VCE Physics Question Thread!

[Adequace]

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] 

Thanks

Edit: In regards to Ex36, I'm still not entirely sure on how doubling the length of the same elastic = a doubled extension length. I can't really apply Hooke's law to this doubled elastic length situation either(since x=compression length).

[lzxnl]

Mathematically, if your force isn't constant, then to find the energy, you need to find the area under a force/displacement graph.

Here's my take on things.

Q22: 1/2 mv^2 + 1/2 kx^2 = constant
So the relationship between KE and x should be quadratic in x; negative parabola.

Q34: as above

Q37: work, by definition, is a force acting on a distance. If your force doesn't move anything, there is no work being done. You can think of work as a change in kinetic energy. If there is no change in displacement, then your object isn't moving any faster or smaller -> no work done.

How to deal with Q36?
What is the origin of this spring force? You have bonds between atoms that make up the spring force. By using Hooke's law, we're approximating this force as being linear in the displacement, which is generally a pretty good approximation. Now, if your spring is twice as long with the same material, you have twice as many of the same identical atoms that pull on each other. The quadratic nature of the energy means that the most stable configuration is when all of the atoms are the same distance from each other (this can be proved using some version of the AM-QM inequality I think but I'm not going to go into the details here).

Now, let's look at the atomic cross section that is attached to the mass. This layer of atoms has to bear the entire weight; by Newton's third law, if it is pulling the mass up, the mass pulls this layer of atoms down. The next layer of atoms therefore has to exert a force equal to the weight force on the first layer of atoms to hold them up. Apply this logic to all of the layers of atoms there and you'll find that the extension = number of layers * extension between layers

Double the length of the spring -> double the number of layers of atoms -> double the total extension

[Adequace]

Thanks lzxnl, helped a lot 

[knightrider]

In order to demonstrate an inelastic collision

is it fine to just show that the initial kinetic energy does not equal the final kinetic energy to receive full marks ?

[odeaa]

In order to demonstrate an inelastic collision

is it fine to just show that the initial kinetic energy does not equal the final kinetic energy to receive full marks ?

yeah, just calculate the initial and final energies (must be correct values, usually a mark here) and then make a quick statement saying that because initial and final are not equal it is inelastic

[knightrider]

yeah, just calculate the initial and final energies (must be correct values, usually a mark here) and then make a quick statement saying that because initial and final are not equal it is inelastic

Thanks for clarifying odeaa 

[knightrider]

How would you do this question attached?

[Syndicate]

How would you do this question attached?

hey knighrider,

formula:
= the subsequent speed
m1 = 10kg
m2 = 10kg
v= 5m/s

50/20 = 2.5m/s

Edited^

hope this helps 

[knightrider]

hey knighrider,

formula:
= the subsequent speed
m1 = 10kg
m2 = 10kg
v= 5m/s

50/20 = 2.5m/s

Edited^

hope this helps 

Thanks Syndicate 

I also found a way of doing it (refer to below  )







speed is

[Syndicate]

Thanks Syndicate 

I also found a way of doing it (refer to below  )







speed is

yes, that is certainly a way of writing the formula down in a different state, but its a good idea, remembering both equations, as you may never know, what comes on the SAC.

[knightrider]

How would you do this question relating to image attached?

Calculate the speed of each ball after the collision.

[Syndicate]

How would you do this question relating to image attached?

Calculate the speed of each ball after the collision.

That questions seems a little inaccurate, as KE = 1/2 x mv^2, which is this case is 18 J and if 20 J is lost, thus, while calculating the final amount of Kinetic Energy, we would get -2 J, which is not possible (as there is nothing like negative kinetic energy)

[nerdgasm]

Hmm, I think the question is saying that the 20J is the total amount of kinetic energy lost, not the energy lost from each ball. So, I think you're meant to work out the amount of kinetic energy 'remaining', then work out how much kinetic energy each ball has after the collision (it should be the same amount for both balls), and use that to work out the final speed of each ball.

[Adequace]

Hmm, I think the question is saying that the 20J is the total amount of kinetic energy lost, not the energy lost from each ball. So, I think you're meant to work out the amount of kinetic energy 'remaining', then work out how much kinetic energy each ball has after the collision (it should be the same amount for both balls), and use that to work out the final speed of each ball.

I did the same as nerdgasm, I got 2m/s as my final answer. If it's correct I can post my working if you want but you're pretty much going E initial = E final as your first line.

[knightrider]

I did the same as nerdgasm, I got 2m/s as my final answer. If it's correct I can post my working if you want but you're pretty much going E initial = E final as your first line.

Yep thats the right answer 

Could you please post your working.