VCE Physics Question Thread!

[knightrider]

In physics how are we expected to give our final answers?

Like how does the rounding work ?(when giving answers)

[JI2015]

In VCE Physics (Units 3 + 4 2016), you are not required to have answers correct to the appropriate amount of significant figures (someone correct me if I'm wrong). You will cover basic data analysis in physics and will be taught how sig. figs. work. You will also learn about sig. figs. in VCE Chemistry where they have a question every year where they randomly check your sig. figs. and marks can be lost, but in physics as long as your answer is reasonable (i.e. 2-4 sig. figs.) then you should be fine. Your answers will usually have 2 decimal places at most, anything more is inappropriate considering the uncertainty present. If you are studying Units 3 + 4 Physics in 2017, then there might be more weight placed on accuracy of answers if I recall correctly.

[knightrider]

In VCE Physics (Units 3 + 4 2016), you are not required to have answers correct to the appropriate amount of significant figures (someone correct me if I'm wrong). You will cover basic data analysis in physics and will be taught how sig. figs. work. You will also learn about sig. figs. in VCE Chemistry where they have a question every year where they randomly check your sig. figs. and marks can be lost, but in physics as long as your answer is reasonable (i.e. 2-4 sig. figs.) then you should be fine. Your answers will usually have 2 decimal places at most, anything more is inappropriate considering the uncertainty present. If you are studying Units 3 + 4 Physics in 2017, then there might be more weight placed on accuracy of answers if I recall correctly.

Thanks so much JI2015 for clarifying 

[knightrider]

For the following image attached the answers for
8a  and b are 6.26ms^-1
8c is 5.42ms^-1

Now for these following questions can someone check if my working out is right?

d Which object had the greater change in speed as it landed? Calculate the speed change of this object.

golf ball-

tomato-

Tomato has greater change in speed

e Which of these objects experienced the greater change in velocity as it landed? Calculate the velocity change of this object.

Let down be positive.

golf ball-

tomato-

golf ball has greater change in velocity

[JI2015]

Yep, your working is fine!

[knightrider]

When resolving vectors into a triangle.

When asked to find the direction. How do you know where to place theta in a triangle (like where does it go? )

Like refer to the image attached.

How do they know theta goes where i have illustrated in red. How do you determine where to place theta?(for all cases)

[knightrider]

Can anyone explain what is happening in this diagram attached.

All the book says is "The forces acting on a bouncing ball."

[JI2015]

When resolving vectors into a triangle.

When asked to find the direction. How do you know where to place theta in a triangle (like where does it go? )

Like refer to the image attached.

How do they know theta goes where i have illustrated in red. How do you determine where to place theta?(for all cases)

Think of it this way: The plane is heading South and is being offset by a wind heading in a West direction. The angle will tell us the amount of deviation that the wind causes in relation to the plane's original path. No wind means that the plane will continue flying South, but when wind is present the plane will be flying in both the South and West directions and the angle informs us on how large the West component of its resultant velocity is. So to summarise for general cases, the angle between the resultant and the original direction of motion (without resistance) is the 'theta' or angle you are after when expressing the direction. Post a question if you need further clarification.

Can anyone explain what is happening in this diagram attached.

All the book says is "The forces acting on a bouncing ball."

Ok, let's start with Picture 1 (noting that air resistance is being neglected):

The ball is in a state of free fall as the only force it is experiencing is the force due to gravity (its weight force). It's acceleration is 9.8 m/s^2 down towards the Earth.

Picture 2:

When the ball comes into contact with the ground, it slows down. This means that it is accelerating up and that the Normal force is larger in size than the weight force (also keep in mind that weight is always constant). Taking up as positive, the Net Force = Normal - Weight and is positive.

Picture 3:

The ball is at maximum compression. It's velocity is 0 m/s and it is about to change its direction of motion. The Normal Force is greatest at the point of maximum compression and therefore its acceleration upwards is greatest at this point as well.

Picture 4:

The ball is about to bounce back up. When it is still in contact with the ground, it is accelerating upwards, and its net force is also upwards. However, once it leaves the ground, THERE IS NO NORMAL FORCE (or any force making it go up!). The ball's velocity will be in the upwards direction until it reaches its maximum height, then it will change direction of motion back towards the Earth. Note that while the ball is not on the ground it is accelerating downwards (towards the Earth) but this does not mean its velocity has to be in that same direction for all its time off the ground.

Picture 5:

Same as Picture 1, but note the explanation for Picture 4 that explains how velocity might be upwards but the ball is still accelerating down towards the Earth.

I hope this helps, it's not easy to explain through typed explanations. If you need further clarification just post again.

[Syndicate]

Think of it this way: The plane is heading South and is being offset by a wind heading in a West direction. The angle will tell us the amount of deviation that the wind causes in relation to the plane's original path. No wind means that the plane will continue flying South, but when wind is present the plane will be flying in both the South and West directions and the angle informs us on how large the West component of its resultant velocity is. So to summarise for general cases, the angle between the resultant and the original direction of motion (without resistance) is the 'theta' or angle you are after when expressing the direction. Post a question if you need further clarification.

Ok, let's start with Picture 1 (noting that air resistance is being neglected):

The ball is in a state of free fall as the only force it is experiencing is the force due to gravity (its weight force). It's acceleration is 9.8 m/s^2 down towards the Earth.

Picture 2:

When the ball comes into contact with the ground, it slows down. This means that it is accelerating up and that the Normal force is larger in size than the weight force (also keep in mind that weight is always constant). Taking up as positive, the Net Force = Normal - Weight and is positive.

Picture 3:

The ball is at maximum compression. It's velocity is 0 m/s and it is about to change its direction of motion. The Normal Force is greatest at the point of maximum compression and therefore its acceleration upwards is greatest at this point as well.

Picture 4:

The ball is about to bounce back up. When it is still in contact with the ground, it is accelerating upwards, and its net force is also upwards. However, once it leaves the ground, THERE IS NO NORMAL FORCE (or any force making it go up!). The ball's velocity will be in the upwards direction until it reaches its maximum height, then it will change direction of motion back towards the Earth. Note that while the ball is not on the ground it is accelerating downwards (towards the Earth) but this does not mean its velocity has to be in that same direction for all its time off the ground.

Picture 5:

Same as Picture 1, but note the explanation for Picture 4 that explains how velocity might be upwards but the ball is still accelerating down towards the Earth.

I hope this helps, it's not easy to explain through typed explanations. If you need further clarification just post again.

Just want to extend on  JI2015's response.

If you also have a look at the force G (gravity), it stays the same, as you may remember, that the gravity on Earth is 9.8 m/s/s.

Due to the acceleration of gravity, all objects have the same rate of acceleration, when they free fall (no matter what their mass is). So if you divide the Netforce by the weight, you will always get 9.8 m/s/s.

I have attached a picture below, hopefuly that helps 

[Syndicate]

When resolving vectors into a triangle.

When asked to find the direction. How do you know where to place theta in a triangle (like where does it go? )

Like refer to the image attached.

How do they know theta goes where i have illustrated in red. How do you determine where to place theta?(for all cases)

If you have a look at Vectors, you see that the theta is placed from the tail of a vector to another tail of another vector. If you have a look at the attached picture below, the angle will be placed in between the vectors a and c (a + b). [look at the second triangle]

[knightrider]

Think of it this way: The plane is heading South and is being offset by a wind heading in a West direction. The angle will tell us the amount of deviation that the wind causes in relation to the plane's original path. No wind means that the plane will continue flying South, but when wind is present the plane will be flying in both the South and West directions and the angle informs us on how large the West component of its resultant velocity is. So to summarise for general cases, the angle between the resultant and the original direction of motion (without resistance) is the 'theta' or angle you are after when expressing the direction. Post a question if you need further clarification.

Ok, let's start with Picture 1 (noting that air resistance is being neglected):

The ball is in a state of free fall as the only force it is experiencing is the force due to gravity (its weight force). It's acceleration is 9.8 m/s^2 down towards the Earth.

Picture 2:

When the ball comes into contact with the ground, it slows down. This means that it is accelerating up and that the Normal force is larger in size than the weight force (also keep in mind that weight is always constant). Taking up as positive, the Net Force = Normal - Weight and is positive.

Picture 3:

The ball is at maximum compression. It's velocity is 0 m/s and it is about to change its direction of motion. The Normal Force is greatest at the point of maximum compression and therefore its acceleration upwards is greatest at this point as well.

Picture 4:

The ball is about to bounce back up. When it is still in contact with the ground, it is accelerating upwards, and its net force is also upwards. However, once it leaves the ground, THERE IS NO NORMAL FORCE (or any force making it go up!). The ball's velocity will be in the upwards direction until it reaches its maximum height, then it will change direction of motion back towards the Earth. Note that while the ball is not on the ground it is accelerating downwards (towards the Earth) but this does not mean its velocity has to be in that same direction for all its time off the ground.

Picture 5:

Same as Picture 1, but note the explanation for Picture 4 that explains how velocity might be upwards but the ball is still accelerating down towards the Earth.

I hope this helps, it's not easy to explain through typed explanations. If you need further clarification just post again.

Just want to extend on  JI2015's response.

If you also have a look at the force G (gravity), it stays the same, as you may remember, that the gravity on Earth is 9.8 m/s/s.

Due to the acceleration of gravity, all objects have the same rate of acceleration, when they free fall (no matter what their mass is). So if you divide the Netforce by the weight, you will always get 9.8 m/s/s.

I have attached a picture below, hopefuly that helps 

If you have a look at Vectors, you see that the theta is placed from the tail of a vector to another tail of another vector. If you have a look at the attached picture below, the angle will be placed in between the vectors a and c (a + b). [look at the second triangle]

Thanks so much JI2015  and Syndicate  . Really helped !!

[Adequace]

Hey guys, I need some help with the theory behind the answers of these questions.

http://imgur.com/a/v8lrM (All uploaded here)

For Ex43, Q22: The answer is B but I wrote C. Could someone clarify why it's B and not C?

For Ex45, Q34: I got the correct answer in the end but I initially just used W=Fx which was wrong. Is there a reason why W=Fx doesn't work? Is it because the force isn't constant?

Ex47, Q36 is the last part of the question above. The answer just said "since the elastic is 2x as long, therefore the mass will extend the original length by twice as much" which is 0.04x2=0.08m. Can someone explain the theory behind this?

For Ex50, Q37: the answer states "0J, because 8N will not alter the compression". I used W=Fx then subtracted 0.25J which gave me an answer >0, wouldn't using 8N to compress the spring require more work regardless?

Cheers, sorry about all the questions. (If you need more context/answers of the previous questions just ask 

[JI2015]

Hi Adequace,

I've prepared some solutions for you in the pdf. Please let me know if you need further explanation.

Some points:

Q34: As you suggested force is not constant so cannot use W=Fx but remember that W=change in Energy and we can use the Energy formulas to solve.

Q36: I used information gained from the structures and materials detailed study, if there were easier ways to explain hopefully someone points it out. However, I'm confident that you will understand my explanation.

Q37: I didn't prepare anything on this in the pdf so I'll comment here. Pay attention to the graph, the maximum extension occurs at a force of 5N. 8N of force will not change the extension and so no work is done. Also some theory, the area under the Force vs Extension graph gives the Work. There is no area between F=5 and F=8 as the extension remains 0.1m and hence no work is done.

Hope this helps you! Let me know if you understand what I have written!

http://jmp.sh/VA5TX1k

[Adequace]

Hi Adequace,

I've prepared some solutions for you in the pdf. Please let me know if you need further explanation.

Some points:

Q34: As you suggested force is not constant so cannot use W=Fx but remember that W=change in Energy and we can use the Energy formulas to solve.

Q36: I used information gained from the structures and materials detailed study, if there were easier ways to explain hopefully someone points it out. However, I'm confident that you will understand my explanation.

Q37: I didn't prepare anything on this in the pdf so I'll comment here. Pay attention to the graph, the maximum extension occurs at a force of 5N. 8N of force will not change the extension and so no work is done. Also some theory, the area under the Force vs Extension graph gives the Work. There is no area between F=5 and F=8 as the extension remains 0.1m and hence no work is done.

Hope this helps you! Let me know if you understand what I have written!

http://jmp.sh/VA5TX1k

You legend, thanks so much!

But yeah with Q36, if anyone could provide an answer with motion theory that would be great since.(regardless thanks for the working/insight JI2015)

[Syndicate]

You legend, thanks so much!

But yeah with Q36, if anyone could provide an answer with motion theory that would be great since.(regardless thanks for the working/insight JI2015)

From my knowledge and a little bit of research, the question simply states that the length of the same elastic band is no 0.4 metres with the same mass hung on it. Last time, the length was .200, so it stretched 0.04 metres, however this time, only the length has doubled, thus the extension would double too. If the mass had been halved, then the extension would have been same as the last one (0.04 metres).

I couldn't exactly find the exact theory behind this, but, you can have a look at Hooke's Law (it's about springs, however, sometimes elasticity applies too).

I also wanted to extend a little on question 34 from JL2015's description (which is amazing  )

As the body is moving towards the spring (compressing it), the kinetic energy is being transformed into potential energy. However, it's not constant as it would require more energy to compress the spring even more. As the body slows down, the kinetic energy is slowly being lost into potential energy, thus, it can't be constant (so it can't be Answer A). Then you may think, about Answer C. It is wrong, due to the slope of the graph in Answer C, as it displays that the amount of kinetic energy being lost accelerates (if you look at the end part of it), however, it should slow down, as it compresses the spring.

Hope the explanation helps [too]