Why Does Calculating the Speed of a Runaway Cable Car Involve Both Masses?

[mullets1200]

Homework Statement

The 1990 kg cable car shown in the figure descends a 200-m-high hill. In addition to its brakes, the cable car controls its speed by pulling an 1780 kg counterweight up the other side of the hill. The rolling friction of both the cable car and the counterweight are negligible.

The trangle that this is on is a 30-20-130 triangle. The counter weight is on the 20 degree side while the cable car is on the 30 degree side

1) How much braking force does the cable car need to descend at constant speed?
3780 N

2) One day the brakes fail just as the cable car leaves the top on its downward journey. What is the runaway car's speed at the bottom of the hill?

The Attempt at a Solution

for part 1 i did:
1990*g*sin(30)-1780*g*sin(20)
3780N which was correct

for part 2 i did:
The height dropped on the AB side is 200m
The height gained on the BC side is
200/L=sin(30)
L=200/sin(30)
hbc*sin(30)/200=sin(20)
hbc=200*sin(20)/sin(30)
137 m

m*g*(200-137)=.5*m*v^2
v=sqrt(2*g*63)
35 m/s but it was wrong.

any help?

 

[willem2]

m*g*(200-137)=.5*m*v^2

any help?

The cable car and the counter weigth have a different mass, so m*g*(200-137) is
wrong.
You also need to account for the increase in kinect energy of both the car and
the counterweight.