VCE Physics Question Thread!

[Syndicate]

You literally just write out all of the angles and go angle-hunting.
Think about it this way. The normal is 90 degrees to the incline. Vertical is 90 degrees to horizontal. It makes sense that the angle between the normal and the vertical is the same as the angle between the horizontal and the incline.

I attached a screenshot with my last post, is that what you mean? But I didn't add the angle between the horizontal (net/ centripetal force) and the incline.

[knightrider]

You literally just write out all of the angles and go angle-hunting.
Think about it this way. The normal is 90 degrees to the incline. Vertical is 90 degrees to horizontal. It makes sense that the angle between the normal and the vertical is the same as the angle between the horizontal and the incline.

Thanks for clarifying !!

[lzxnl]

I attached a screenshot with my last post, is that what you mean? But I didn't add the angle between the horizontal (net/ centripetal force) and the incline.

Yeah, but you need to relate the two angles in your diagram.

[sweetiepi]

Can somebody please explain how to do the question attached? 

[lzxnl]

Can somebody please explain how to do the question attached? 

In 1 second there are 2 revolutions.
A period is the time for 1 revolution.
How many seconds needed?

[sweetiepi]

Thanks for the help!

[Maz]

hey guys...i kinda somewhere in my brain know how to do this question and it has to do with centripetal and centrifugal forces...
the question: Explain why curved railway tracks are usually banked towards the inside of the curve? and I'm supposed to use a force vector diagram to illustrate

some help would be much appreciated
Thankyou sooooo much in advance

[Syndicate]

hey guys...i kinda somewhere in my brain know how to do this question and it has to do with centripetal and centrifugal forces...
the question: Explain why curved railway tracks are usually banked towards the inside of the curve? and I'm supposed to use a force vector diagram to illustrate

some help would be much appreciated
Thankyou sooooo much in advance

Hey mq123,

The train tracks are usually banked (on a curve), as fast moving trains has a great amount of Inertia, it tends to move tangentially off the track (contrifugal force), however, when the tracks are banked, it produces the necessary amount of centripetal force to keep the train moving in the curvature at the same speed. Friction can't provide the required amount of Centripetal force, and thus, the tracks must be banked for the reason being.

The vertical forces (Normal and Gravitation force) balances out the weight, while the Horizontal forces will provide the necessary centripetal force.

Your vector diagram would look like:

Net force (centripetal force) facing towards the center of the curvature.
Normal force facing directly vertical from the banked curve (on a tilt)
Gravitation force facing directly downwards

[lzxnl]

Hey mq123,

The train tracks are usually banked (on a curve), as fast moving trains has a great amount of Inertia, it tends to move tangentially off the track (contrifugal force), however, when the tracks are banked, it produces the necessary amount of centripetal force to keep the train moving in the curvature at the same speed. Friction can't provide the required amount of Centripetal force, and thus, the tracks must be banked for the reason being.

The vertical forces (Normal and Gravitation force) balances out the weight, while the Horizontal forces will provide the necessary centripetal force.

Your vector diagram would look like:

Net force (centripetal force) facing towards the center of the curvature.
Normal force facing directly vertical from the banked curve (on a tilt)
Gravitation force facing directly downwards

Centrifugal forces are a little problematic because they're fictitious and don't actually exist. A centrifugal force only appears if you're considering the forces felt by an accelerating object. In other words, they come about because the object doesn't have the right force to move with that acceleration.

Your explanation hasn't mentioned where this centripetal force comes from. Its origin is the tilt of the banked curve; as now the normal to the banked curve has a horizontal component, the normal reaction force will partially push vertically and partially push horizontally. The vertical component balances its weight force (it's a horizontal circle, no vertical motion) and the horizontal component is equal to the centripetal force.

[TheAspiringDoc]

Hey, here's a real life physics application for you guys

I just found the Mazerunner #2 DVD in our freezer.. (LOL mum)
I don't have my proper computer with me at the moment (i.e. can't play the DVD yet) but will the DVD still work? :O

[Syndicate]

Hey, here's a real life physics application for you guys

I just found the Mazerunner #2 DVD in our freezer.. (LOL mum)
I don't have my proper computer with me at the moment (i.e. can't play the DVD yet) but will the DVD still work? :O

well offcourse... xD (unless it's fully scratched  )

Your explanation hasn't mentioned where this centripetal force comes from. Its origin is the tilt of the banked curve; as now the normal to the banked curve has a horizontal component, the normal reaction force will partially push vertically and partially push horizontally. The vertical component balances its weight force (it's a horizontal circle, no vertical motion) and the horizontal component is equal to the centripetal force.

yea, but he only asked if why the train tracks are banked (when going through a curve)

Centrifugal forces are a little problematic because they're fictitious and don't actually exist. A centrifugal force only appears if you're considering the forces felt by an accelerating object. In other words, they come about because the object doesn't have the right force to move with that acceleration.

It's good to learn something new everyday 

[Maz]

Hey, here's a real life physics application for you guys

I just found the Mazerunner #2 DVD in our freezer.. (LOL mum)
I don't have my proper computer with me at the moment (i.e. can't play the DVD yet) but will the DVD still work? :O

I'm still confused on how a situation like that happenes -LOL
it will work- water doesn't do anything to a DVD...the actual data burned on the dvd is a couple of layers deep- not on the surface...which is why scratches sometimes harm and sometimes don't harm the disc...but the ice may make a few marks (possibly)
Good Luck 

[bts]

Hello!
is someone able to help me here please

A bullet is fired from a gun with a barrel of 50cm, towards a target exactly 100m away. To fire a bullet, gunpowder is ignited and the expanding gasses that result from the burning gunpowder applies a constant force to the bullet while it travels down the barrel of the gun. The force that is applied to the bullet is 100N, for each cm2 of surface area of the bullet, per gram of gunpowder used. The bullet is 9mm in diameter and has a mass of 5g. The bullet leaves the gun at a velocity of 500m/s and experiences negligible air friction.
6. How much gunpowder, rounded to the nearest gram, was used to fire the bullet?

thank you in advance

[lzxnl]

Hello!
is someone able to help me here please

A bullet is fired from a gun with a barrel of 50cm, towards a target exactly 100m away. To fire a bullet, gunpowder is ignited and the expanding gasses that result from the burning gunpowder applies a constant force to the bullet while it travels down the barrel of the gun. The force that is applied to the bullet is 100N, for each cm2 of surface area of the bullet, per gram of gunpowder used. The bullet is 9mm in diameter and has a mass of 5g. The bullet leaves the gun at a velocity of 500m/s and experiences negligible air friction.
6. How much gunpowder, rounded to the nearest gram, was used to fire the bullet?

thank you in advance

I'll simplify the wording of the question.
1. Bullet experiences constant force from gas.
2. This gas acts over 50 cm
3. Constant force is given by 100 N/cm^2 * area of bullet (calculate assuming bullet is a sphere) * mass of gunpowder (need to work this out)
4. Assume bullet is initially at rest, and after the force acts on it 50 cm later it's moving at 500 m/s
5. What is the force acting on the bullet, and consequently what is the mass of gunpowder used? (calculate using point 3)

[bts]

To calculate force do I use F=(mv^2)/r
where m is 5g, v is 500m/s and the r is 50cm?

the answer is apparently around twenty but i cant seem to get it using this

(0.005x500^2)/0.5 = 2500

2500 = 100 x 4pi(0.45^2) x gram of gunpowder used?

the answer wouldve been 9.8 if that was the case... what am i doing wrong?