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Author Topic: VCE Physics Question Thread!  (Read 603639 times)  Share 

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Conic

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Re: VCE Physics Question Thread!
« Reply #795 on: January 19, 2015, 05:26:46 pm »
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When the ball rebounds it is moving upwards at a speed of 9 m/s. If we define upwards to be positive, the velocity is 9 m/s, the acceleration is -10 m/s/s. When the ball reaches its maximum height it has a velocity of 0 m/s. Now we use one of the constant acceleration formulas.



so the ball reaches a height of 4.05 m. Now the second question asks for the average force on the ball. We can work out the change of momentum, and use this to find the average force. Firstly, the change in momentum is given by



so the change in momentum is 3.15 kgm/s, and we are given that the contact time is 0.12 s. Now we use this in the impulse equation.



so the average reaction force is 26.25 N.
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JackSonSmith

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Re: VCE Physics Question Thread!
« Reply #796 on: January 19, 2015, 06:02:23 pm »
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Thanks Conic.

A box of mass 15kg is pulled 5m along a rough surface by a constant force of 80N.
 
The coefficient of sliding friction for the box over this surface is 0.4.

a) What is the magnitude of the friction force acting on the box? Ans = 60N

b) What is the change in kinetic energy of the box over its 5 m journey? Ans = 100J
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Conic

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Re: VCE Physics Question Thread!
« Reply #797 on: January 19, 2015, 06:16:40 pm »
+1
I'm pretty sure you aren't expected to know about the coefficient of friction in VCE Physics, but it does appear in Spesh.  The coefficient of friction is 0.4, and the normal force is 150 N (to balance the weight force mg = 150 N). Now the friction is given by



so the frictional force is 60 N in the direction opposing motion. Therefore there is a net force of 20 N acting in he direction of motion. Now, we want to find the change in kinetic energy. This will be equal to the net work on the system, as no energy is being lost or converted to potential energy. The net work done is



so the box gains 100 J of kinetic energy over its journey.
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FarAwaySS2

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Re: VCE Physics Question Thread!
« Reply #798 on: January 19, 2015, 06:20:49 pm »
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I keep getting this question wrong.

A bicycle accelerates from rest, covering 16m in 4 s. The total mass of the bicycle and its rider is 90 kg. What is its average acceleration?


Thanks!
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Kel9901

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Re: VCE Physics Question Thread!
« Reply #799 on: January 20, 2015, 08:31:02 am »
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I keep getting this question wrong.

A bicycle accelerates from rest, covering 16m in 4 s. The total mass of the bicycle and its rider is 90 kg. What is its average acceleration?


Thanks!

u=0, t=4, s=16, a=?
s=ut+1/2at^2
16=0+1/2*a*16=8a
a=2 ms^-2
s=change in displacement for physics
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AirLandBus

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Re: VCE Physics Question Thread!
« Reply #800 on: January 20, 2015, 12:53:30 pm »
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u=0, t=4, s=16, a=?
s=ut+1/2at^2
16=0+1/2*a*16=8a
a=2 ms^-2

Just out of curiosity, do you have to use the motion equations?
Why couldnt you do something along the lines of:
V = 16/4 = 4m/s
Acceleration = DeltaV/Time
= 4-0/4
=1m/s^2
Why doesnt that work?

Conic

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Re: VCE Physics Question Thread!
« Reply #801 on: January 20, 2015, 01:23:50 pm »
+1
The problem is that you are assuming that the speed at the end of the 4 seconds (i.e., v) is equal to the average speed of the journey. This will not be true if there is non-zero constant acceleration.
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Kel9901

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Re: VCE Physics Question Thread!
« Reply #802 on: January 20, 2015, 04:54:14 pm »
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Just out of curiosity, do you have to use the motion equations?
Why couldnt you do something along the lines of:
V = 16/4 = 4m/s
Acceleration = DeltaV/Time
= 4-0/4
=1m/s^2
Why doesnt that work?

Because the change in velocity is not equal to the average velocity (which you calculated to be 4 m/s).

Yes, you have to use the constant acceleration equations, as work-energy can't make use of the time figure (it needs the final velocity rather than time) and momentum-impulse can't make use of the change in displacement figure (it needs the final velocity rather than change in displacement)
s=change in displacement for physics
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alchemy

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Re: VCE Physics Question Thread!
« Reply #803 on: January 22, 2015, 03:26:54 pm »
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A softball of mass 250 g is thrown with an initial velocity of 16 m s–1 at an angle θ to the horizontal. When the ball reaches its maximum height, its kinetic energy is 16 J.
a What is the maximum height achieved by the ball from its point of release?
b Calculate the initial vertical velocity of the ball.
c What is the value of θ?
d Whatisthespeedoftheballafter1.0s?
e What is the displacement of the ball after 1.0 s?
f How long after the ball is thrown will it return to
the ground?
g Calculate the horizontal distance that the ball will
travel during its flight.

Need help with part b and part e specifically.
Thanks.

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Re: VCE Physics Question Thread!
« Reply #804 on: January 22, 2015, 04:05:51 pm »
+2
A softball of mass 250 g is thrown with an initial velocity of 16 m s–1 at an angle θ to the horizontal. When the ball reaches its maximum height, its kinetic energy is 16 J.
a What is the maximum height achieved by the ball from its point of release?
b Calculate the initial vertical velocity of the ball.
c What is the value of θ?
d Whatisthespeedoftheballafter1.0s?
e What is the displacement of the ball after 1.0 s?
f How long after the ball is thrown will it return to
the ground?
g Calculate the horizontal distance that the ball will
travel during its flight.

Need help with part b and part e specifically.
Thanks.

I'll just answer the whole Q systematically because answers to previous parts are probably needed for the later parts

a) When it was thrown KE=1/2 mv^2=32 J, at the top KE=16 J therefore GPE at the top=16 J=mgh
h=6.4 m

b) vertically:
v=0, a=-10, s=6.4, u=?
v^2=u^2+2as
0=u^2-128
u=11.3 ms^-1

c) sin(θ)=11.3/16=0.71
θ=45°

d) horizontal velocity=16 cos(45)=11.3
vertically:
u=11.3, t=1, a=-10, v=?
v=u+at=11.3-10=1.3
Using Pythagoras, speed=11.4 ms^-1

e) After 1 second:
horizontal: s=ut=11.3 m
vertically:
u=11.3, t=1, a=-10, s=?
s=ut+1/2 at^2=11.3-5=6.3 m
11.3 m to the right of (assuming it was launched to the right) and 6.3 m above the launch point
OR 13.0 m (Pythagoras) from the launch point at an angle of tan⁻¹(6.3/11.3)=29° from the horizontal

f) vertically:
u=11.3, a=-10, s=0 (as ball returns to the ground at the same height as where it was launched), t=?
s=ut+1/2 at^2
0=11.3t-5 t^2
0=t(11.3-5t)
t=0, 11.3-5t=0 --> t=2.26
t>0 therefore t=2.26 s

g) horizontal:
u=11.3, t=2.26, s=?
s=ut=25.6 m
s=change in displacement for physics
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Re: VCE Physics Question Thread!
« Reply #805 on: January 23, 2015, 12:24:51 am »
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Need help with this question:

A 0.50kg puck rests on a level air table and is connected by a light thread passing through a hole in the table to support a hanging mass of 3.0 kg. A stable orbit is achieved when the puck is sent into a circular path of radius 0.15 metres around the hole.
(A) neglecting friction at the edge of the hole, calculate the period of revolution of the puck in its orbit.

(B) Suppose that the speed of the puck in its orbit is now doubled, while the radius remains fixed. what central mass will be needed to achieve a stable orbit.

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Re: VCE Physics Question Thread!
« Reply #806 on: January 23, 2015, 09:13:07 pm »
+2
Need help with this question:

A 0.50kg puck rests on a level air table and is connected by a light thread passing through a hole in the table to support a hanging mass of 3.0 kg. A stable orbit is achieved when the puck is sent into a circular path of radius 0.15 metres around the hole.
(A) neglecting friction at the edge of the hole, calculate the period of revolution of the puck in its orbit.

(B) Suppose that the speed of the puck in its orbit is now doubled, while the radius remains fixed. what central mass will be needed to achieve a stable orbit.

I took a while to understand what this question was asking and the directions of the forces. The tension in the string counteracts the weight of the 3.0 kg and provides the centripetal force for the 0.50 kg

a) On 3.0 kg: Fnet=0=Tension-mg=Tension-30
Tension=30
On 0.50 kg: Fnet=Tension=30=ma
a=60=v^2/r=v^2/0.15
v^2=9
v=3=2πr/T
T=0.31 s

b) Let the new velocity, acceleration and net force on the 0.50 kg be v', a' and Fnet'
0.50 kg:
v'=2*3=6
a'=v'^2/r=240
Fnet'=ma'=120=T'
Other mass:
120=mg
m=12 kg
s=change in displacement for physics
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Cosec

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Re: VCE Physics Question Thread!
« Reply #807 on: January 27, 2015, 07:24:53 pm »
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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 ^.
« Last Edit: January 27, 2015, 09:20:11 pm by Cosec »

odeaa

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Re: VCE Physics Question Thread!
« Reply #808 on: January 27, 2015, 07:59:38 pm »
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I tried the first one but I'm pretty sure i was wrong ahaha
Our teacher has forced us to do the second type of question to death though so I'll give it a crack

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?

First of all I would convert to m/s
Car=22.2m/s
Cop=27.8m/s
It takes the cop 15 seconds to accelerate to 27.8m/s, during which time the car has travelled 333m
In the first 10 seconds the cop travels
In the following 5 seconds the cop travels bringing him to a total of 291.5m
The cop is travelling 5.5m/s faster than the car, and has to catch up 41.5m, which will take him 7.5 seconds, bringing the time up to 22.5 seconds
Usually it isnt that messy and you have a nice 20m/s and 30m/s values to work with (classic VCE Physics)

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







I hope this is right ahaha
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Cosec

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Re: VCE Physics Question Thread!
« Reply #809 on: January 27, 2015, 09:24:48 pm »
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I tried the first one but I'm pretty sure i was wrong ahaha
Our teacher has forced us to do the second type of question to death though so I'll give it a crack

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?

First of all I would convert to m/s
Car=22.2m/s
Cop=27.8m/s
It takes the cop 15 seconds to accelerate to 27.8m/s, during which time the car has travelled 333m
In the first 10 seconds the cop travels
In the following 5 seconds the cop travels bringing him to a total of 291.5m
The cop is travelling 5.5m/s faster than the car, and has to catch up 41.5m, which will take him 7.5 seconds, bringing the time up to 22.5 seconds
Usually it isnt that messy and you have a nice 20m/s and 30m/s values to work with (classic VCE Physics)

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







I hope this is right ahaha

Ive done the car one to death to. But this ones weird. When i added the values for the distance initially covered by the copy im getting a different value to you.
I think its where you calculated


Shouldnt it be 22.2*5 + (27.8-22.2 *5/2)?
Take a look at the graph attached.
Imgur: http://imgur.com/wmdtr5g

And thats what i got for the second equation. Book must be wrong.
thanks buddy.

Edit: The way I got taught is a bit differnt for the first question
Where told to make two equations for each vehicle, find the distance both will cover in terms of time and then let them equal each other and solve for time.
Not sure.
« Last Edit: January 27, 2015, 09:27:44 pm by Cosec »