A (trolley + child) of total mass 200 kg is moving with a uniform speed of 36 km/h on a frictionless track. The child of mass 20 kg starts running on the trolley from one end to the other (10 m away) with a speed of 10 `ms^(–1)` relative to the trolley in the direction of the trolley's motion and jumps out of the trolley with the same relative velocity. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run?

Answer

To solve the problem, we will follow these steps:

### Step 1: Convert the initial speed of the trolley
The initial speed of the trolley is given as 36 km/h. We need to convert this speed into meters per second (m/s).

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

### Step 2: Calculate the initial momentum of the system
The total mass of the trolley and child is 200 kg. The mass of the child is 20 kg. Therefore, the total mass of the system is:

\[ \text{Total mass} = 200 \text{ kg} + 20 \text{ kg} = 220 \text{ kg} \]

The initial momentum \( P_1 \) of the system is given by:

\[ P_1 = \text{Total mass} \times \text{Initial speed} = 220 \text{ kg} \times 10 \text{ m/s} = 2200 \text{ kg m/s} \]

### Step 3: Determine the final speed of the trolley
Let \( V \) be the final speed of the trolley after the child runs to the other end and jumps off. The child runs at a speed of 10 m/s relative to the trolley, so the speed of the child relative to the ground when he jumps off will be:

\[ \text{Child's speed} = V + 10 \text{ m/s} \]

The final momentum \( P_2 \) of the system when the child is running is given by:

\[ P_2 = \text{Mass of trolley} \times V + \text{Mass of child} \times (V + 10) \]

Substituting the masses:

\[ P_2 = 200V + 20(V + 10) = 200V + 20V + 200 = 220V + 200 \]

### Step 4: Apply the conservation of momentum
Since there are no external forces acting on the system, we can apply the conservation of momentum:

\[ P_1 = P_2 \]

Substituting the values we calculated:

\[ 2200 = 220V + 200 \]

### Step 5: Solve for \( V \)
Rearranging the equation gives:

\[ 2200 - 200 = 220V \]
\[ 2000 = 220V \]
\[ V = \frac{2000}{220} \approx 9.09 \text{ m/s} \]

### Step 6: Calculate the distance moved by the trolley
The child runs a distance of 10 m at a relative speed of 10 m/s. The time \( t \) taken for the child to run from one end of the trolley to the other is:

\[ t = \frac{\text{Distance}}{\text{Relative speed}} = \frac{10 \text{ m}}{10 \text{ m/s}} = 1 \text{ s} \]

Now, we can calculate the distance moved by the trolley during this time at its final speed \( V \):

\[ \text{Distance moved by trolley} = V \times t = 9.09 \text{ m/s} \times 1 \text{ s} \approx 9.09 \text{ m} \]

### Final Answers
- The final speed of the trolley is approximately **9.09 m/s**.
- The trolley has moved approximately **9.09 m** from the time the child begins to run.