A trolley of mass 10 kg carries 16 kg grain and moves on a horizontal smooth and straight track at `20ms^(-1)`. If the grain start lenking out of a hole at the bottom at time t = `0_(sec)`. At the rate of 0.5 `kgs^(-1)` the speed of the trolley at ` t= 22s` will be nearly

Options

A. `13.5 ms^(-1)`
B. `15ms^(-1)`
C. `24ms^(-1)`
D. `26ms^(-1)`

Answer

To solve the problem step-by-step, we will use the principle of conservation of momentum. Here’s how we can approach it:

### Step 1: Identify the initial conditions
- Mass of the trolley (m_trolley) = 10 kg
- Mass of the grain (m_grain) = 16 kg
- Initial velocity (v_initial) = 20 m/s

### Step 2: Calculate the total initial mass of the system
The total mass of the system (m_initial) is the sum of the mass of the trolley and the mass of the grain:
\[ m_{\text{initial}} = m_{\text{trolley}} + m_{\text{grain}} = 10 \, \text{kg} + 16 \, \text{kg} = 26 \, \text{kg} \]

### Step 3: Determine the rate of grain leaking
The grain is leaking at a rate of 0.5 kg/s. Over a time period of 22 seconds, the total mass of grain that has leaked out is:
\[ \text{mass leaked} = \text{leak rate} \times \text{time} = 0.5 \, \text{kg/s} \times 22 \, \text{s} = 11 \, \text{kg} \]

### Step 4: Calculate the final mass of the system
The final mass of the system (m_final) after 22 seconds is:
\[ m_{\text{final}} = m_{\text{initial}} - \text{mass leaked} = 26 \, \text{kg} - 11 \, \text{kg} = 15 \, \text{kg} \]

### Step 5: Apply the conservation of momentum
According to the conservation of momentum:
\[ m_{\text{initial}} \cdot v_{\text{initial}} = m_{\text{final}} \cdot v_{\text{final}} \]
Substituting the known values:
\[ 26 \, \text{kg} \cdot 20 \, \text{m/s} = 15 \, \text{kg} \cdot v_{\text{final}} \]

### Step 6: Solve for the final velocity
Calculating the left-hand side:
\[ 520 \, \text{kg m/s} = 15 \, \text{kg} \cdot v_{\text{final}} \]
Now, solve for \( v_{\text{final}} \):
\[ v_{\text{final}} = \frac{520 \, \text{kg m/s}}{15 \, \text{kg}} = \frac{520}{15} \approx 34.67 \, \text{m/s} \]

### Step 7: Round the final answer
Rounding to two decimal places, we get:
\[ v_{\text{final}} \approx 34.67 \, \text{m/s} \approx 35 \, \text{m/s} \]

### Final Answer
The speed of the trolley at \( t = 22 \, \text{s} \) will be nearly \( 35 \, \text{m/s} \).

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