In this question, not that the sand will exert a force on the boat even if the sand is dropped onto the boat. Also, this will accelerate the boat and change the momentum of the sand-boat system.

When the sand is dropped onto the boat, the initial horizontal velocity of the sand is zero as it is at the rest position on the boat. The net horizontal velocity is equal to the velocity of the boat for the sand-boat system.

The initial momentum of the system is\({m_{\rm{b}}}{v_{\rm{b}}}\), where\({m_{\rm{b}}}\)is the mass of the boat and\({v_{\rm{b}}}\)is the initial velocity of the boat.

The sand is accelerated from its rest position to the final speed of the boat. The frictional force provides this acceleration to the sand. From Newton's third law, the sand applies force with the same magnitude on the boat from different directions, inducing the boat to decrease its speed.

Also, both the boat and the sand move with the same final speed v.

The final momentum of the system becomes\(\left( {{m_{\rm{s}}} + {m_{\rm{b}}}} \right)v\).

Here,\({m_{\rm{s}}}\)is the mass of sand.

From the momentum conservation principle, you get:

\(\begin{aligned}{l}{\rm{Initial momentum}} = {\rm{Final momentum}}\\{m_{\rm{b}}}{v_{\rm{b}}} = \left( {{m_{\rm{s}}} + {m_{\rm{b}}}} \right)v\\v = \frac{{{m_{\rm{b}}}{v_{\rm{b}}}}}{{\left( {{m_{\rm{s}}} + {m_{\rm{b}}}} \right)}}.\end{aligned}\)

Here, the final speed of the boat has reduced.

Thus, option (b) is the correct answer.