Irodov Problem 1.184

The part of chain that reaches the ground looses tension and stops accelerating. Suppose that at some instant of time the length of chain that has still not fallen is given by x as shown in the figure. Let the total mass of the chain be m. The mass of x length of the chain will be mx/l. The length of hanging part of the chain is always constant and equal to h. This hanging part of the chain is responsible to pulling the rest of chain downwards. The force of gravity acting on the chain at any instant is given by mgh/l. Let velocity of the chain at some instant be v. So we have,





Now v is also the rate of decrease of length of the remaining part of the chain and so,
















Note that the upper limit of x was chosen as h since that will be the remaining part of the chain when its just slipping out of the tube.

Irodov Solution 1.183

If there were no force acting, as the sand falls momentum would be conserved. The force however acts on both the sand and the cart to change the momentum of the (cart + sand) system. The falling sand retards the cart by increasing its mass while the force accelerates it.
Suppose that at some instant of time t the combined mass of the sand and the cart is m and its velocity is v. The mass m is given by ,

Force is rate of change of momentum and so we have,

Irodov Solution 1.182

To answer some questions and clarifications I decided to provide a bit more explanation to the solution of this problem. I will solve this problem using the basic Newton's laws .

Consider two instants of time t and t + dt, where dt is an infinitesimally small interval of time. Suppose that the mass of the cart at time t was m and that at time t + dt was m-dm. Further suppose that the speed of the cart at time t was v and that at t + dt was v + dv.

The momentum of the cart at time t was mv. The momentum of the cart at time t+dt was (m-dm)(v+dv). Now the dm mass of sand that left the cart in the dt interval would be still moving in the horizontal direction with a speed v. Hence the momentum of the sand that fell off the cart is given by vdm. The total momentum of the (sand + cart) system at time t+dt is sum of the momentum of the cart and the momentum of the sand that fell off which is given by (m-dm)(v+dv) + vdm. The change in momentum of the (cart + sand) system due to the force on the (cart + sand) system during this interval is given by Fdt. Hence, we have,







Basically, as the sand falls out, the force F has to pull a progressively lighter cart and so the acceleration of the cart increases with time. The mass of the cart varies as . So we have,














Some students have asked as to why we do not see the term as we saw in problem 1.181. I hope that the reason is clear from the initial explanation above. Basically, the force F does not act on the dm part of sand that has left the cart and so its momentum remains intact. When unsure please use basic principles and you will not go wrong.