Let at time **t** velocity of trolley is **v**_{t}

mass of trolly=(m+ut)

**momentum of the system=v(ut) + (m+ut)v**_{t}

At time **t+dt, corresponding velocity of trolley v**_{t} + dv_{t} , mass=m+u(t+dt)

**momentum of the system=vu(t+dt) +(m+ut+udt)(****v**_{t} + dvt )

**Since there is no external forces acting on the system, hence momentum is conserved:**

**v(ut) + (m+ut)v**_{t =}**vu(t+dt) +(m+ut+udt)(****v**_{t} + dvt )

**Dividing both sides by dt we get:**

**(m+ut)dv**_{t} /dt= -u(v+v_{t} )

**Solving above dirrerential equation for v**_{t} , we get:

**(m+ut)(v+v**_{t} ) = K( a constant)

**when t=o, v**_{t} =0

**Hence k=mv**

**Hence v**_{t} = - uvt/(m+ut)^{ }

**dv**_{t} / dt= acc^{n} = -muv/(m+ut)^{2}