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Why isn't the force 1N times the gravitational constant?

  1. Feb 22, 2004 #1
    For a fabrication (*sniff*), I certainly am having trouble in a very real way. These are some upcoming test questions. I don't understand the answers. *sniff*

    Q. An astronaut in outer space wants to play a solitary game of "throw, bounce, catch" by tossing a ball as massive as he is against a massive and perfectly elastic wall. What will occur?
    A. the astronaut will never catch the first bounce.

    I don't understand why he won't catch it at least once. Because it bounced? It knocks him down? Or he flies away backwards and it won't ever catch up?
    Q. A 1-N apple falls to the ground. The apple hits the ground with an impact force of about:
    A. 1 N.

    Why isn't the force 1N times the gravitational constant?
    Q. A golf ball moves forward with 1 unit of momentum, strikes and bounces off a bowling ball that is initally at rest and free to move. The bowling ball is set in motion with a momentum of:
    A. More than 1 unit.

    I thought momentum was conserved. Why is the bowling ball getting more?

    Thanking anyone in advance.
  2. jcsd
  3. Feb 22, 2004 #2
    Here's a question: if an astronaut is in space and he has a ball with exactly the same mass as him and he throws the ball, how fast is he going and how fast is the ball going?

    A newton is a unit of weight (force), not mass.

    After the golf ball hits the bowling ball, which direction is the golf ball going? Does it still contribute any momentum?

  4. Feb 23, 2004 #3
    To understand the first question, you need to understand the concepts of Newton's Third Law: Whenever one body exerts a force on a second body, the second body exerts an oppositely directed force of equal magnitude on the first body.

    When the astronaut exerts a force on the ball at the wall, the ball exerts a force back on the astronaut equal in magnitude but opposite in direction. So, the ball travels towards the wall at some velocity while the astronaut travels in the opposite direction. When the ball rebounds, it is traveling at the same velocity (speed and direction) as the astronaut and will never catch up.

    The key here is that the ball's mass is equal to that of the astronaut.

    F=ma and mball = mastronaut
    Fball = -Fastronaut
    mballaball = -(mastronautaastronaut)
    aball = - aastronaut

    Second Question:

    The weight of an object is the gravitational force exerted on that object by an astronomical body. This force is always pointed downward towards the center of the body.

    Since weight is equal to the force exerted, Weight = F = ma.

    Third question

    I don't agree with the answer and if anyone disagrees, please post.

    Assuming no other forces act on the balls and the collision is linear, then the momentum is conserved. Momentum conservation occurs whether or not the collision is elastic.

    m v = p momentum defined
    p0 = pf
    mgvog + mbv0b= mgvfg + mbvfb

    Since the initial velocity of the bowling ball (b) is 0 the equation can be simplified:

    mgvog = mgvfg + mbvfb

    Rearrange the equation to solve for the final momentum of the bowling ball:

    mbvfb = mgvog - mgvfg

    Since the bowling ball does not add any intial momentum to the system, the final momentum of the bowling ball will always be less than the intial momentum of the golf ball.

    The answer should be less than 1 unit, not more than 1 unit.
    Last edited: Feb 23, 2004
  5. Feb 23, 2004 #4
    Thank you both very much for the answers. The bowling ball/golf ball still has me confused a bit, but the other two are now perfectly clear. Thanks again.
  6. Feb 23, 2004 #5
    The bowling ball has more than 1 unit of momentum.

    What I was getting at by asking what the golf ball was doing is to point out that after it contacts the bowling ball, it's going in the opposite direction. The golf ball plus bowling ball system will have 1 unit of momentum, but the golf ball's momentum is negative (consider the -v term, since m is always positive), so the bowling ball's must be greater than 1 to counter the negative momentum.

    What happens:

    Golf ball comes to the right (arbitrarily selected direction). It contacts a bowling ball at rest. After the collision, the bowling ball is traveling right and the golf ball is traveling left.

    However, since momentum is conserved and the golf ball has "negative" velocity, the equation can be shown to be

    [tex]-|p_{\textrm{golf ball}}| + |p_{\textrm{bowling ball}}| = p_{\textrm{initial}}[/tex]

    From this, it's pretty easy to see that the bowling ball's velocity will be greater than 1 by whatever magnitude of momentum the golf ball has after the collision.

  7. Feb 23, 2004 #6


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    Re: Momentum

    From where did you get this Q/A? This looks like a bit of unconventional termonology that could be interpretted to yield a result different from 1 N. I think that the others (cookie and game) gave good explanations consistent with the Q/A indicated, but you have my sympathy that this seems to be a rather poorly worded question.

    To qualify, I would consider the question: "Would the impact force of the apple incident at 100 mph on a wall in outer space be zero (negligibly small)?" That doesn't seem correct, but it seems to be the (unreasonable) underlying assumption (that is, that the "impact force" is being defined as the force on the apple immediately before it makes contact, which almost seems contradictory).
  8. Feb 23, 2004 #7
    These questions and answers are from our text. It's a horrible text. To wit, the author is grinding this axe for several pages: You feel weightless, you are weightless. Falling down in an elevator and your feet aren't touching the floor? You're weightless. In outer space, riding around in the Shuttle? You're weightless. In the space-station, trying to stuff a sock in a hole? Weightless again. He defines weight as "the weight of something is the force it exerts against a supporting floor or a weighing scale."

    That just ain't right. That just can't be right. Can it? I have trouble enough when things are explained clearly. I'm just plain doomed with this text.

    Thx Cookie for the clarification on the bowling ball & golf ball. Thx turin for the apple help.
  9. Feb 23, 2004 #8


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    Homework Helper

    Well, it sounds like the authors of your text are neglecting their target audience (or it was poorly chosen for your course). The axe grinding that you mention here is called the "Equivalence Principle." Albeit technically correct (except for that sock thing, I cannot comment on that), it further supports the ambiguity of the apple impulse force question. Again, I sympathize. Good luck with the course, and good for you for taking the initiative (by using this HW forum).
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