Physics HL/SL Help

Rigel · June 7, 2011

Need help on this urgently!!

A body starting from rest moves along a straight-line under the action of a constant force. After

travelling a distance d the speed of the body is v.

The speed of the body when it has travelled a distance d/2

from its initial position is:

A. v/4

B. v/2

C. v/sqrt2

D. v/2sqrt2

Thanks!

Rigel · June 7, 2011

Need help on this one:

10. A fireman is holding a hosepipe so that water leaves the pipe horizontally. The hosepipe has a constant cross-sectional area. The magnitude of the force that the fireman exerts to hold the hosepipe stationary is F.

The volume of water delivered by the hose per second doubles, the force that the fireman must now exert is

A. sqrt2F.

B. 2F.

C. 4F.

D. 8F.

Thanks!

Drake Glau · June 7, 2011

Need help on this urgently!!

A body starting from rest moves along a straight-line under the action of a constant force. After

travelling a distance d the speed of the body is v.

The speed of the body when it has travelled a distance d/2

from its initial position is:

A. v/4

B. v/2

C. v/sqrt2

D. v/2sqrt2

Thanks!

constant force means a constant acceleration according to f=ma

Following this CONSTANT acceleration when you get to d you have a speed of v...so if you only go half way, you only speed up half way giving you v/2

I don't know the 2nd, sorry

Edited June 7, 2011 by Drake Glau

Rigel · June 7, 2011

Need help on this urgently!!

A body starting from rest moves along a straight-line under the action of a constant force. After

travelling a distance d the speed of the body is v.

The speed of the body when it has travelled a distance d/2

from its initial position is:

A. v/4

B. v/2

C. v/sqrt2

D. v/2sqrt2

Thanks!

constant force means a constant acceleration according to f=ma

Following this CONSTANT acceleration when you get to d you have a speed of v...so if you only go half way, you only speed up half way giving you v/2

I don't know the 2nd, sorry

Lol yeah i also got that, but the answer is C.

Drake Glau · June 7, 2011

Need help on this urgently!!

A body starting from rest moves along a straight-line under the action of a constant force. After

travelling a distance d the speed of the body is v.

The speed of the body when it has travelled a distance d/2

from its initial position is:

A. v/4

B. v/2

C. v/sqrt2

D. v/2sqrt2

Thanks!

constant force means a constant acceleration according to f=ma

Following this CONSTANT acceleration when you get to d you have a speed of v...so if you only go half way, you only speed up half way giving you v/2

I don't know the 2nd, sorry

Lol yeah i also got that, but the answer is C.

I honestly don't know then. Everything I get is v=sqrt(ad)...

v²=u²+2ad

u²=0

v²=2ad

if it's d/2, the 2's will then cancel leaving v²=ad so v=sqrt(ad)...idk...

For this I would just know that velocity is a square function of d (immediately ruling out A and B) and since the d is havled that will cancel that extra two you see in D, leaving you C. I just don't know where they are getting the 2 from to sqrt it...

Alright last edit. After I went to bed after this I kind of realized how we both came up with v/2 as our initial answer, and then remember I said it was a squared function, well to get back to just v you would end up with v/sqrt2 I think that's how they got to their answer...

Edited June 7, 2011 by Drake Glau

Rigel · June 9, 2011

Need help on this urgently!!

A body starting from rest moves along a straight-line under the action of a constant force. After

travelling a distance d the speed of the body is v.

The speed of the body when it has travelled a distance d/2

from its initial position is:

A. v/4

B. v/2

C. v/sqrt2

D. v/2sqrt2

Thanks!

constant force means a constant acceleration according to f=ma

Following this CONSTANT acceleration when you get to d you have a speed of v...so if you only go half way, you only speed up half way giving you v/2

I don't know the 2nd, sorry

Lol yeah i also got that, but the answer is C.

I honestly don't know then. Everything I get is v=sqrt(ad)...

v²=u²+2ad

u²=0

v²=2ad

if it's d/2, the 2's will then cancel leaving v²=ad so v=sqrt(ad)...idk...

For this I would just know that velocity is a square function of d (immediately ruling out A and B) and since the d is havled that will cancel that extra two you see in D, leaving you C. I just don't know where they are getting the 2 from to sqrt it...

Alright last edit. After I went to bed after this I kind of realized how we both came up with v/2 as our initial answer, and then remember I said it was a squared function, well to get back to just v you would end up with v/sqrt2 I think that's how they got to their answer...

Wow lol that's weird... I'l ask my teacher tomorrow!

I don't understand why the v/2 turns into a v/sqrt2.

Wouldn't the acceleration be in there also?

Drake Glau · June 9, 2011

a was a constant due to the constant force on a constant mass so i just kind of removed it

Rigel · June 9, 2011

a was a constant due to the constant force on a constant mass so i just kind of removed it

Hehe gotcha!

I just did my Physics exam yesterday.

Didn't go as well as i expected, Paper 1 was too hard.

Paper 2 was so-so.

Sublime Sunshine · June 9, 2011

I was procrastinating so I figured I'd do this for kicks

at $gif.latex?d \over 2$ , speed =

at , speed =

using the equation

$gif.latex?v_1^2 = 2a \left({d \over 2}\r$ - since as it starts at rest.

$gif.latex?v_1 = \sqrt{ad}$

$gif.latex?v_2 = \sqrt{2ad}$

you are trying to find the speed at $gif.latex?d \over 2$ in terms of , so break up and substitute.

$gif.latex?v_2 = \sqrt{2} \times \sqrt{ad$ - split up

$gif.latex?v_2 = \sqrt{2} \times v_1$ - since $gif.latex?v_1 = \sqrt{ad}$

then solve for

$gif.latex?v_1 = \left v_2 \over \sqrt{2}$

answer C

Edit: changed it to tex, sooo much better

Edited June 9, 2011 by Sublime Sunshine

Rigel · June 9, 2011

I was procrastinating so I figured I'd do this for kicks

at $gif.latex?d \over 2$ , speed =

at , speed =

using the equation

$gif.latex?v_1^2 = 2a \left({d \over 2}\r$ - since as it starts at rest.

$gif.latex?v_1 = \sqrt{ad}$

$gif.latex?v_2 = \sqrt{2ad}$

you are trying to find the speed at $gif.latex?d \over 2$ in terms of , so break up and substitute.

$gif.latex?v_2 = \sqrt{2} \times \sqrt{ad$ - split up

$gif.latex?v_2 = \sqrt{2} \times v_1$ - since $gif.latex?v_1 = \sqrt{ad}$

then solve for

$gif.latex?v_1 = \left v_2 \over \sqrt{2}$

answer C

Edit: changed it to tex, sooo much better

WOW TYVM!

Looks too complicated for a P1, doesn't it? P:

I used s=ut... Lol. I forgot that as a force was present, the acceleration had to be somewhere.. TY!

Rigel · June 9, 2011

10. A fireman is holding a hosepipe so that water leaves the pipe horizontally. The hosepipe has a constant cross-sectional area. The magnitude of the force that the fireman exerts to hold the hosepipe stationary is F.

The volume of water delivered by the hose per second doubles, the force that the fireman must now exert is

A. sqrt2F.

B. 2F.

C. 4F.

D. 8F.

Anyone got this?

I can only get 2F, unless the volume per second makes the velocity double...

Ezeh · June 10, 2011

10. A fireman is holding a hosepipe so that water leaves the pipe horizontally. The hosepipe has a constant cross-sectional area. The magnitude of the force that the fireman exerts to hold the hosepipe stationary is F.

The volume of water delivered by the hose per second doubles, the force that the fireman must now exert is

A. sqrt2F.

B. 2F.

C. 4F.

D. 8F.

Anyone got this?

I can only get 2F, unless the volume per second makes the velocity double...

The important parts to realise are that the hose pipe end only lets a certain amount of water out at the same time. So if the volume of water delivered is doubled, then this means that the velocity must have increased.

From the data booklet:

$gif.latex?F={\Delta p \over \Delta t}$

$gif.latex?F={{\Delta m \Delta v} \over \$

If there is more water coming out per second, then the mass will also double. So both velocity and mass doubles, and thus the force will have to change by a factor of 4.

$gif.latex?F_{new}=4F$

puyol9 · June 17, 2011

Thank you very much people!

More questions?

[HL, Gravitation and Orbital Motion] A satellite is orbiting the earth in a circular orbit. Which one of the following properties of the satellite does not remain constant?

A. Kinetic energy

B. Gravitational potential energy

C. Angular momentum

D. Velocity

I think A and B should be both constant since the R is not changing. So it's either C or D. What is angular momentum, though? How is it different from momentum? I think velocity should be constant but may be not due to acceleration and air resistance... (even though I think air resistance is negligible)

Answer is D

Always remember in circular motion: speed may be constant, but velocity is changing! As velocity is a scalar product, measure in direction and magnitude. The direction is always changing (the direction at some point is the tangent of its path at that point), thus the velocity is changing.

This also leads into the idea that there is a force being applied upon the satellite (as there is change in velocity, meaning there is acceleration, and thus there is a force). This is called the centripetal (not centrifugal) force.

Thus, its great to know that the connection between velocity and force is that if there is a change in velocity, there is always a force being applied to that mass, and if there is force (provided mass is constant) there is a change in velocity.

Angular momentum:

I have never heard of it, and I am doing HL physics, so I would assume that you don't need to know this for the course. However, I wikied it, and the general idea seems to be:

It is a vector (cross) product of the position of the body relative to its origin by the linear momentum (that is p=mv). Ie, L = r x (cross multiplication) p, where r is the body's position from the origin and p is the linear momentum.

This would be constant in cicular motion, as when r2 is in the opposite direction (ie. same magnitude opposite direction) to r1, v2 (and thus p2) would be opposite (same magnitude opposite direction), thus resulting in the same angular momentum (L2=L1). This is essentially vectors (maths) not physics.

However, I still don't really understand the concept of angular momentum.

Hope that helps.

Drake Glau · June 17, 2011

Thank you very much people!

More questions?

[HL, Gravitation and Orbital Motion] A satellite is orbiting the earth in a circular orbit. Which one of the following properties of the satellite does not remain constant?

A. Kinetic energy

B. Gravitational potential energy

C. Angular momentum

D. Velocity

I think A and B should be both constant since the R is not changing. So it's either C or D. What is angular momentum, though? How is it different from momentum? I think velocity should be constant but may be not due to acceleration and air resistance... (even though I think air resistance is negligible)

Answer is D

Always remember in circular motion: speed may be constant, but velocity is changing! As velocity is a vector(?) product, measure in direction and magnitude. The direction is always changing (the direction at some point is the tangent of its path at that point), thus the velocity is changing.

Procrastination · June 22, 2011

This question is about wind power.

a) A wind turbine produces 15 kW of electric power at a wind speed v.

- Assusming a constant efficiency for the wind turbine, determine the power output of the turbin for a wind speed of 2v.

-Suggest two reasons why all of the kinetic energy of the incident wind cannot be converted into mechanical energy in the turbine.

b)State and explain one advantage of using wind power to generate electrical energy as compared to using fossil fuels.

I think B is clearly obvious, it doesn't pollutes as it doesn't burn elements which generate CO2 that go to the atmosphere, anyways... I need serious help with number a. Thank you

Drake Glau · June 23, 2011

First part of A is asking if you can manipulate the equation given to you in topic 8 in your equation packet.

0.5Apv³=15kW

Now if I double that v³ to 2v³ you would get 8 times the power since the 2 gets cubed in that so the answer would be 15*8 or 120kW

This is a perfect example of what WILL BE (99% sure) on your paper1 btw. Some question just like this in style. It'll likely have an answer that is obvious, 2 weird ones, and the right one, the "obvious" one would likely be (in this case) 30kW, simply double it, so don't fall for it, think it through

According to one of the laws of thermodynamics (I think) it's impossible to have 100% efficiency. Assuming the incident wind is the wind that passes through the area formed by the propellers there is a lot of wind that doesn't actually give kinetic energy to the turbine at all, it's the wind that passes through.

Edited June 23, 2011 by Drake Glau

July 3, 2011

Hey guys can you please explain to me ( a Physics SL student) how one gets the potential divider formula in terms of R2 and R1. Please show me a step-by-step derivation of the formulas,(I really appreciate it!) Thanks SO much!

Note: In the wiki formula replace VL with V out and Vs with VIn

puyol9 · July 8, 2011

Hey guys can you please explain to me ( a Physics SL student) how one gets the potential divider formula in terms of R2 and R1. Please show me a step-by-step derivation of the formulas,(I really appreciate it!) Thanks SO much!
Note: In the wiki formula replace VL with V out and Vs with VIn

From Wikipedia (http://en.wikipedia.org/wiki/Voltage_divider):

Applying Ohm's Law, the relationship between the input voltage, V_in, and the output voltage, V_out, can be found:

Proof:

If we have a circuit such as: http://en.wikipedia.org/wiki/File:Impedance_Voltage_divider.png

V_in = I*(Z₁+Z₂)

and

V_out = I*Z₂

Thus,

I = V_in/(Z₁+Z₂)

Finally,

V_out=V_in*Z₂/(Z₁+Z₂)

Drake Glau · July 8, 2011

It was already answered, sorry =/ Apparently he asked in two places...

Peanut Butter Jelly · July 8, 2011

Does the mass of a pendulum affecft its period?

Edited July 8, 2011 by fut

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