Question

A 675 kg train car coasts slowly (v = 2.4m/sec) under a hopper that fills it with coal. If the car's velocity slows to 1 m/sec, what is the combined mass of the train car and the coal?

Answer 1

The rate of coal entering from the hopper is not given?

Answer 2

No. Is the question broken?

Answer 3

maybe not :)

Answer 4

I think the hopper is irrelevant. So the question would be:

Answer 5

I think we have to assume the addition of coal slows down the train.

Answer 6

The train car weighs 675 kg and moves at a velocity of 2.4m/sec while empty, and with the addition of coal slows down to 1 m/sec.

Answer 7

Is that right?

Answer 8

correct, so im thinking we can use momentum here somehow

Answer 9

can we assume conservation of momentum. in which case we have
p1 = p2
m1*v1 = m2*v2
675 * 2.4 = m2 * 1

Answer 10

which evaluates to 1620 kg.
that seems too easy though

Answer 11

Hmmm. It does, doesn't it?

Answer 12

we should add some more details
before the coal enters the train, we have a system
of two masses. the train and the coal. the train moves at velocity v1 = 2.4
the coal has velocity zero (before it enters the train).
After the coal gets loaded we have a combined mass m1 + m2
which moves at velocity v final

p1 before = p2 after
m1v1 + m2 v2 = (m1 + m2) * v
675 * 2.4 + m2 * 0 = (675 + m2) * 1

from which we derive m1 + m2 = 1620 kg

so that is a better solution

Answer 13

That makes more sense. Sorry for the delay, I got distracted for a bit.