Desperately Need Help With Questions

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Desperately Need Help With Questions!!!!!

I havent done any physics work in about half a year n now i need to do sooo many questions in about a months time! ='( pleasee help me with these questions! I would HIGHLY HIGLYY appreciate it sooooooo much for anybody who would help me solve these questions.

http://dsmarketing.com/samples/phys_smpl1.pdf [Broken]

Also the website above is to the multiple choice questions...i kno they are pretty easy but I forgot how to do them too! ='( I would HIGLLYYYYYYYYYY appreciate if somebody could explain how to get the answers to me PLEASE!!!
(you need to scroll down to Physics C Mechanics and Physics Electricity and Magnetism Sample Questions)


THANKYOU SOOOOOOOOOOO MUCHH FOR ALL OF YOU WHO CAN HELP ME!!!!!!!!!! =D!
 

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  • #2
Integral
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You had to ok the site guidelines when you registered. Did you read them? We do not do your homework for you, you have to put in some effort.


Please show what you know about each of your questions.

You may find it more effective to devote a thread to each question, showing your work of course.
 
  • #3
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I WOULD attempt them but its been like 6 mnths since i have done anyy work related to this watsoever...and i basically forgot everything...i dont even kno how to attack the problem besides writing down the info given ='(
 
  • #4
Vanadium 50
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And if we were to do them for you, how would that help you?
 
  • #5
hage567
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I WOULD attempt them but its been like 6 mnths since i have done anyy work related to this watsoever...and i basically forgot everything...i dont even kno how to attack the problem besides writing down the info given ='(
Where's your textbook? Have you tried studying it over thoroughly to refresh your understanding? Do that first. If you knew how to do them before, it shouldn't take you long to pick it up again, right?
 
  • #6
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i dont want any body to do these problems for me...i jus need some assistance like which formulas should i use as suggestions ....and i have looked in my books...ive spent hours looking in my books AND lookin online and everytime i think i get somewhere i get an answer that is COMPLETELYYY OFF...i mean by like the 100s =S
 
  • #7
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thanku for the help...so far these r wat i got.. but idk if its right =S

(a)
J=FΔt=mΔv=Δp
J=(2000N)(.010s)
J=20Nxs

(b)
m1v1+m2v2=m1v1'+m2v2'
(.5kg)(26m/s)+(5kg)(0m/s)=(.5kg)(26m/s)+(5kg)v2'
(.5kg)(26m/s)=(.5kg)(26m/s)+(5kg)v2'
13kg x m/s = 13kg x m/s +5kg x v2'
13kg x m/s = 18kg x m/s x v2'
-5m/s = v2'

(c)
(i)
26m/s
(ii)
left

(d)
K=1/2mv2
which speed n mass am i supposed to use?

(e) d=v0t + (1/2)at2
would this be the correct dist formula?
 
  • #8
alphysicist
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Hi HELP_786,

thanku for the help...so far these r wat i got.. but idk if its right =S

(a)
J=FΔt=mΔv=Δp
J=(2000N)(.010s)
J=20Nxs
This answer would be the impulse if the force was a constant 2000 N. However, the force is a varying force over the time period. How can you use the graph itself to find the impulse?


(b)
m1v1+m2v2=m1v1'+m2v2'
(.5kg)(26m/s)+(5kg)(0m/s)=(.5kg)(26m/s)+(5kg)v2'
(.5kg)(26m/s)=(.5kg)(26m/s)+(5kg)v2'
13kg x m/s = 13kg x m/s +5kg x v2'
13kg x m/s = 18kg x m/s x v2'
-5m/s = v2'
The problem with the above work is that m1 is slowed down by the collision, so v_1' is not 26 m/s. So instead of this, if the momentum of something is changing, what is the change in momentum equal to?
(c)
(i)
26m/s
(ii)
left
Once you find the answer to part b, you can use the approach you used in part b to find the final speed of the cube.

(d)
K=1/2mv2
which speed n mass am i supposed to use?
You need to take into account the change in kinetic energy of the entire (cube+ball) system due to the collision.

(e) d=v0t + (1/2)at2
would this be the correct dist formula?
The motion after the collision is a two dimensional free fall problem.
 
  • #9
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for part 'a', if i was to find the impulse for each millisecond n then add them up?
 
  • #10
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for part 'b', the change in momentum is also equal tot he impulse right?
sooo......would i use the impulse answer i got from part 'a' n set it equal to m1'v1' so i can get the correct velocity for v1'? n then from there use tht answer n substitute it into the whole formula i did previously of m1v1 + m2v2 = m1v1' + m2v2'?
i hope im rite lol
 
  • #11
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ORRRR lol
for part 'a' would i use:

(linear impulse-momentum equation)

ΣFΔt=Mv2-Mv1

and in this formula what does M represent? just mass or total mass? or...neither? lol
 
  • #12
alphysicist
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for part 'a', if i was to find the impulse for each millisecond n then add them up?
That's the right kind of idea; do you see how to find the impulse for each millisecond? How do you find that from the graph? In other words, since the graph is force vs. time, what property of the graph would be the impulse?


for part 'b', the change in momentum is also equal tot he impulse right?
sooo......would i use the impulse answer i got from part 'a' n set it equal to m1'v1' so i can get the correct velocity for v1'?
The impulse on the object is equal to the change in momentum of the object. So if you apply the impulse-momentum theorem to m1, you'll need to take into account its initial momentum, too. Also, since the impulse from the graph is the impulse on m2, you'll need to give it a minus sign to get the impulse for m1. It's not that difficult, you'll just need to make sure you get the details right.

Instead, since part b asks for the velocity of m2, why not apply that procedure to m2 directly to find v2'? Since m2 is not moving initially, your procedure would work just as it is.





n then from there use tht answer n substitute it into the whole formula i did previously of m1v1 + m2v2 = m1v1' + m2v2'?
i hope im rite lol
Once you find v2', this would be a good way to find v1'.
 
  • #13
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for part 'e' im confused as to which formula i am supposed to use! =S:confused::cry:
 
  • #14
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ok so this is what ive done so far...

(a)
J=FΔt=mΔv=Δp
J=(500n x .001s)+(1000n x .002s)+(1500n x .003s)+(2000n x .004)+(2000n x .005)+(2000n x .006)+(1500n x .007)+(1000n x .008)+(500n x .009s)+(0n x .010s)
J=60n x s

(b)
ball
Δpb=Δpa
FΔtb=FΔta
Jb=Δpa
Jb=mΔva

60n x s=(5kg)v'
6.1m/s=v'

(c)
m1v1+m2v2=m1v1'+m2v2'
(5kg)(0m/s)+(.5kg)(26m/s)=(5kg)(6.1m/s)+(.5kg)v2'
(.5kg)(26m/s)=(5kg)(6.1m/s)+(.5kg)v2'
26kg x m/s=30.5kg x m/s + v2'
-4m/s=v2'<<<<<<<<<(i)
left<<<<<<<<<<<(ii)

(d)
Δvcube=13m/s
Δvball=5.55m/s

Kball=1/2mv2
Kball=1/2(5kg)(5.6m/s)^2
Kball=78.4J

Kcube=1/2(.5kg)(13m/s)^2
Kcube=42.25J


is any of this right?
 
  • #15
alphysicist
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Looking at part a:

ok so this is what ive done so far...

(a)
J=FΔt=mΔv=Δp
J=(500n x .001s)+(1000n x .002s)+(1500n x .003s)+(2000n x .004)+(2000n x .005)+(2000n x .006)+(1500n x .007)+(1000n x .008)+(500n x .009s)+(0n x .010s)
J=60n x s
No, this has a couple of errors repeated several times.

If you'll look at what you have for the first millisecond: 500 N x 0.001 seconds. This is not correct, because the force is not 500 N for the entire millisecond. You have to take into account that during that millisecond, the force starts at zero, and increases during the time interval.

There are several ways to take this into account, but if you have a force vs. time graph like you have here, the way to get the impulse is straightforward: The impulse is the area under the curve of the force vs. time graph.

Do you see how that takes into account the changes when the force is not constant? What do you get for part a?
 
  • #16
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o yea! that makes sence =)
can i get the area by counting the squares so the area under the curve would be 24n x s ?

if I was to do it mathematically Im confused as to what equation i should use.
would i take the integral of f(t)dt from .010 to 0 ?
im pretty rusty on my calculus as well =(
 

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