To tackle this problem, we need to consider the principles of motion and the exponential decay of mass due to disintegration. The scenario describes a body of mass \( m_0 \) that is placed on a smooth horizontal surface, and as it disintegrates, its mass decreases exponentially. The key here is to understand how this change in mass affects the velocity of the body over time.

Understanding Mass Disintegration

The mass of the body decreases exponentially, which can be mathematically expressed as:

• \( m(t) = m_0 e^{-\lambda t} \)

Here, \( \lambda \) is the disintegration constant, and \( t \) is the time elapsed. This equation tells us how the mass changes over time.

Applying the Concept of Momentum

When mass is ejected from the body, it moves backward with a relative velocity \( u \). According to the law of conservation of momentum, the momentum before and after the mass is ejected must remain constant, assuming no external forces are acting on the system.

Momentum Before and After Ejection

Initially, the momentum of the system is:

• \( P_{\text{initial}} = m_0 v_0 \)

Assuming the body starts from rest, \( v_0 = 0 \), so:

• \( P_{\text{initial}} = 0 \)

After a small time \( dt \), a mass \( dm \) is ejected backward with velocity \( u \). The remaining mass of the body is \( m(t) = m_0 e^{-\lambda t} \), and its velocity at this moment is \( v(t) \). The momentum after the mass is ejected can be expressed as:

• \( P_{\text{final}} = m(t) v(t) + dm (-u) \)

Setting the initial momentum equal to the final momentum gives us:

• \( 0 = m(t) v(t) - dm \cdot u \)

Relating Mass and Velocity

Rearranging the equation, we find:

• \( m(t) v(t) = dm \cdot u \)

Now, substituting \( m(t) \) into the equation:

• \( e^{-\lambda t} m_0 v(t) = dm \cdot u \)

To find \( dm \), we can express it as:

• \( dm = -\lambda m_0 e^{-\lambda t} dt \)

Substituting \( dm \) back into the momentum equation gives:

• \( e^{-\lambda t} m_0 v(t) = -(-\lambda m_0 e^{-\lambda t} dt) \cdot u \)

After simplifying, we can cancel \( m_0 e^{-\lambda t} \) from both sides (assuming \( m_0 \) is not zero):

• \( v(t) = \lambda u dt \)

Final Velocity Expression

Integrating this expression over time gives us the velocity of the body after time \( t \):

• \( v(t) = u t \)

This result indicates that the velocity of the body increases linearly with time, scaled by the relative velocity \( u \) of the ejected mass. Thus, as time progresses, the body continues to gain velocity due to the continuous ejection of mass.

In summary, the velocity of the body after time \( t \) is directly proportional to the time elapsed and the relative velocity of the ejected mass, leading us to the final expression:

• \( v(t) = u t \)