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March 28, 2024, 07:39:01 pm

Author Topic: VCE Physics Question Thread!  (Read 603139 times)  Share 

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odeaa

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Re: VCE Physics Question Thread!
« Reply #810 on: January 27, 2015, 09:47:09 pm »
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Yeah my bad, I ahould have drawn a graph but I was too lazy
I'm gonna have a crack at the first one but no promises
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Cosec

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Re: VCE Physics Question Thread!
« Reply #811 on: January 27, 2015, 09:55:27 pm »
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Yeah my bad, I ahould have drawn a graph but I was too lazy
I'm gonna have a crack at the first one but no promises

Haha no worries. The problem is when i try to solve for t im getting a negative value. Which isnt right. My dads a engineer and cant work it out either.  :o :)

Cosec

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Re: VCE Physics Question Thread!
« Reply #812 on: January 27, 2015, 10:24:12 pm »
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All good dude. Found my error with some help. The distance covered in the first 15 seconds travelled i calculated was right, when i was solving for time i forgot it should be (for the cop car) t-15 and not just t. Cause the distance added to the first 15 seconds is the time after 15 seconds (hence the t-15 bit. All good. Thanks buddy.

odeaa

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Re: VCE Physics Question Thread!
« Reply #813 on: January 27, 2015, 10:27:46 pm »
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All good dude. Found my error with some help. The distance covered in the first 15 seconds travelled i calculated was right, when i was solving for time i forgot it should be (for the cop car) t-15 and not just t. Cause the distance added to the first 15 seconds is the time after 15 seconds (hence the t-15 bit. All good. Thanks buddy.

Gave this one a go, have no idea if it's right but it's feasible
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dankfrank420

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Re: VCE Physics Question Thread!
« Reply #814 on: January 27, 2015, 10:35:27 pm »
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Edit: fucked it up
« Last Edit: January 27, 2015, 10:48:33 pm by dankfrank420 »

Kel9901

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Re: VCE Physics Question Thread!
« Reply #815 on: January 28, 2015, 09:45:29 am »
+1
Can someone answer these couple of questions. Ive answered them and my questions different ot the BOB. Checked over them and they appear right (probs wrong).

1) A child rolls a 50 gram marble up a playground slide that is inclined at 15 degrees to the horizontal. The slide is 3.5m long and the marble is launched with a speed of 4.8m/s.
How fast is the marble travelling when it is halfway up the slide?

2) A car travelling with a constant speed of 80km/hr passes a stationary motorcycle policeman. The cop sets off in pursuit, acclerating uniformly to 80km/hr in 10secs, and 100km/hr after a further 5 secs. he stays at 100km/hr for the rest of the journey. At what time will the policeman catch up with the car?
I know the distance of both need to be the same and you solve for time, but i keep getting a different answer.

3) Cassie starts from rest at the top of a 3.2m long smooth playground slide and slides to thwe bottom with a constant acceleration. If she takes 2.4 secs to reach the bottom. Calculate her average acceleration?

Last bit. If your asked to find the speed of a golf ball which has fallen 2 meters and find the speed when it has rebounded. How is this done. I got the right answer but im trying to think why?
For speed would it be just pick a point that is jsut above the point of impact (ie. 0.5 meters) assuming that down is engative? So youve picked a point on its upwards journey.
And for velocity, the displacement would be say 1.5m from the top?
Im just kind of making a educated guess here ^.

1. u=4.8, a=10sin(15)=-2.6, s=1.75, v=?
v^2=u^2+2as=23.04-9.06=13.98
v=3.74 ms^-1

You can also use energy conservation like this:
Bottom: GPE=0, KE=1/2 mv^2=0.576
1/2 way up slide: GPE=mgh=mg(1.75sin(15))=0.23
KE=0.576-0.23=0.35=1/2 mv^2
v^2=14
v=3.74 ms^-1

2. 80 km/hr=22.2 ms^-1
100 km/hr=27.8 ms^-1
In the first 15 seconds:
Car travels 22.2*15=333.3 m
Policeman travels 1/2*10*22.2 + 1/2*5*(22.2+27.8)=236.1 m

After exactly 15 seconds the policeman is behind by 97.2 m, and the difference in velocity is 27.8-22.2=5.6 ms^-1

The policeman takes a further 97.2/5.6=17.5 s to catch up, so total time taken is 15+17.5=32.5 s

3. u=0, t=2.4, s=3.2, a=?
s=ut+1/2 at^2
3.2=2.88a
a=1.11 ms^-2

For the last question, you can find out the KE just before the ball hits the ground (=GPE at top=mgh=20m). Assuming an elastic collision (I think that's what you're supposed to assume), the KE just after the ball bounces is also 20m. This means that in both cases, the speed can be calculated via:
20m=KE=1/2 mv^2
v^2=40
v=6.32 ms^-1
s=change in displacement for physics
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Kel9901

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Re: VCE Physics Question Thread!
« Reply #816 on: January 29, 2015, 10:18:29 am »
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(Image removed from quote.)
Gave this one a go, have no idea if it's right but it's feasible

You can't just add initial speed in; remember that energy is proportional to speed squared, so you have to calculate kinetic energy initially, gravitational potential energy at the end and subtract and calculate
s=change in displacement for physics
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odeaa

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Re: VCE Physics Question Thread!
« Reply #817 on: January 29, 2015, 10:37:53 am »
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You can't just add initial speed in; remember that energy is proportional to speed squared, so you have to calculate kinetic energy initially, gravitational potential energy at the end and subtract and calculate
I get it now, thanks so much! I asked the class genius on facebook how to do that, I didn't even doubt his answer for one second ahaha I can't believe he was wrong
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JackSonSmith

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Re: VCE Physics Question Thread!
« Reply #818 on: February 04, 2015, 05:32:14 pm »
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A speeding motorbike travels past a stationary police car at a speed of 35 m/s. The police car starts accelerating immediately at 4 m/s^2, and keeps accelerating at this rate until it has passed the bike.

a) How far does the police car travel before it overtakes the motorbike?

b) At what time does the police car overtake the motorbike? 
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odeaa

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Re: VCE Physics Question Thread!
« Reply #819 on: February 04, 2015, 05:33:20 pm »
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A speeding motorbike travels past a stationary police car at a speed of 35 m/s. The police car starts accelerating immediately at 4 m/s^2, and keeps accelerating at this rate until it has passed the bike.

a) How far does the police car travel before it overtakes the motorbike?

b) At what time does the police car overtake the motorbike?
Almost this exact same question got answered a few days ago if you scroll back, different values but same concepts
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JackSonSmith

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Re: VCE Physics Question Thread!
« Reply #820 on: February 04, 2015, 05:35:27 pm »
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I am having trouble working this one out as I can not seem to find time taken. I had read the previous question before asking.
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odeaa

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Re: VCE Physics Question Thread!
« Reply #821 on: February 04, 2015, 05:37:42 pm »
+1
Sorry if I came across rude
The time taken if you arrange d=vt is t=d/v
If you sub in 35m/s from the constant speed of the bike you will get an answer
I'm assuming you got the distance?
« Last Edit: February 04, 2015, 05:53:17 pm by odeaa »
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Kel9901

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Re: VCE Physics Question Thread!
« Reply #822 on: February 04, 2015, 07:49:32 pm »
+1
A speeding motorbike travels past a stationary police car at a speed of 35 m/s. The police car starts accelerating immediately at 4 m/s^2, and keeps accelerating at this rate until it has passed the bike.

a) How far does the police car travel before it overtakes the motorbike?

b) At what time does the police car overtake the motorbike?

a) , b)
Motorbike: s=ut=35t
Car: u=0, a=4, s=?, t=?
s=ut+1/2 at^2
s=2t^2

So the equations are s=35t and s=2t^2
35t=2t^2
t=35/2 (as t>0)=17.5 s
s=35t(=2t^2)=612.5 m
s=change in displacement for physics
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Cosec

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Re: VCE Physics Question Thread!
« Reply #823 on: February 04, 2015, 07:56:44 pm »
+3
I am having trouble working this one out as I can not seem to find time taken. I had read the previous question before asking.

Here:
Imgur: http://imgur.com/qCeuwvq

Basically, when the cop pass the car? When their displacement is the same.
Therefore, its easiest to draw a simple graph and set up two equations like shown and solve for time.

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Re: VCE Physics Question Thread!
« Reply #824 on: February 06, 2015, 05:25:03 pm »
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Hey Guys, is anyone selling the Nelson or Jacaranda 3/4 Physics book?
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