Wow, its a sad day when you can't sleep because all you can think about is how to explain a question without audio or visual to a forum. Gotta get this down so I can sleep.
😴
Oh boy, where to begin. Might not be pleasant.
This is a horrendously incorrect statement. You are looking at this problem like it is a perfect elastic conservation of momentum simplification and that is not what is asked by the question.
If you drop the package
it will fall accelerating at 9.8m/s^2 regardless of weight. The balloon system will begin rising at half that speed of 4.9m/s^2 because it was previously in equilibrum and now half of it's mass is removed but it's volume remains constant so the buoyant force will accelerate the balloon at half the acceleration of gravity.
If you need a more obviously slap to wake you up ,,,
/facepalm
Ok, basic kinematics...
MCAT expects us to know that if you use a catapult to launch a rock through the air it will have some positive initial velocity. T=0; V=Positive... So your comparison of a gun and nanoseconds of time does not make sense. That is like asking what is happening while the catapult is undergoing the launch motion. It doesn't matter because that is all before T=0. Same with a gun, Time doesn't start until the bullet would leave the barrel and begin it's Newtonian trajectory.
MCAT also expect you to know that when you drop an object from rest THE INITIAL VELOCITY IS ZERO. I have no idea how you can try and make an argument otherwise. The "instant" (gerrr) that the objects is dropped Time=0 and Velocity=0.
For the sake of reputation don't try and argue anything contrary to the very basic physics kinematics MCAT problems. This is common knowledge for most people reading this forum.
Semantics maybe but you shouldn't try and say that the instant you drop a ball it has a velocity. If that were true all the equations we learn and memorize for the exam would be wrong because they assume T=0.
In my previous post I actually said "you can't argue this" assuming nobody would, because it is so fundamental to basic physics motion problems.
Again, this is where you are missing basic MCAT reasoning. For example when you throw a ball up into the air, at it's highest point what is the velocity? Zero.
Yes it will immediately begin to increase but if you only look at the latter half of that kinematics problem you will see that at initial max height (drop scenario), the velocity is zero despite being subjected to gravitational acceleration.
For some reason it seems like you are rounding and unable to believe things happen concurrently.
You release the ball, gravity begins accelerating the ball, and the ball has zero velocity simultaneously. There is no NEED for any time to pass. It all happens in an "instant"... Maybe you can argue that
"in the real world" gravitational waves propagate at the speed of light so there is an infinitesimal time delay but COME ON!~ This is only the MCAT.
I don't think you understand the concept of "initial" or the concept of when Time = Zero.
Perhaps you are thinking that no dropping mechanism is capable of releasing an object quick enough to instantly release it before it starts accelerating. MCAT writers have amazing technology that is up to the challenge and you should assume that the instant after release, no time has passed and velocity is zero.
To respond to this I'm just going to quote
@Chrisz because it was probably the most concise and "MCAT accurate" post in the thread.
I really hope anyone reading this understands that when you drop something the initial velocity is zero. It is fundamental to scoring high marks on the MCAT.
Lastly at T=0; Vi=Vo; so we only have to interpret what is considered "the system".
If you interpret the balloon at it's own system as was suggested earlier then momentum is unchanged because mass and velocity do not change.
If you interpret the package/balloon system together and that mass is lost (as stated in the answer key), then Velocity is unchanged and mass is decreased. (p=M*V) so P decreases.
What the book is trying to do is say that: (1), the package/balloon system was together and as you release the package the balloon loses mass... And (2) you should
use the same mass when calculating the momentum before and after release?? No.
Those two things inherently disagree with one another. The only way for the balloon to accelerate up is to decrease it's mass but the argument is that initial momentum is ([Mpackage+Mballoon]*Velocity), and final momentum is still ([Mpackage+Mballon]*Velocity) but this second calculation ignores the decrease in mass that was the cause of the initial acceleration.
Lastly, my previous post addressed the issue that
@BerkReviewTeach was trying give by saying that if you lose mass, some
near-infinitely small amount of time must have passed.. OK Sure, even with that belief a
near-infinitely small velocity change has occurred, while a very measurable change in mass results in the net change in momentum of the balloon after release to have decreased.
Simple proof with ridiculous numbers. Say the balloon was "super"-helium and floating up at a constant 98 meters per second. You then drop HALF your weight. How long would it take before the acceleration due to buoyancy allows the momentum of the balloon after dropping to equal the momentum before dropping.
ANSWER: 20 SECONDS, how is that instantly? Sure eventually the acceleration would allow the balloons velocity to become the more important factor but for the first 20 seconds the loss in mass is more important. This scales down to realistic numbers but the bottom line is NO MATTER HOW SLOW THE BALLOON WAS INITIALLY TRAVELING, THE LOSS IN MASS IS ALWAYS GOING TO BE THE FIRST THING TO HIT MOMENTUM. done.
Regardless of if you believe no time has passed, or some almost infinitely small time has passed, either way, around the "instant" of release, Momentum initially decreases by dropping weight.
@BerkReviewTeach No disrespect intended. I hope the forum remains a place where interpretation can be challenged and discussed for overall improved understanding. I will try not to take offence to your allegations of misinformation. You should do the same.
The question states that the package is "released", as in let go with no force exerted. So it does not have momentum downward. In fact the instant before release the balloon and package were moving at a constant velocity up, so it reasons that at the instant of release the package is still moving up and therefore, the momentum of the package at release is UP!. Again answer is wrong, wow.
