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Various Physics 12 problems

  • Thread starter seiferseph
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EDIT: 1 question left

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

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!
 
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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
 
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ascky said:
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
i get them both now, thanks!
 
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I have a couple more problems

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

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 [Broken]

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 [Broken]

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 [Broken]

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 [Broken]

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

Thanks again!
 
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OlderDan

Science Advisor
Homework Helper
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seiferseph said:
I have a couple more problems

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

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 [Broken]

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 [Broken]

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 [Broken]

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 [Broken]

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

Thanks again!
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.
 
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OlderDan said:
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?
wow i feel stupid. i used what is given as the kinetic energy for the velocity :cry:

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

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?
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)?

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.
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
 
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thanks for the help! i have two more questions (provincial exam coming up)

http://i2.photobucket.com/albums/y15/seiferseph/11.jpg [Broken]

not sure how to tackle this one, i know the forces are balanced, but there appear to be too many unknowns

http://i2.photobucket.com/albums/y15/seiferseph/12.jpg [Broken]

tried a couple things, can't see how to solve it.
 
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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.
 

OlderDan

Science Advisor
Homework Helper
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seiferseph said:
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
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.
 

ek

180
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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.
 
4.jpg is correct.
 

Pyrrhus

Homework Helper
2,160
1
You can find F by taking a sum of torques at the beam's contact point with the ground.
 

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