Answer:
A=1
B=-2
Explanation:
Part A and B of the question wasn't given, however, I attached the relevant parts to solve this question as follows.
From part B as attached, it shows that the right option is C which is
2A+3B=-4
Substituting B with 3A-5 then we form the second equation as shown
2A+3(3A-5)=-4
By simplifying the above equation, we obtain
2A+9A-15=-4
Re-arranging, then
11A=-4+15
Finally
11A=11
A=1
To obtain B, we already know that 3A-5 so substituting the value of A into the above then we obtain
B=3(1)-5=-2
Therefore, required values are 1 and -2
Answer:
a) m=20000Kg
b) v=0.214m/s
Explanation:
We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.
For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is,
, were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as
, and the mass of the first and second coals as
and
respectively
We start with the transition between parts A and B, so we have:

Which means

And since we want the mass of the first coal thrown (
) we do:



Substituting values we obtain

For the transition between parts B and C, we can write:

Which means

Since we want the new final speed of the car (
) we do:

Substituting values we obtain
