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Various Physics 12 problems

  1. Jun 25, 2005 #1
    EDIT: 1 question left

    http://i2.photobucket.com/albums/y15/seiferseph/4.jpg

    tried a couple different things, nothing worked. i'm not sure what to do with the force at the top, where to take the torques, how to use the friction (which way does the force go?) and which angle to use for the friction.

    thanks!
     
    Last edited: Jun 26, 2005
  2. jcsd
  3. Jun 25, 2005 #2
    The first: Yes, momentum is conserved. The guys do go away at different velocities. You want to find the speed of separation (ie. the relative speed). Because they'll be moving away from each other, just add their speeds for their speed of separation. Try splitting the problem into two halves, because there are two collisions happening.

    The second: Energy is conserved. When the weight is on the ground, it has no bulk energy. Because the graph is F against x, you can use the area under the graph to find the work done by the lifting. At the top, the weight will have grav. potential energy, as well as kinetic energy. So Fx=mgh+0.5mv^2
     
  4. Jun 25, 2005 #3
    i get them both now, thanks!
     
  5. Jun 26, 2005 #4
    I have a couple more problems

    http://i2.photobucket.com/albums/y15/seiferseph/6.jpg

    I solved this one (got the right answer) using the fact that Fn = 0 at the top, so Fc = mg. is this right? Is that what i means by "does not fall off the track"?

    http://i2.photobucket.com/albums/y15/seiferseph/7.jpg

    I solved this problem a couple months ago, but can't now. i use the fact that kinetic energy is converted to potential enegry, and i try to solve for the potential at the end. i use 1/2mv^2 = qV and solve for V, but i get a really small number. what am i doing wrong?

    http://i2.photobucket.com/albums/y15/seiferseph/8.jpg

    not sure what to do with the velocity, power = W/t, right? so how do i get time?

    http://i2.photobucket.com/albums/y15/seiferseph/9.jpg

    using m*g*h, why do you use the height of 9 m? that gets the right answer.

    http://i2.photobucket.com/albums/y15/seiferseph/10.jpg

    really not sure where any of this comes from, what equations/laws? :confused:

    Thanks again!
     
  6. Jun 26, 2005 #5

    OlderDan

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    You are good on the first one. When the normal force just goes to zero, the centripetal acceleration required for circular motion is being provided by gravity.

    Your approach looks correct for the second one. Are you using negative exponents on your powers of 10. What is the charge of a proton?

    For the third one, remember that work is force times distance moved in the direction of the force. In this problem force is constant. What is distance moved divided by time interval?

    For the fourth, how much has the jumper's gravitational potential energy changed between where he left the platform and the maximum extension of the trampoline?

    For the last, compare the times from the end of the ramp to the ground for the two cases. Use energy conservation to compare the velocities at the end of the ramp in the two cases. d is the horizontal velocity at the end of the ramp times the time from the end of the ramp to the ground.
     
  7. Jun 26, 2005 #6
    wow i feel stupid. i used what is given as the kinetic energy for the velocity :cry:

    i used P = W/t, and solved for t using v = d/t. thanks

    why do you use the change in energy and not just his energy at that height (which would be m*g*h using the height of 1 m)?

    so the velocity for 2h is sqrt(2) times v. is the time the same for both? if not, how do you get d from the velocity
     
    Last edited: Jun 26, 2005
  8. Jun 26, 2005 #7
    Last edited: Jun 27, 2005
  9. Jun 27, 2005 #8
    11.jpg: Because, it's in static equilbrium the horizontal components of all the forces add to zero, and the vertical components of all the forces add to zero. So you can form a simultaneous equation- there is enough information.

    12.jpg: Draw the internal resistance as a separate resistor, and go around the circuit adding up the voltage drops. Eg. 'emf of battery'-0.5i-5.0i=0... etc. So you can find the emf. Then write another equation for when the resistor is 10 ohms instead of 5 and substitute in the emf.
     
  10. Jun 30, 2005 #9

    OlderDan

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    Yes, the time is the same for both. That time depends only on the acceleration due to gravity and the height of the end of the slide above the ground. There is no vertical velocity when the object leaves the ramp. The distance, d, is the horizontal velocity due to sliding from a height h times the time. The new distance will be the horizontal velocity due to sliding from the height 2h times the same time.
     
  11. Jun 30, 2005 #10

    ek

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    Ah...Physics 12...how was the provincial on Monday?

    Hope it went well. I remember thinking I shanked the thing up badly (Jan/04) but I ended up getting an 85 so it wasn't too bad.
     
  12. Jul 3, 2005 #11
    4.jpg is correct.
     
  13. Jul 3, 2005 #12

    Pyrrhus

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    You can find F by taking a sum of torques at the beam's contact point with the ground.
     
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