An open-top railroad car (initially empty and of mass Mo) rolls with negligible friction along a straight horizontal track and passes under the spout of a sand conveyor. When the car is under the conveyor, sand is dispensed from the conveyor in a narrow stream a steady rate Change(M)/Change(T)=C and falls vertically from an average height h above the floor of the railroad car. The car has an initial speed Vo and sand is filling it from time t=0 to t=T.
1. Determine the mass of M of the car plus the sand that it catches as a funtion of time t for 0

Question

An open-top railroad car (initially empty and of mass Mo) rolls with negligible friction along a straight horizontal track and passes under the spout of a sand conveyor. When the car is under the conveyor, sand is dispensed from the conveyor in a narrow stream a steady rate Change(M)/Change(T)=C and falls vertically from an average height h above the floor of the railroad car. The car has an initial speed Vo and sand is filling it from time t=0 to t=T. 
1. Determine the mass of M of the car plus the sand that it catches as a funtion of time t for 0<t<T
2. Determine the speed v of the car as

anonymous · Answer

The first part is straightforward,  M=Mo+Ct

Now consider the situation at time t. Mass of the car is Mo+Ct and let its velocity is u.
An additional amount of mass dm falls in time time dt.
We focus on the mass dm. Its initial velocity in direction of motion of cart is zero. And final velocity is u. So change in its momentum is dm(u-0)=udm
Rate of change of momentum= udm/dt=Cu
This is the force on the sand of mass dm. Force on the cart of mass Mo+Ct will be -Cu .

Write the eqn of motion as   (Mo+Ct)du/dt=-Cu

Solve the D.E. and use the initial condition t=0,u=Vo