VCE Physics Question Thread!

[Syndicate]

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 am still a little short of this

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.

Can you kindly show the essential parts of your working out?

Thanks   

[Adequace]

E initial = E final
EKi = EKf + Mechanical Energy(sound)
0.5mu^2 + 0.5mu^2 = 0.5mv^2 + 0.5mv^2 + 20
Sub in values then you'll get: 36 = 4v^2 + 20
Solve for v: =2m/s

[knightrider]

For this question and solution attached.

How is the normal force down ? Shouldn't the normal force be up as it opposes gravity?

I have written out my working  below shouldn't  this be correct  ?

let down be positive

[Syndicate]

For this question and solution attached.

How is the normal force down ? Shouldn't the normal force be up as it opposes gravity?

I have written out my working  below shouldn't  this be correct  ?

let down be positive

Normal force doesn't always oppose gravitational force. In this case, the toy truck will basically be weightless, thus normal force will be in the same direction as the gravitation force.

Look at this website and scroll down: http://www.ux1.eiu.edu/~cfadd/1350/06CirMtn/VertCircle.html

If you don't understand it, don't hesitate to reply back 

EDIT: you gained a negative normal force, so your working out is fine. (negative force = same direction as the gravity)

[knightrider]

Normal force doesn't always oppose gravitational force. In this case, the toy truck will basically be weightless, thus normal force will be in the same direction as the gravitation force.

Look at this website and scroll down: http://www.ux1.eiu.edu/~cfadd/1350/06CirMtn/VertCircle.html

If you don't understand it, don't hesitate to reply back 

EDIT: you gained a negative normal force, so your working out is fine. (negative force = same direction as the gravity)

Thanks Syndicate 

why will the toy truck  be weightless ?

also in my working out my final answer had the normal force as up what did i do wrong (working out is below) ?



let down be positive

[Syndicate]

Thanks Syndicate 

why will the toy truck  be weightless ?

also in my working out my final answer had the normal force as up what did i do wrong (working out is below) ?



let down be positive

You did everything right, except the last line. You already got a negative normal force (-7.31 N), which means that the normal force will face in the direction of gravity (which will always be downwards, so it should be 7.31 Ndown)

As for the toy truck being weightless.

As both forces (gravity and normal force) will be facing downwards, the toy truck will seem to be weightless.

Think it about this way:

When you freefall, you would feel weightless, as there is no force impacting in the opposite direction of gravity ( I think, there is no normal force, when an object is free falling).

[knightrider]

You did everything right, except the last line. You already got a negative normal force (-7.31 N), which means that the normal force will face in the direction of gravity (which will always be downwards, so it should be 7.31 Ndown)

As for the toy truck being weightless.

As both forces (gravity and normal force) will be facing downwards, the toy truck will seem to be weightless.

Think it about this way:

When you freefall, you would feel weightless, as there is no force impacting in the opposite direction of gravity ( I think, there is no normal force, when an object is free falling).

Thanks Syndicate 

[knightrider]

How would you do this question attached.

[Syndicate]

How would you do this question attached.

Hey Knightrider,

You can workout the angle by simply using the formula:

where v = velocity (18 m/s)
r = radius (80)
g = gravity (9.8 m/s)

so if you sub in the quantities


=

=

which can be rounded to a whole number of 22 degrees

hopefully this helps 

[knightrider]

Hey Knightrider,

You can workout the angle by simply using the formula:

so if you sub in the quantities
=
=
=

which can be rounded to a whole number of 22 degrees

hopefully this helps 

Thanks Syndicate 

Where did you get this formula from  ?

[Syndicate]

Thanks Syndicate 

Where did you get this formula from  ?

I discovered it last year, while reading over my cousin's Physics notes, however, I am sure, it can be found on a webpage 

[JI2015]

These are the equations from which the banked corner formula is derived from. Fc is the centripetal force (net force). I'll try to find a sheet I have that explains this in detail but here are the equations.

[lzxnl]

The idea is:

Object in a banked curve is not moving vertically (horizontal plane) so the vertical acceleration has to be zero. Vertical component of net force is zero. Through geometry, the incline angle happens to also be the angle between the normal and the vertical. It makes sense as the incline and normal are at 90 degrees and the horizontal/vertical are at 90 degrees.

Hence, resolving the forces vertically gives N cos theta = mg and N sin theta = mv^2/r (centripetal acceleration requirement as the net force is directed radially towards the centre, much like the horizontal force component)

[knightrider]

These are the equations from which the banked corner formula is derived from. Fc is the centripetal force (net force). I'll try to find a sheet I have that explains this in detail but here are the equations.

The idea is:

Object in a banked curve is not moving vertically (horizontal plane) so the vertical acceleration has to be zero. Vertical component of net force is zero. Through geometry, the incline angle happens to also be the angle between the normal and the vertical. It makes sense as the incline and normal are at 90 degrees and the horizontal/vertical are at 90 degrees.

Hence, resolving the forces vertically gives N cos theta = mg and N sin theta = mv^2/r (centripetal acceleration requirement as the net force is directed radially towards the centre, much like the horizontal force component)

Thanks JI2015 and lzxnl. Really helped 

[knightrider]

How would you do this question attached?