A clean body of mass 100 g starts with a velocity of 2 m/s on a smooth horizontal plane, accumulating dust at the rate of 5 g/s. Find the velocity at the end of 20 seconds and the distance travelled during that period.

Answer

To solve the problem step by step, we will follow the principles of conservation of momentum and kinematics.

### Step 1: Understand the problem
We have a body of mass \( m_0 = 100 \, \text{g} = 0.1 \, \text{kg} \) moving with an initial velocity \( v_0 = 2 \, \text{m/s} \). The body accumulates dust at a rate of \( \frac{dm}{dt} = 5 \, \text{g/s} = 0.005 \, \text{kg/s} \). We need to find the final velocity after \( t = 20 \, \text{s} \) and the distance traveled during that time.

### Step 2: Determine the mass as a function of time
The mass of the body increases over time due to the accumulation of dust. The mass at time \( t \) can be expressed as:
\[ m(t) = m_0 + \frac{dm}{dt} \cdot t = 0.1 \, \text{kg} + 0.005 \, \text{kg/s} \cdot t \]
Thus, after \( t = 20 \, \text{s} \):
\[ m(20) = 0.1 + 0.005 \cdot 20 = 0.1 + 0.1 = 0.2 \, \text{kg} \]

### Step 3: Apply conservation of momentum
Since there are no external forces acting on the system, the momentum of the system remains constant. The momentum \( p \) is given by:
\[ p = m \cdot v \]
Initially, the momentum is:
\[ p_0 = m_0 \cdot v_0 = 0.1 \, \text{kg} \cdot 2 \, \text{m/s} = 0.2 \, \text{kg m/s} \]
At time \( t = 20 \, \text{s} \), the momentum will be:
\[ p = m(20) \cdot v(20) \]
Setting the initial momentum equal to the final momentum:
\[ 0.2 = 0.2 \cdot v(20) \]

### Step 4: Solve for the final velocity
From the equation:
\[ 0.2 = 0.2 \cdot v(20) \]
We can simplify:
\[ v(20) = \frac{0.2}{0.2} = 1 \, \text{m/s} \]

### Step 5: Calculate the distance traveled
To find the distance traveled, we can use the relationship between velocity and time. The velocity of the body as a function of time can be derived from the conservation of momentum:
\[ v(t) = \frac{m_0 \cdot v_0}{m(t)} \]
Substituting for \( m(t) \):
\[ v(t) = \frac{0.1 \cdot 2}{0.1 + 0.005t} \]
Now, we need to integrate this to find the distance \( x \):
\[ x = \int_0^{20} v(t) \, dt = \int_0^{20} \frac{0.2}{0.1 + 0.005t} \, dt \]
This integral can be solved as:
\[ x = 0.2 \int_0^{20} \frac{1}{0.1 + 0.005t} \, dt \]
Let \( u = 0.1 + 0.005t \), then \( du = 0.005 \, dt \) or \( dt = \frac{du}{0.005} \). Changing the limits accordingly:
- When \( t = 0 \), \( u = 0.1 \)
- When \( t = 20 \), \( u = 0.1 + 0.1 = 0.2 \)

Thus, the integral becomes:
\[ x = 0.2 \cdot \frac{1}{0.005} \int_{0.1}^{0.2} \frac{1}{u} \, du = 40 \cdot [\ln(u)]_{0.1}^{0.2} \]
Calculating the integral:
\[ x = 40 \cdot (\ln(0.2) - \ln(0.1)) = 40 \cdot \ln(2) \approx 40 \cdot 0.693 = 27.72 \, \text{m} \]

### Final Answers
- Final velocity after 20 seconds: \( v(20) = 1 \, \text{m/s} \)
- Distance traveled during that period: \( x \approx 27.72 \, \text{m} \)