Let me state from the start that I am not looking for help answering my daughter’s homwork problems. I’m looking more for a reality check.
My daughter asked me to help her with her 8th-grade physics, in preparation for a test. The teacher handed out some practice questions, and she was having a hard time reconciling her answers with those on the handout. I took a look, and was equally flummoxed. Even though I graduated from the trade school just down the river from Harvard, the “correct” answers and their explanations have me stumped.
I am going to repeat the original questions, verbatim, and would like to hear your answers and reasoning to see if it jibes with mine, or if I am indeed, dumb as a stump.
Question 1 :
Question 2:
Question 3 ( a two-parter):
If I knew how, I’d give the teacher’s answers in a spoiler box. Since I can’t, I’ll hold off for the time being
If we assume that each ball has enough distance to reach terminal velocity (the speed where the drag force equals weight), the iron ball will experience more force because it takes more force to balance its weight.
But for the same sized balls, wind resistance (drag) depends only on velocity. So during the early phase of their descent, when they’re traveling at virtually the same speed, they would experience the same force from the air.
But then both balls accelerate until each hits its terminal velocity. At that point, we already know the drag on the iron ball will be greater.
Answer, it depends on what phase of the descent you’re worried about.
Question 2:
Her acceleration decreases. Her downward acceleration is at a maximum (equal to gas soon as she leaves the plane, when she has zero downward velocity. Her acceleration keeps approaching zero until she hits terminal velocity, when her velocity remains constant, and her acceleration is therefore zero.
Question 3:
A) equal to P (because of reasoning in part b… net force equals zero, so *f ** and P must be equal to cancel each other out)
B) Equal to zero. All you neet to know is that the box has constant velocity. Therefore acceleration equals 0, and so does Force by F=ma.
Of course, this assumes P is opposite in direction from f. If there is a component of P perpendicular to the ground, that part will be cancelled by the weight of the box or the normal force applied on it by the ground. If so, it would technically be correct to say that the total magnitude of P is greater than f
For part 1, I’d note that technically, as soon as the steel ball is moving even just a little faster than the wooden ball, it’s likewise encountering more air resistance, notwithstanding that it’s still moving faster.
Yeah, the first question is ill-posed — it depends on whether the question is trying to ask “which ball will experience a greater air resistance when it’s at a given velocity?” (answer: the force is the same), or “which ball will experience the largest momentum change (i.e. integral of force w.r.t. time) due to air resistance on its way down?” (answer: hmm, I’ll have to think about that one.) I would wager that the teacher’s trying to ask the first question, since it’s much simpler; but as it stands it’s not crystal-clear what’s being asked.
Q1: C - both the same. I think. Same size, same shape, same air resistance. Or at least, that’s the answer I’m sticking with.
Q2: C - remains the same. Her acceleration = gravity and gravity remains constant.
Q3: part a - f is very slightly less than P, P needs to be greater than F for it to move.
part b - once the thing is moving, net force=0 because it’s pulling at a constant velocity, no accelleration means no force. But from standstill to movement, it’s greater than 0.
Q1) The velocity and acceleration of the spheres would be identical in a vacuum. Air resistance is proportional to cross-section and velocity squared. A quick shortcut to the answer is to note that terminal velocity increases with density. At no time will a lighter sphere fall more quickly than a denser sphere of ientical dimension. They both start at zero velocity, and the denser sphere will accelerate slightly faster, and reach a higher erminal velocity because the two spheres would experience identical air resistance at any given speed, but that air resistance would a smaller fraction of the gravitational force acting on the denser sphere than the lighter sphere.
Answer B) Iron sphere falls faster
Q2) In a vacuum, a falling parachutist would experience a constant acceleration of 1 G (32 ft/sec[sup]2[/sup] or 9.8 m/sec [sup]2[/sup])due to the constant force of gravity (actually increasing very slightly as she approached the Earth due to the inverse square law). In air, however, air resistance will increase with the square of the velocity as she falls, meaning that there will be an increasing force counteracting the constant gravitational force, and her acceleration will be steadily decreasing. Guess what? That’s how parachutes work: the air resistance (which increases with velocity) balances against the gravitational force (which is independent of velocity) to slow your initially rapid free-fall to a survivable velocity. Above that terminal velocity the force of air resistance is higher than gravity, and you decelerate. Below terminal velocity, the force of air resistance is less than gravity (but increasing) so your velocity increases, but your acceleration is actually decreasing. After all, AT terminal velocity, you have a constant spped, and therefore zero acceleration
Answer B) acceleration decreases (until she achieves terminal velocity and zero acceleration)
Q3) Newton’s law says that an object in motion will remain in the same linear motion unless acted on by an external force. Since the box is travelling in a straight line at a constant velocity, all forces on it must exactly cancel out (no net external force). Therefore, ignoring small forces like air resistance:
ANSWER [a]: P equals F
ANSWER **: Net force equals zero (i.e. steady state motion)
The problem with question three is that P varies. When the box isn’t moving, P<f, obviously. When it starts to move, it accelerates, so P>f. But then, P has to decrease to P=f in order to make it move at a constant velocity.
While I agree that P>f when the box is accelerating, the part I bolded is incorrect.
You may prove this in many ways:
Newton: an object at rest will remain at rest unless acted on by a [net] external force. A box remaining at rest MUST have zero net force action on it. so P=f
You stated that that P=f at a constant velocity. This means that P=f when the box isn’t moving, because that is a constant velocity (of zero)
What is the direction of the excess Force of f? Under F=ma, the box must move in that direction -there’s nothing to cancel it out. Yet the box is immobile
Friction CAN’T be greater than the applied force, though a given frictional condition may be capable of canceling more applied force. When you jam the brakes on your car, they may be capable of opposing more force than when you tap them, but if slam your brakes when you’re already stopped, it has no additional effect.
This is a common misconception regarding ‘resisting’ forces: just because a situation is capable of generating a certain maximum resisting force, doesn’t mean it IS APPLYING that amount of force. Friction can only oppose; it can never exceed what it opposes.
It is somewhat easier to see in the case of a 1-ton box on a floor: the floor can’t exert more force on the box than the box exerts on the floor. [Equal and opposite reaction] The floor may be capable of supporting 10 tons if you crank up the artificial gravity, but at any time it only exerts force equal to the box’s weight (if it exerted more, the box would fly up in the air, a effect floors rarely produce)
b, iron ball. The iron ball will fall a bit faster than the wooden ball, so it will “encounter the greater force of air resistance”
b, decreases. Her acceleration starts at g, but decreases because of air resistance. The key is “as she falls faster”–which removes the time when she reaches terminal velocity from consideration.
3a) f is less than P. By a small amount, because P must be greater than f in order for the crate to move. The pull P will not be the same as the force that causes the box to move with constant velocity across the floor, since dynamic friction is usually less.
3b) Net force is equal to zero. The crate is moving with a constant velocity.
Hmm. I see that I disagree with Shagnasty’s 1, 3b; aerodave’s 3a (and 1 since he covered all bases); Civil Guy’s 3a since he agreed with aerodave (but not 1 since he narrowed down aerodave’s response); MikeS’s 1 (since he seems to prefer a different interpretation); amarinth’s 1, 2; and KP’s 3a. That means I disagree with everybody, especially on 3a. Perhaps the teacher did intend the answer to be f=P, since the difference could be as small as we want, in theory. On the other hand, aerodave’s comment that P doesn’t have to be parallel to the floor would make P strictly greater than f, so I’ll stick with my answers.
Thanks to all for your replies. It’s heartening to see that there is some divergence in the answers from people I consider to be the brightest.
For the record; here are the answers as provided by the teacher:
Q1: B. “Air resistance depends on both the size and speed of a falling object. Both balls have the same size, but the heavier iron ball falls faster through the air and encounters more air resistance in its fall”
Q2: B. “Her acceleration decreases because the net force on her decreases. net force is equal to her weight minus her air resistance, and since air resistance increases with increasing speed, net force, and hence acceleration decrease.”
Q3A: Friction is equal to P. (no explanation given)
Q3B: Net force on the Crate is equal to zero. (no explanation given).
I agree that Q 1 and 3 are very poorly stated.
Thanks again for the reality check.
1.) air resistant is dependent on size, shape, surface texture and speed. The first three are the same for the two ball. The fourth is slightly more complex. If both balls are travelling and 10m/s, then the upward force of the air resistance is the same, but the downward force is different, causing them to accelerate at different rates, which means that very soon, the speeds will be different, and so will the air resistance.
2.) Decreases. At first, your acceleration is -9.8 m/s[sup]2[/sup]As you fall in any fluid (including air) you are accelerating towards your terminal velocity, at which point, your acceleration is 0.
3.) F=ma. If a = 0, then net force = 0, therefore friction f = force P
I don’t like question 3 at all. First off, it suggests that P does not change once the box is moving. P must be greater than f in order to accelerate the box above 0 velocity. If P is greater than f when the box is motionless, it will be greater than f when the box is moving, and the box will accelerate.
At the point at which the box begins to move, P>f and there is a net force on the box, if it weren’t, the box would just sit there.
At the point where the box is sliding across the floor with constant velocity, P=f and there is no net force on the box otherwise the box would accelerate.
Question 1 is somewhat poorly stated, but there is only one point at which the force of air resistance will be the same for wood and iron, at the very start of the drop, when the force is 0. After that point, the iron ball will accelerate downward faster and encounter greater air resistance forces.
KP seems to have nailed the answers. Looking back over the posts, I’d imagine anyone who was in the class and paid attention should be able to answer the questions easily.
I disagree that question 1 is poorly stated. That one point at which the air resistance is the same is disallowed by the wording of the problem: “on the way down”.
It mentions that P is slight then increasing until the box moves, but doesn’t mention P at any point thereafter. We are then expected to make statements about P, presuming that P has been adjusted downward to finally equal f. Asking questions about a force that has at least 3 different values without clearly defining when to make the comparison is confusing.
So too with the 1st one, though it’s somewhat less ambiguity, one can state that at the point of release, the balls are “on the way down”. It takes little additional effort to more clearly define when you want the comparison made. If I take the wooden ball at time T+5 and compare that to the iron ball at T+1, then the wooden ball likely has the higher force. Is it a valid comparison? They are both “on the way down” at the time, so why wouldn’t it be?
Questions should be designed to test your knowledge of the subject, not your ability to decipher questions and make the right assumptions.
Not explicitly, but it does say " It is pulled so that it moves with constant velocity across the floor." That’s good enough.
Plus, f itself will decrease significantly. Dynamic friction is usually a lot less than static friction.
The question asks “The ball to encounter the greater force of air resistance on the way down is the: ??”
As near as I can tell, no matter whether you interpret that as a comparison at a particular time T, or as a cumulative, maximum force encountered over the entire trip, the answer is still the same.
Question 1 is stooopid. It makes a bad assumption about the mass of the respective balls – almost as bad as “which is heavier, a pound of feather or a pound of lead?” If the balls are the same size, then either the wood is extraordinarily dense, or the iron ball is hollow. If the iron ball is hollow, then its shell can be arbitrarily thin. I could easily imagine a hollow iron ball with a skin as thin as that of a Christmas tree ornament and a wooden ball made out of something dense like Lignum Vitae. Or I could fiddle about until the two balls had equivalent mass.
Whoops – slight correction. It isn’t stated that they have the same mass, so the above statement is wrong. However, it’s still possible for the iron ball to be hollow and mass less than the wooden ball.
With air resistance in play (parachutist, remember), there’s a term proportional to the negative of velocity. Thus the faster the body is moving down, the more upward acceleration from resistance counteracts the force of gravity.