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Author Topic: VCE Physics Question Thread!  (Read 603326 times)  Share 

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skrt skrt

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Re: VCE Physics Question Thread!
« Reply #1950 on: January 01, 2018, 12:21:02 am »
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This is question 10 (7.1)in the hinemann year 12 book.A tennis ball of mass 57.5g is tested for compliance with tennis regulations by being dropped from a height of 251cm onto concrete. A bounce height of 146cm is deemed acceptable. Find the magnitude of the average force on the ball if it is in contact with the floor for 0.0550s.


Answer is 12.9N
Thanksss
« Last Edit: January 01, 2018, 12:23:04 am by skrt skrt »
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kalopsia

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Re: VCE Physics Question Thread!
« Reply #1951 on: January 01, 2018, 09:23:34 am »
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This is question 10 (7.1)in the hinemann year 12 book.A tennis ball of mass 57.5g is tested for compliance with tennis regulations by being dropped from a height of 251cm onto concrete. A bounce height of 146cm is deemed acceptable. Find the magnitude of the average force on the ball if it is in contact with the floor for 0.0550s.






Answer is 12.9N
Thanksss

Please find my working out attached.

The average force is f=mΔv/Δt. So you have to find the Δv, which is the change in velocity the ball experiences when it is in contact with the ground.
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skrt skrt

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Re: VCE Physics Question Thread!
« Reply #1952 on: January 02, 2018, 11:27:34 pm »
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Hi guys need help with this question. A skier wishes to have skiing equipment of mass 200kg transported tot he top of the downhill ski slope. The items are pulled on a cart by a rope with a force Fa and in doing so, a constant friction force of 100N is encountered.
For the cart to be towed at a constant speed, what must be the magnitude of the applied force?
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chooby

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Re: VCE Physics Question Thread!
« Reply #1953 on: January 02, 2018, 11:35:53 pm »
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Hi guys need help with this question. A skier wishes to have skiing equipment of mass 200kg transported tot he top of the downhill ski slope. The items are pulled on a cart by a rope with a force Fa and in doing so, a constant friction force of 100N is encountered.
For the cart to be towed at a constant speed, what must be the magnitude of the applied force?
Here you would need to use Newton's second law (F=ma) and forces addition. Through newtons second law, we know the net force is 0 as there is a constant speed and there for 0 acceleration. Through force addition we know that net force = Fa - 100 (assuming that Fa is the positive direction). Then you equate so that 0 =Fa - 100 , then Fa = 100N. Direction not needed as only magnitude was asked for.
« Last Edit: January 02, 2018, 11:38:14 pm by chooby »
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skrt skrt

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Re: VCE Physics Question Thread!
« Reply #1954 on: January 02, 2018, 11:54:04 pm »
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Crap, I forgot to mention there's an incline of 20 degrees to the horizontal. Sorry for the inconvenience
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chooby

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Re: VCE Physics Question Thread!
« Reply #1955 on: January 03, 2018, 12:05:34 am »
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All good. Most of is it the same but it does add another force that must be taken into account and that is the gravitational force of the load (Fg) which must be accounted for.

Fg = mg, so that is ( 20 )*( 9.8 ) which means Fg = 196N.

We can break Fg down into its vertical and horizontal component. However, we only need to find the horizontal compenent as it is parallel to the direction of motion. So using trig, the horizontal component is 67.04N in the same direction as the frictional force.

The rest is the same with net force = 0 due to newtons second law but now net force is also Fa - 100 - 67.04. So when you equate these, Fa = 167.04N
« Last Edit: January 03, 2018, 12:12:04 am by chooby »
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skrt skrt

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Re: VCE Physics Question Thread!
« Reply #1956 on: January 14, 2018, 01:35:17 am »
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Hey people, just wondering why the time is the same for horizontal and vertical component for motion.
Thanks
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Calebark

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Re: VCE Physics Question Thread!
« Reply #1957 on: January 14, 2018, 02:09:21 am »
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Hey people, just wondering why the time is the same for horizontal and vertical component for motion.
Thanks

Even though we use horizontal and vertical components, you need to remember that they're still measuring the same thing overall - the motion of a single object. A single object can't have two different time measurements. It might be easier to answer your question (as I may have missed the point) if I explain why we only use the vertical component of motion to measure time.

In VCE Physics, we assume that air resistance in projectile motion is negligible; it doesn't matter. This means that there is nothing slowing down the object in the horizontal component. However, overall, it DOES slow down, which shouldn't happen if we just consider the horizontal component (can you recall Newton's First Law of motion?), meaning the vertical component is acting upon the object to slow it down. What's a force that's always acting on an object in the vertical component? Gravity. Gravity is the only thing slowing an object down in the vertical component, meaning gravity is the only thing slowing down an object in projectile motion.

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Re: VCE Physics Question Thread!
« Reply #1958 on: January 14, 2018, 09:19:23 am »
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Hey people, just wondering why the time is the same for horizontal and vertical component for motion.
Thanks

Once the body hits the ground/other surface it stops moving

Edit: (We don't think about skidding, bouncing etc. in this course)

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Re: VCE Physics Question Thread!
« Reply #1959 on: January 19, 2018, 11:39:30 pm »
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I'm not sure how to do this question: (attached)

I tried finding the change in gravitational field strength (which will be equal to the kinetic energy). However, that graph only goes up to 10,000km, and in order to find enough information for the field strength, I need to go to about 14,000Km (I think the question assumes you know that the surface is around 6000Km from the core).

Is there another approach or am I missing something?

EDIT:I think I'm onto something

EDIT 2: I got it.
« Last Edit: January 19, 2018, 11:46:32 pm by A TART »
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Re: VCE Physics Question Thread!
« Reply #1960 on: January 19, 2018, 11:47:01 pm »
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Hey people, just wondering why the time is the same for horizontal and vertical component for motion.
Thanks
It is because the entire flight of the projectile will take a given time, and the horizontal and vertical components are simply breaking that individual motion up into parts so that it can be analysed more easily.
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Re: VCE Physics Question Thread!
« Reply #1961 on: January 19, 2018, 11:50:32 pm »
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I'm not sure how to do this question: (attached)

I tried finding the change in gravitational field strength (which will be equal to the kinetic energy). However, that graph only goes up to 10,000km, and in order to find enough information for the field strength, I need to go to about 14,000Km (I think the question assumes you know that the surface is around 6000Km from the core).

Is there another approach or am I missing something?

EDIT: I think I'm on something


Reread the question. The maximum distance is 8000 km from the center  (the starting position is 600km + radius of Earth away).  Thus all of the needed information is provided. 

Finding the area under the graph (between the two distances) effectively finds you g * h  so then you just need to multiply that by m to get the change in gravitational potential energy and therefore the initial kinetic energy

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Re: VCE Physics Question Thread!
« Reply #1962 on: January 20, 2018, 12:26:17 am »
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Reread the question. The maximum distance is 8000 km from the centre (the starting position is 600km + radius of Earth away).  Thus all of the needed information is provided. 

Finding the area under the graph (between the two distances) effectively finds you g * h  so then you just need to multiply that by m to get the change in gravitational potential energy and therefore the initial kinetic energy


Yeah... I actually found the Mass of Earth then did the question. Then I realised the question stated "8000Km from the CENTRE of eath" not surface XD




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Re: VCE Physics Question Thread!
« Reply #1963 on: January 20, 2018, 07:32:02 am »
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Yeah... I actually found the Mass of Earth then did the question. Then I realised the question stated "8000Km from the CENTRE of eath" not surface XD






Not sure if this was a typo or not,  but you should be using the mass of the communications satellite (not earth) in your equation.
As you are finding the change in max GPE of the satellite all the variables you're using in the equation should be relevant  to it.

The only impact Earth's mas here is that it influences g - but that's been given to you anyway

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Re: VCE Physics Question Thread!
« Reply #1964 on: January 22, 2018, 07:55:25 pm »
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Hi, I'm struggling with two physics questions, hoping for a little bit of help please!

1) 0.5 kg of ice at 0 degrees is mixed with 0.1kg of steam at 100 degrees. What will be the final temperature?

2) A student attempts to identify metal by measuring the specific heat capacity. 100g of the metal is heated to 75 degrees and then transferred to a 70g copper calorimeter containing 200g of water at 20 degrees. The temperature of the final mixture is 25 degrees? What metal is the student testing? ( given a bunch of metals with their approximate specific heat capacity)

Thanks in advance!