An ∃mazing Physics thread (Unit 2)

[Guest]

A few problems that come to mind here.

Firstly, where does the horizontal momentum go? Very unrealistic.

Secondly, an analysis using conservation of energy is incomplete the way you guys have done it. What you've done is work out the height at which the person will have a speed of 1.5 ms^-2 WITH NO ROTATIONAL MOTION.
Now...I don't ever recall seeing a pole vault event where someone clears the bar without rotating their body.

Dude, I'm pretty sure this is the Unit 1&2 Thread.....and I don't remember covering rotational motion.
You should know by now that the VCE curriculum itself isn't anywhere near ideal. Even the teachers at ours rant about it.

[RKTR]

Ahh right. Makes much more sense. However, I don't get what you've done with your negative values. Isn't acceleration due to gravity -10, since he is jumping upwards and opposing the acceleration of gravity? In order to find the velocity change at the top isn't it 1.5²-9.1²?

Therefore: mgh=0.5 x 84 x (1.5² -9.1²) 
                     84 x 10 x h =0.5 x 84 x (1.5² -9.1²)
Equals 4.03m, but now is the working out how it should be (with appropriate negatives and all)?

i think a better way to say what i did is

the decrease in k.e = the increase in g.p.e

the change in k.e is negative but the decrease in k.e is positive ya?

[SocialRhubarb]

where does the horizontal momentum go?

The earth.

[Guest]

i think a better way to say what i did is

the decrease in k.e = the increase in g.p.e

the change in k.e is negative but the decrease in k.e is positive ya?

Yes, that's true as well. Might as well jut consider gravity as an opposition to upward motion and set it to -10. Less confusion. Moving on.

[lzxnl]

The earth.

OK, if you're telling me the guy can push the pole into the ground and exert a force to convert exactly all of his horizontal momentum to vertical, fine. Seems a bit idealistic to me.

Dude, I'm pretty sure this is the Unit 1&2 Thread.....and I don't remember covering rotational motion.
You should know by now that the VCE curriculum itself isn't anywhere near ideal. Even the teachers at ours rant about it.

I know this is the 1/2 thread. You can appreciate, however, that a ball moving at 10 m/s but non-rotating has less energy than a ball moving at 10m/s and spinning. The example isn't the best.

[Guest]

Came across a weird question on the 'Area of Study Review' section. I don't understand the relevance of Hz to time or speed.

The following information applies to questions 1–2. During a physics experiment a student sets a multi-flash timer at a frequency of 10 Hz. a nickel marble is rolled across a horizontal table. The first four flashes are in the order: A,B,C and D. The distance from A to B is 1.0cm. The distance from B to C is 3.0cm. The distance from C to D is 5.0cm.
Assume that when flash a occurred t = 0, at which time the marble was at rest.

1) Determine the average speed of the marble as it traveled from A to B.
2) Determine the instantaneous speeds of the marble when t=0.05 s.

Can anyone see the link between Hz and time yet? Was my connection between both correct?

For Question 1, I designated the following values: u=o, x=0.01, t=1/6, v=?
Notice t=1/6. I'm not entirely sure if it should be 1/6. I simply got that by dividing 10 by 60. Is this correct?
My answer is close (0.12) to the book's answer (0.1). Need to know what the time value should be and why.

Thanks

[Professor Polonsky]

Frequency is a measure of how often (frequent) something occurs. The unit of measure hertz signifies how many times it occurs per second. Therefore, if something has a frequency of 10 Hz it occurs once every 1/10 s.

[Guest]

Another question here.

A rifle of mass 2.2kg is attached to a free rolling trolley of mass 1.5kg. A bullet of mass 32 g is fired from the rifle and the rifle recoils with a speed of 4.5 ms/s. the friction caused by the movement of the trolley can be considered to be negligible.

10) What is the speed of the bullet as it leaves the rifle?
11) What is the kinetic energy of the bullet immediately after it leaves the rifle?

[RKTR]

Another question here.

A rifle of mass 2.2kg is attached to a free rolling trolley of mass 1.5kg. A bullet of mass 32 g is fired from the rifle and the rifle recoils with a speed of 4.5 ms/s. the friction caused by the movement of the trolley can be considered to be negligible.

10) What is the speed of the bullet as it leaves the rifle?
11) What is the kinetic energy of the bullet immediately after it leaves the rifle?

momentum before = momentum after
0 = (2.2+1.5)x(-4.5) + 0.032 v

[Guest]

momentum before = momentum after
0 = (2.2+1.5)x(-4.5) + 0.032 v

Ohhhh right!
Thanks

[Guest]

I faced a difficult question on the test yesterday. It was:

A bullet is shot and it hits a crate which is on the ground but hanging from a cable. The crate is 20kg and is hanging from a 1 meter cable. As soon as the bullet hits the crate, the crate moves upwards 55 degrees, with the bullet in it.

a) Calculate the velocity of both the bullet and the crate after the bullet has hit the crate.
b) Calculate the initial velocity of the bullet before it hit the crate.

I have attached a diagram that helps clarify the question.

[lzxnl]

I faced a difficult question on the test yesterday. It was:

A bullet is shot and it hits a crate which is on the ground but hanging from a cable. The crate is 20kg and is hanging from a 1 meter cable. As soon as the bullet hits the crate, the crate moves upwards 55 degrees, with the bullet in it.

a) Calculate the velocity of both the bullet and the crate after the bullet has hit the crate.
b) Calculate the initial velocity of the bullet before it hit the crate.

I have attached a diagram that helps clarify the question.

The first bit is just conservation of energy. Let your initial speed be v, right after the collision. You can work out the height, namely cos 55 degrees, that the crate + bullet travel. Then, at max height, speed = 0. So 1/2 mv^2 = mgh
v=sqrt(2gh). Work out v.

The second bit is conservation of momentum. Instantaneously before and after the collision, the bullet moves in the same direction. With what speed? Let bullet mass be m. mu=(m+20)*v from conservation of momentum. Work out u.

[Guest]

So I've finished motion for now and have since begun on Electricity. Is it just me or does the Heinemann books make this chapter sound really winding and tedious, when it really isn't meant to be? Take the 2 page "useful analogy" which could have been said clearly enough in possibly two paragraphs at most. With that being said, I'm probably going to flunk this chapter, as I'm already having difficulties at the 2.3 questions. I've ended up making up my own methods to tackle these questions as this chapter is so badly explained. Now I don't even know if their correct or not. If anyone could just verify if what I've done is correct, that'd be great.
Thanks, in advance, for helping me out for the rest of this chapter.

Question:
3a) In a solution of salt water a total positive charge of +15C was seen to move past a point to the right in 5s, and in the same time a total negative charge of -30C was seen to move to the left.  What was the current through the solution during this time?
3b) Some time later it was found that in 5s a total of +5C had moved to the right while -15C had moved to the right as well. What was the current during this time?
My solution:
3a) right
      left = right
      right.
3b) right
      right = left
      = left

[Guest]

Confused with a question on the textbook. I have attached a photo of it, since it refers to a graph.
I'm particular confused on part (b).
I don't understand how internal resistance can be calculated on a non-ohmic device. I am assuming it's non ohmic because it does not pass through point (0,0).

[SocialRhubarb]

Okay, so you've got a battery, but the problem with batteries is that the battery itself has some resistance, which is internal resistance, which you probably already know.

So what you've really got is a circuit which looks like this:

.

R1 is the internal resistance of your battery, and the 'voltage' you're measuring is the voltage across R2.

You can work out the EMF of the cell by drawing a line from the points given on the graph and looking at the x-intercept, which turns out to be 1.6V.

For b.), if you look at the y-intercept of the graph, you get a current when there is no voltage across the output; that is, when all the voltage is going across the battery's internal resistance. Using , we get a value of for R.