Roller coaster cart gaining weight into a loop

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A roller-coaster car starts from a height twice that of a loop and loses 25% of its speed when a 100 lb bag of bricks falls on it just before entering the loop. The potential energy at the starting height is calculated as 39.24Rm, which equals the kinetic energy at the bottom of the loop. After the bag falls, the car's kinetic energy is reduced to 9/16 of what it would have been without the added weight. The discussion revolves around whether this reduced energy is sufficient for the car to make it over the loop. The conclusion hinges on comparing the effective height the loaded car can achieve against the loop's height.
almightyfoon
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Homework Statement


A roller-coaster car starts from rest at a high point 2 times as high as the top of the loop-the-loop. When it is just about to enter the loop from the bottom, a 100 lb bag of bricks fall on the car, causing it to lose 25% of its speed. Neglecting friction, do you think the car will make it over the top of the loop wihtout falling off?
H=4R

Homework Equations


PE=mgh
KE=(1/2)mv^2
v=rω
a=rα


The Attempt at a Solution


m= mass of the cart
R= radius of the loop

I figured out the potential energy = 39.24Rm (9.81*4R*m)

Which means the KE at the bottom of the loop is the same thing. However the mass of the cart goes up by 100 lb (45.6 kg roughly) once the cart hits the loop. SO the KE=(1/2)(m+100)v^2.

I can solve for either of the variables, but that's a lot of algebra to do, and I don't really think its necessary. So... people of physics forum's, I have come from my lurking to beg for your help. This is the last problem of my homework set that's due at 9 AM.
 
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almightyfoon said:

Homework Statement


A roller-coaster car starts from rest at a high point 2 times as high as the top of the loop-the-loop. When it is just about to enter the loop from the bottom, a 100 lb bag of bricks fall on the car, causing it to lose 25% of its speed. Neglecting friction, do you think the car will make it over the top of the loop wihtout falling off?
H=4R

Homework Equations


PE=mgh
KE=(1/2)mv^2
v=rω
a=rα


The Attempt at a Solution


m= mass of the cart
R= radius of the loop

I figured out the potential energy = 39.24Rm (9.81*4R*m)

Which means the KE at the bottom of the loop is the same thing. However the mass of the cart goes up by 100 lb (45.6 kg roughly) once the cart hits the loop. SO the KE=(1/2)(m+100)v^2.

I can solve for either of the variables, but that's a lot of algebra to do, and I don't really think its necessary. So... people of physics forum's, I have come from my lurking to beg for your help. This is the last problem of my homework set that's due at 9 AM.

We are now asking the cart plus sack to go through the loop.

The velocity has reduced to 3/4 of what it would have been had the sack been added before coming from the start.
if v is only 3/4, then the KE is only 9/16 - slightly more than half.
Relating KE at the bottom to PE at the start, as you did, you can find the height the loaded car has effectively started at.

Is that high enough for the car to make it through?
 
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