.5 \text{ s} \]

During this time, the trolley moves a distance of:

\[ d = v_t \cdot t = 10 \text{ m/s} \cdot 2.5 \text{ s} = 25 \text{ m} \]

### Final Answer
The trolley has moved 25 meters from the time the child begins to run.

---