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What Am I Doing Wrong?

  1. Mar 13, 2005 #1
    I have a few questions that I got wrong, and I'm trying to find out what I did. All help is appreciated.

    1. A cube of aluminum has side length .10 m. It is dropped into a deep swimming pool of water withd ensity 1.00x10^3 kg/m^3. The cube comes gently to rest at the bottom of the pool. If the density of aluminum is 2.7x10^3 kg/m^3, the magnitude of the bouyant force acting on the aluminum is ________.
    The answer is 9.8 N, but I keep getting 2.7 by multiplying (1.00x10^3) by(2.7x10^3). Is there a certain formula I must use?

    2. A cement truck of mass 16,000 kg moving at 15 m/s slams into a cement wall and comes to a halt. What is the force of impact on the truck?
    Is this question just asking me for the force? F=ma, so I got F=(16,000 kg)(15 m/s). But, that was wrong on my paper... What exactly is it asking me for? Those are the only two variables stated in there, so what else could I use?

    3. A 40-kg footbal player leaps through the air to colide with and tackle a 60-kg player heading toward him, also in the air. If the 40-kg player is heading to the right at 7 .0 m/s, and the 60-kg player is heading toward the left at 3.0 m/s, what is the speed and direction of the tangled players?
    The vectors would look like -------> <---. The equation I use was:
    (40 kg)(7 m/s) + (60 kg)(3 m/s) = (40 + 60 kg)(final velocity) =
    final velocity = 4.6 meters per second. That was counted wrong, and I don't know "what direction" the players are in... How would you know?

    And lastly,

    4. A 5.0-kg blob of clay moving horizontally at 2.0 m/s has a head-on collision with a 6.0-kg blob of clay that moves toward it at 1.0 m/s. What is the speed of the two blobs stuck together immediately after the collision?
    I used the same equation as above and got 1.5 m/s. But, it's also wrong.

    Are there two types of equations, one for where one of the forces is not moving, and another for where both forces are moving? Is that what is causing the problems? If so, what is the other equation? Oh, and thanks for all help in advance.
  2. jcsd
  3. Mar 13, 2005 #2
    alright. The problem is that velocity is magnitude and direction (a vector) That means the velocity can tell you speed as well as direction.

    For example... #3....
    (40 kg)(7 m/s) + (60 kg)(-3 m/s) = (40 + 60 kg)(final velocity)
    v final=1m/s

    in this case, the 60kg football player is heading in the opposite direction of the 40 kg player so it has a negative velocity. Positive velocity usually means going to the right. Negative usually means going to the left. You can apply this to many of your problems.

    for #4:
  4. Mar 13, 2005 #3
    Oh, thanks. I forgot to use the negative sign for force heading in the opposite direction. Thanks again for that help.
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