Question

A 4100 kg open railroad car coasts along with a constant speed of 9.90 m/s on a level track. Snow begins to fall vertically and fills the car at a rate of 5.00 kg/min . Ignoring friction with the tracks, what is the speed of the car after 80.0 min ?

Answer 1

As the snow falls into the car, the mass of the car is increasing. This is similar to colliding with one 5.00kg object at rest, each minute. Since the snow falls into the car, the total mass after the collision equals the mass of the car plus the mass of the snow.

Momentum is always conserved.
Initial Momentum of car, before colliding with snow = 4100 * 9.90
Momentum of car after 1 minute of snow fall = (4100 + 9.90) * v
4100 * 9.90 = (4100 + 5.00) * v1
v1 = (4100 * 9.90) ÷ (4100 + 5.00) =

v1 = 9.90 * [4100 ÷ (4100+ 5.00)]
This is velocity at the end of the first minute.
AND
This is the velocity at the beginning of the 2nd minute.
Momentum at beginning of 2nd minute = (4100 + 5.00) * 9.90 * [4100 ÷ (4100 + 5.00)]
Momentum at end of 2nd minute = (4100 + 2 * 5.00) * v2
(4100 + 2 * 5.00) * v2 = (4100 +5.00) * 9.90 * [4100 ÷ (4100 + 5.00)]
v2 = (4100 + 5.00) * 9.00 * [4100 ÷ (4100 + 5.00)] ÷ (4100 + 2 * 5.00)

v2 = 9.90 * (4100 + 5.00) * [4100 ÷ (4100 +5.00)] ÷ (4100 + 2 * 5.00)
v2 = 9.90 * 4100 ÷ (4100 + 2 * 5.00)
This is the velocity at the end of the 2nd minute.

The equation for the velocity at the end of n minutes:
vn = 9.90 * 4100 ÷ (4100 + n * 5.00)
The velocity at end of 80minutes = (9.90 * 4100) ÷ (4100 + 80 *5.00)
The velocity at end of 80 minutes = 9.02 m/s