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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: olly_s15 on February 07, 2010, 10:26:48 pm

Title: Olly_s15's Physics Question Thread
Post by: olly_s15 on February 07, 2010, 10:26:48 pm: Two physics students are trying to determine the instantaneous speed of a bicycle 5.0m from the start of a 1000m sprint. They use a stopwatch to measure the time taken for the bicycle to cover the first 10m. If the acceleration was constant, and the measure time was 4.0s, what was the instantaneous speed of the bicycle at the 5.0m mark?

I've done this several ways and got 2.5 m/s for the answer each time but the answer is supposedly 3.5 m/s.

Can anyone clear this up if the answer is wrong or not.

Thanks :)
Title: Re: Olly_s15's Physics Question Thread
Post by: the.watchman on February 07, 2010, 10:30:04 pm: It certainly appears to be 2.5ms^-1
Then again, I could be wrong as well ;)
Title: Re: Olly_s15's Physics Question Thread
Post by: appianway on February 08, 2010, 05:10:05 pm: Let's go through it.

Solve for a:

d= ut + (at^2)/2
Taking u=0, a=2d/t^2
d=10
t^2=16
a=20/16 = 10/8

v^2 = u^2 + 2ad
d=4
u=0
a=10/8
v^2 = 20/8*4
=80/8
=10

Thus v is approx 3.1

And I'm pretty sure that's not right... (desperately checks arithmetic...)
Title: Re: Olly_s15's Physics Question Thread
Post by: the.watchman on February 08, 2010, 05:13:05 pm: I was thinking that if the acceleration is constant, then the velocity must be linear.
Therefore, the velocity at 5 is the midpoint of the velocity at 0 and at 10, thus it is 2.5ms^-1

Or is this reasoning flawed?
Title: Re: Olly_s15's Physics Question Thread
Post by: appianway on February 08, 2010, 05:20:43 pm: Velocity varies directly with acceleration in respect to time. The time value should be a bit greater than half the total time (and hence not the midpoint), as the average velocity in the first 5 metres is slower than the average velocity in the last five.
Title: Re: Olly_s15's Physics Question Thread
Post by: the.watchman on February 08, 2010, 05:23:16 pm: Good point! Then you're probably right! I'm so :uglystupid2:
Title: Re: Olly_s15's Physics Question Thread
Post by: appianway on February 08, 2010, 05:24:33 pm: But my answer's not even the same anyway...
Title: Re: Olly_s15's Physics Question Thread
Post by: superflya on February 08, 2010, 05:28:08 pm: appians working for the first part is correct as she got $a=\frac{10}{8}$ which is 1.25 ms^-2.

then use $v^{2}=u^{2}+2ad$

$v^{2}=2\cdot (1.25)\cdot (5)$

$\Rightarrow v^{2}=12.5$

$\therefore v=3.5ms^{-1}$
Title: Re: Olly_s15's Physics Question Thread
Post by: TrueTears on February 08, 2010, 05:28:58 pm: Quote from: superflya on February 08, 2010, 05:28:08 pm
appians working for the first part is correct as she got $a=\frac{10}{8}$ which is 1.25s.

then use $v^{2}=u^{2}+2ad$

$v^{2}=2\cdot (1.25)\cdot (5)$

$\Rightarrow v^{2}=12.5$

$\therefore v=3.5ms^{-1}$
watta pro
Title: Re: Olly_s15's Physics Question Thread
Post by: the.watchman on February 08, 2010, 05:30:34 pm: Quote from: TrueTears on February 08, 2010, 05:28:58 pm
Quote from: superflya on February 08, 2010, 05:28:08 pm
appians working for the first part is correct as she got $a=\frac{10}{8}$ which is 1.25s.

then use $v^{2}=u^{2}+2ad$

$v^{2}=2\cdot (1.25)\cdot (5)$

$\Rightarrow v^{2}=12.5$

$\therefore v=3.5ms^{-1}$
watta pro

Ahhh $d\neq 4$ but rather $5$ (from appianway's working)

That makes sense!
Title: Re: Olly_s15's Physics Question Thread
Post by: appianway on February 08, 2010, 05:31:39 pm: Ah, I subbed in the wrong value of d. Whoops! I told you, the.watchman, that something was wrong :P
Title: Re: Olly_s15's Physics Question Thread
Post by: the.watchman on February 08, 2010, 05:32:53 pm: Quote from: appianway on February 08, 2010, 05:31:39 pm
Ah, I subbed in the wrong value of d. Whoops! I told you, the.watchman, that something was wrong :P

Yeah, I guess I was wrong about you being right about being wrong then! :P
Title: Re: Olly_s15's Physics Question Thread
Post by: superflya on February 08, 2010, 05:33:30 pm: Quote from: the.watchman on February 08, 2010, 05:30:34 pm
Quote from: TrueTears on February 08, 2010, 05:28:58 pm
Quote from: superflya on February 08, 2010, 05:28:08 pm
appians working for the first part is correct as she got $a=\frac{10}{8}$ which is 1.25s.

then use $v^{2}=u^{2}+2ad$

$v^{2}=2\cdot (1.25)\cdot (5)$

$\Rightarrow v^{2}=12.5$

$\therefore v=3.5ms^{-1}$
watta pro

Ahhh $d\neq 4$ (from appianway's working)

That makes sense!

yea $d=10$ and $a$ is constant.
Title: Re: Olly_s15's Physics Question Thread
Post by: appianway on February 08, 2010, 05:34:25 pm: d=5 for that section :P As you've already written...
Title: Re: Olly_s15's Physics Question Thread
Post by: superflya on February 08, 2010, 05:35:31 pm: Quote from: appianway on February 08, 2010, 05:34:25 pm
d=5 for that section :P As you've already written...

lol i meant for the first part when finding $a$
Title: Re: Olly_s15's Physics Question Thread
Post by: olly_s15 on February 16, 2010, 07:15:20 pm: Okay, a bit stuck on this one.

A 4.0kg magpie flies towards a very tight plastic wire on a clothes line. The wire is perfectly horizontal and is stretched between poles 4.0m apart. The magpie lands on the center of the wire, depressing it by a vertical distance of 4.0cm. What is the magnitude of the tension in the string?

Thanks.
Title: Re: Olly_s15's Physics Question Thread
Post by: QuantumJG on February 17, 2010, 12:47:43 am: The best I can think of is that the shape of the wire can be expressed as:

$y = 50cosh( \dfrac{x}{50} ) - 50.04$

c, tension at the lowest point is represented as:

$c = \lambda a g$ , note: a = 50 here!

I thought that λ could be represented as $\lambda = \dfrac{M}{L} = \dfrac{4}{4} = 1kg/m$

c = 1 x 50 x 9.8 = 490N

Something tells me that my resoning is flawed!

Btw: Where did you get this question?
Title: Re: Olly_s15's Physics Question Thread
Post by: mark_alec on February 17, 2010, 01:52:18 am: Quote from: QuantumJG on February 17, 2010, 12:47:43 am
The best I can think of is that the shape of the wire can be expressed as:

$y = 50cosh( \dfrac{x}{50} ) - 50.04$
The wire will not be curved, it will be made up of straight line segments.

In order to calculate the tension, resolve the forces in the vertical direction.
Title: Re: Olly_s15's Physics Question Thread
Post by: Mao on February 17, 2010, 01:54:27 am: All the forces acting on the magpie: there is gravity force downwards, and normal reaction force acting upwards, that is equal and opposite to gravity (40N)

Hence, the net force exerted by the string (tension) must be a 40N upwards force. Tension force must travel in the direction of the rope, hence, a force diagram can be constructed.

The rest should be fairly straight-forward, application of trigonometry. [1000.2N is the answer I got]
Title: Re: Olly_s15's Physics Question Thread
Post by: /0 on February 17, 2010, 02:11:34 am: Quote from: mark_alec on February 17, 2010, 01:52:18 am
Quote from: QuantumJG on February 17, 2010, 12:47:43 am
The best I can think of is that the shape of the wire can be expressed as:

$y = 50cosh( \dfrac{x}{50} ) - 50.04$
The wire will not be curved, it will be made up of straight line segments.

In order to calculate the tension, resolve the forces in the vertical direction.

I think (dunno for sure) that cosh models a wire hanging under only its own weight
Title: Re: Olly_s15's Physics Question Thread
Post by: mark_alec on February 17, 2010, 02:38:29 am: http://en.wikipedia.org/wiki/Catenary
Title: Re: Olly_s15's Physics Question Thread
Post by: olly_s15 on February 17, 2010, 11:07:40 am: The correct answer is 1000N so Mao's method was correct. Thanks for the help.
Title: Re: Olly_s15's Physics Question Thread
Post by: QuantumJG on February 17, 2010, 12:29:19 pm: Quote from: Mao on February 17, 2010, 01:54:27 am
All the forces acting on the magpie: there is gravity force downwards, and normal reaction force acting upwards, that is equal and opposite to gravity (40N)

Hence, the net force exerted by the string (tension) must be a 40N upwards force. Tension force must travel in the direction of the rope, hence, a force diagram can be constructed.

The rest should be fairly straight-forward, application of trigonometry. [1000.2N is the answer I got]

So it doesn't curve but basically make that shape you showed?

It becomes much simpler now and getting 1000N is easy to find.
Title: Re: Olly_s15's Physics Question Thread
Post by: Cthulhu on February 17, 2010, 04:04:35 pm: Quote from: QuantumJG on February 17, 2010, 12:29:19 pm
Quote from: Mao on February 17, 2010, 01:54:27 am
All the forces acting on the magpie: there is gravity force downwards, and normal reaction force acting upwards, that is equal and opposite to gravity (40N)

Hence, the net force exerted by the string (tension) must be a 40N upwards force. Tension force must travel in the direction of the rope, hence, a force diagram can be constructed.

The rest should be fairly straight-forward, application of trigonometry. [1000.2N is the answer I got]

So it doesn't curve but basically make that shape you showed?

It becomes much simpler now and getting 1000N is easy to find.
When have you ever come across hyperbolic functions in 3/4 physics? and as that link from mark_alec explained, the first equation you gave was for a Catenary hanging under its own weight and doesn't really apply in this case.
Title: Re: Olly_s15's Physics Question Thread
Post by: olly_s15 on February 23, 2010, 11:03:06 pm: A car of mass 1500kg traveling due west at a speed of 20m/s on an icy road collides with a truck of mass 2000kg traveling at the same speed in the opposite direction. The vehicles lock together after impact. Which vehicle experiences the greatest force.

Now from just looking at this, it seems the car experiences the biggest force (truck hitting car = owned car) but, I find that the truck experiences a greater impulse than the car.

Using the formula $I=F_{av}\vartriangle v$ , i am assuming that the change in time is fixed for both vehicles and thus the vehicle experiencing the greatest impulse will experience the greatest force. Because: $F_{av}=\frac{I}{\vartriangle v}$

Can anyone tell me if my reasoning is wrong :-\. I have a feeling it is. But it seems to make sense to me. Lol.
Title: Re: Olly_s15's Physics Question Thread
Post by: superflya on February 23, 2010, 11:18:06 pm: another way to look at it, $I=\Delta \rho$
by finding the momentum of both the car and truck before and after the collision, u can work out which of the 2 has experienced a greater change in momentum. The one that has experienced the greater change in momentum will also experience a greater impulse.

correct me if im mistaken :)
Title: Re: Olly_s15's Physics Question Thread
Post by: QuantumJG on February 23, 2010, 11:23:36 pm: Quote from: olly_s15 on February 23, 2010, 11:03:06 pm
A car of mass 1500kg traveling due west at a speed of 20m/s on an icy road collides with a truck of mass 2000kg traveling at the same speed in the opposite direction. The vehicles lock together after impact. Which vehicle experiences the greatest force.

Now from just looking at this, it seems the car experiences the biggest force (truck hitting car = owned car) but, I find that the truck experiences a greater impulse than the car.

Using the formula $I=F_{av}\vartriangle v$ , i am assuming that the change in time is fixed for both vehicles and thus the vehicle experiencing the greatest impulse will experience the greatest force. Because: $F_{av}=\frac{I}{\vartriangle v}$

Can anyone tell me if my reasoning is wrong :-\. I have a feeling it is. But it seems to make sense to me. Lol.

Doesn't Newton's third law say that both experience the same force? Just an idea I threw out there.
Title: Re: Olly_s15's Physics Question Thread
Post by: olly_s15 on February 23, 2010, 11:25:00 pm: Quote from: superflya on February 23, 2010, 11:18:06 pm
another way to look at it, $I=\Delta \rho$
by finding the momentum of both the car and truck before and after the collision, u can work it out which of the 2 has experienced a greater change in momentum. The one that has experienced the greater change in momentum will also experience a greater impulse.

correct me if im mistaken :)

Yes I've found the impulse of both car and truck and $I_{truck}>I_{car}$ . But we're looking at which experiences a greater force. Not greater impulse.
Title: Re: Olly_s15's Physics Question Thread
Post by: olly_s15 on February 23, 2010, 11:26:17 pm: Quote from: QuantumJG on February 23, 2010, 11:23:36 pm
Quote from: olly_s15 on February 23, 2010, 11:03:06 pm
A car of mass 1500kg traveling due west at a speed of 20m/s on an icy road collides with a truck of mass 2000kg traveling at the same speed in the opposite direction. The vehicles lock together after impact. Which vehicle experiences the greatest force.

Now from just looking at this, it seems the car experiences the biggest force (truck hitting car = owned car) but, I find that the truck experiences a greater impulse than the car.

Using the formula $I=F_{av}\vartriangle v$ , i am assuming that the change in time is fixed for both vehicles and thus the vehicle experiencing the greatest impulse will experience the greatest force. Because: $F_{av}=\frac{I}{\vartriangle v}$

Can anyone tell me if my reasoning is wrong :-\. I have a feeling it is. But it seems to make sense to me. Lol.

Doesn't Newton's third law say that both experience the same force? Just an idea I threw out there.

Hmm.. in a collision.. I'm not sure.
Title: Re: Olly_s15's Physics Question Thread
Post by: Cthulhu on February 23, 2010, 11:28:54 pm: $\frac{I_{truck}}{\Delta v} > \frac{I_{car}}{\Delta v}$

$\therefore F_{truck} > F_{car}$
.. dunno.
Title: Re: Olly_s15's Physics Question Thread
Post by: olly_s15 on February 23, 2010, 11:30:32 pm: Quote from: Cthulhu on February 23, 2010, 11:28:54 pm
$\frac{I_{truck}}{\Delta v} > \frac{I_{car}}{\Delta v}$
$\therefore F_{truck} > F_{car}$
.. dunno.

Hmm...
Title: Re: Olly_s15's Physics Question Thread
Post by: superflya on February 23, 2010, 11:31:46 pm: Quote from: olly_s15 on February 23, 2010, 11:30:32 pm
Quote from: Cthulhu on February 23, 2010, 11:28:54 pm
$\frac{I_{truck}}{\Delta v} > \frac{I_{car}}{\Delta v}$
$\therefore F_{truck} > F_{car}$
.. dunno.

It appears so.

yea if the time is equal then that would be true.
Title: Re: Olly_s15's Physics Question Thread
Post by: olly_s15 on February 23, 2010, 11:32:22 pm: Quote from: superflya on February 23, 2010, 11:31:46 pm
Quote from: olly_s15 on February 23, 2010, 11:30:32 pm
Quote from: Cthulhu on February 23, 2010, 11:28:54 pm
$\frac{I_{truck}}{\Delta v} > \frac{I_{car}}{\Delta v}$
$\therefore F_{truck} > F_{car}$
.. dunno.

It appears so.

yea if the time is equal then that would be true.

But is the time definately equal?
Title: Re: Olly_s15's Physics Question Thread
Post by: superflya on February 23, 2010, 11:35:29 pm: not too sure, however i dont think its an assumption we should be making anyway :S
Title: Re: Olly_s15's Physics Question Thread
Post by: olly_s15 on February 23, 2010, 11:36:54 pm: In most (if not all) impulse related questions in VCE physics there is usually only one specified value of time which you use to determine impulse on both objects. So going by that I think it's a fair assumption.
Title: Re: Olly_s15's Physics Question Thread
Post by: superflya on February 23, 2010, 11:39:43 pm: Quote from: olly_s15 on February 23, 2010, 11:36:54 pm
In most (if not all) impulse related questions in VCE physics there is usually only one specified value of time which you use to determine impulse on both objects. So going by that I think it's a fair assumption.

okay then the answer should be correct.

shed some light on this quantum :P
Title: Re: Olly_s15's Physics Question Thread
Post by: olly_s15 on February 23, 2010, 11:40:28 pm: I think since the F_av on the truck is greater, the force applied by the truck to car is greater and thus the car experiences the greater force. Which makes sense because the truck will obviously be less damaged than the car. I think I understand this now.
Title: Re: Olly_s15's Physics Question Thread
Post by: QuantumJG on February 24, 2010, 12:01:17 am: m_caru_car + m_trucku_truck = (m_car + m_truck)v

1,500*20 - 2,000*20 = -10,000 = 3,500v

$\therefore$ v = -2.86m/s

Δp_truck = 2,000*(-2.86-(-20)) = 34,286 kgm/s

Δp_car = 1,500*(-2.86-20) = -34,286kgm/s

So both have the same impulse 'magnitude' and contact time implying that the net force on both is equal.

Since Newton's second law states:

F = $\dfrac{dp}{dt}$
Title: Re: Olly_s15's Physics Question Thread
Post by: kamil9876 on February 24, 2010, 12:30:07 am: A case of conservation of momentum implying newton's third law hehe

Traditionally, you can go the other way for the general case, ie: use 3rd law to prove conservation of momentum.
Title: Re: Olly_s15's Physics Question Thread
Post by: olly_s15 on February 24, 2010, 03:52:25 pm: Quote from: QuantumJG on February 24, 2010, 12:01:17 am
m_caru_car + m_trucku_truck = (m_car + m_truck)v

1,500*20 - 2,000*20 = -10,000 = 3,500v

$\therefore$ v = -2.86m/s

Δp_truck = 2,000*(-2.86-(-20)) = 34,286 kgm/s

Δp_car = 1,500*(-2.86-20) = -34,286kgm/s

So both have the same impulse 'magnitude' and contact time implying that the net force on both is equal.

Since Newton's second law states:

F = $\dfrac{dp}{dt}$

Thanks Quantum, my physics teacher confirmed this today.
Title: Re: Olly_s15's Physics Question Thread
Post by: olly_s15 on February 24, 2010, 07:38:44 pm: The ancient Egyptians relied on a knowledge of the physics of energy transformations to build the Great Pyramids at Giza. They used ramps to push limestone blocks with an average mass of 2300kg to heights of almost 150m. The ramps were sloped at about 10degrees to the horizontal. Friction was reduced by pumping water onto the ramps.
a) How much work would have to be done to lift an average limestone block vertically through a height of 150m?

Is this just $W=E_{GP}=m*g*\vartriangle h$ ??

b) How much work would have been done to push an average limestone block to the same height along a ramp inclned at 10degrees to the horizontal? Friction is negligible.
Title: Re: Olly_s15's Physics Question Thread
Post by: Cthulhu on February 24, 2010, 08:07:07 pm: Quote from: olly_s15 on February 24, 2010, 07:38:44 pm
The ancient Egyptians relied on a knowledge of the physics of energy transformations to build the Great Pyramids at Giza. They used ramps to push limestone blocks with an average mass of 2300kg to heights of almost 150m. The ramps were sloped at about 10degrees to the horizontal. Friction was reduced by pumping water onto the ramps.
a) How much work would have to be done to lift an average limestone block vertically through a height of 150m?

Is this just $W=E_{GP}=m*g*\vartriangle h$ ??

b) How much work would have been done to push an average limestone block to the same height along a ramp inclned at 10degrees to the horizontal? Friction is negligible.

a) Yes. Work is $W = F_{g} \cdot d = mgh = E_{gp}$

b) Um.... I think I did something wrong here? Fuck this shit. I'm going back to Quantum Mechanics
You need to find the force on the block as its going up the slope which is:
$F = mg \sin(\theta)$
Then you have to find the distance the brick has to travel to get to this height:
$\sin (\theta) = \frac{h}{d}$
Where h = 150m.
So
$d = \frac{h}{\sin(\theta)}$
and then:
$W = F_g \cdot d$
$W = mg\sin(\theta) \cdot \frac{h}{\sin(\theta)}$
$W = mg \cdot h$

:-\ nega karma me for failing if you wish.
EDIT: I don't think I did anything wrong...... >.> I can't work on this forum text editor...
The definition of work is:
$W = F \cdot d$
Using the definition of the dot product we get:
$W = Fd\cos(\phi)$
$\phi = \theta + 90$
$W = mgd\cos(\theta + 90)$
$\cos(\theta + 90) = -\sin(\theta)$
$W = -mgd\sin(\theta)$
Using MAGIC:
$\sin(\theta) = \frac{h}{d}$
$h = d \sin (\theta)$
so:
$W = -mgh$ .
I think my other working out makes more sense though
Title: Re: Olly_s15's Physics Question Thread
Post by: olly_s15 on March 22, 2010, 07:27:41 pm: Can anyone confirm this:

My answer is 4000 Nm-1 however the answer in the solutions is 160 Nm-1.

See file below.
Title: Re: Olly_s15's Physics Question Thread
Post by: superflya on March 22, 2010, 07:32:20 pm: hmm i get 4000 aswell :/
Title: Re: Olly_s15's Physics Question Thread
Post by: Twenty10 on March 22, 2010, 07:34:48 pm: I got 4000N/m as well o.0
Title: Re: Olly_s15's Physics Question Thread
Post by: qshyrn on March 22, 2010, 07:55:01 pm: i got that as well
Title: Re: Olly_s15's Physics Question Thread
Post by: olly_s15 on March 22, 2010, 08:01:30 pm: I guess that's good enough for me. 4000 Nm-1 it is.
Title: Re: Olly_s15's Physics Question Thread
Post by: olly_s15 on April 21, 2010, 06:22:57 pm: Hey is the sinusoidal AC voltage stuff still in the course or should I not worry about these questions?
Title: Re: Olly_s15's Physics Question Thread
Post by: Juddinator on April 21, 2010, 08:13:49 pm: Quote from: olly_s15 on April 21, 2010, 06:22:57 pm
Hey is the sinusoidal AC voltage stuff still in the course or should I not worry about these questions?
Yes it is, we were just going over it today
Title: Re: Olly_s15's Physics Question Thread
Post by: olly_s15 on April 21, 2010, 10:45:37 pm: Hmm I can't find it in my textbook
Title: Re: Olly_s15's Physics Question Thread
Post by: Juddinator on May 07, 2010, 05:56:39 pm: How do you know when to use 'sine' or 'cos' when dealing with inclined planes. I always seem to find myself getting it wrong even though what I was intending to achieve would be correct. For example, when the 'normal force' is asked for, how do we know this is cos? The same goes with the 'net force'. How do we know how to use sine?

Thanks
Title: Re: Olly_s15's Physics Question Thread
Post by: Blakhitman on May 08, 2010, 07:34:30 pm: SOH CAH TOA!!!!

It's just basic trig, just gotta draw the triangle correctly!
Title: Re: Olly_s15's Physics Question Thread
Post by: Juddinator on May 08, 2010, 10:27:00 pm: oh yeah.. sorry had a retarded moment :|
Title: Re: Olly_s15's Physics Question Thread
Post by: Blakhitman on May 08, 2010, 11:16:05 pm: Quote from: Juddinator on May 08, 2010, 10:27:00 pm
oh yeah.. sorry had a retarded moment :|

Oh don't I know those kind of moments!