 October 22, 2019, 09:19:46 am AuthorTopic: VCE Physics Question Thread!  (Read 255375 times) Tweet Share

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dream chaser

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« Reply #2220 on: December 20, 2018, 04:34:55 pm »
+1
honestly don't know, I've never encountered something like this in physics, probs cause the question is flawed. I'd go with the one that matched my v=u+at formula.

Okay. Cheers for the help

lzxnl

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« Reply #2221 on: December 20, 2018, 07:13:18 pm »
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Let's clear up some issues here.

Just a quick question. When you get t=3.79sec and t=5.04sec, which one would you choose as the answer?
If you get two answers, NORMALLY that's because the first time is on the way up and the second time is on the way down. However, the question is about maximum height so you should only have one time. So, there is another possibility that the OP has hinted at here:

I am using the vertical component of the initial velocity. it is 43.3m/s(given as part of the question). Plus, no angle was given. No diagram was given either.

What 2 answers did you get anyway.

When I did it, I got 10.30s using x=ut+1/2at^2 and 4.42s using v=u+at. However, for the first approach, I got 10.30 with a negative infront. Did you have the same problem?

In this question, you're given FOUR pieces of information:
y = +93.75 m
u = 43.3 m/s
v = 0 m/s
a = -9.8 m/s2
t = ?

Given that the question says 'initial vertical velocity' and makes no mention of the horizontal velocity, this is, in fact, a one-dimensional problem.
The issue is, a 1D constant (but nonzero) acceleration problem is uniquely defined by any three of the five quantities. You're given four, so there's a decent chance you have too much information. And indeed, we see that the information you have does not satisfy the equation v2 = u2 + 2ay. Therefore, your question itself is physically impossible, explaining why you can't solve the problem.

Next time you encounter a constant acceleration problem with four pieces of information, check, using the equation that involves all four, that the four pieces of information make sense. Not all questions have answers. A simple one is trying to solve x = x + 1 for x. Obviously, this won't work.
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dream chaser

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« Reply #2222 on: January 03, 2019, 08:47:39 am »
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Hi Guys,

Just a quick question. If you include air resistance, would the horizontal velocity of an object in projectile motion, eg. a ball have constant velocity still?

Also, why does the vertical velocity of an object change, ignoring air resistance, throughout projectile motion?
« Last Edit: January 03, 2019, 08:55:35 am by dream chaser »

FelixHarvey

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« Reply #2223 on: January 03, 2019, 09:12:07 am »
+4
Just a quick question. If you include air resistance, would the horizontal velocity of an object in projectile motion, eg. a ball have constant velocity still?

No it would not have a constant velocity. The net force in the horizontal direction would be non-zero and thus deceleration would occur, and therefore the object changes velocity.

Also, why does the vertical velocity of an object change, ignoring air resistance, throughout projectile motion?
The net force in the vertical direction is mg (assuming no air resistance), thus the forces are unbalanced and a change in acceleration will occur, therefore, the object changes velocity.
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dream chaser

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« Reply #2224 on: January 03, 2019, 09:27:45 am »
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No it would not have a constant velocity. The net force in the horizontal direction would be non-zero and thus deceleration would occur, and therefore the object changes velocity.
The net force in the vertical direction is mg (assuming no air resistance), thus the forces are unbalanced and a change in acceleration will occur, therefore, the object changes velocity.

Okay. Thank you for the help. Much appreciated.

dream chaser

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« Reply #2225 on: January 03, 2019, 07:50:33 pm »
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Hi Guys,

Are my reasoning's for these questions okay? The question is related to projectile motion.

Question: A parcel is dropped from a height of 500m from a helicopter travelling at a speed of 20m/s.

(a) Describe the effects of air resistance on:
(i) the horizontal component of the motion of the parcel.
(ii) the vertical component of motion of the parcel.

(i)
- Air resistance of an object increases when the velocity of the object increases as well(Air Resistance ∝ Velocity).
- As the object descends more and more, the air resistance would be applied to a larger effect on the horizontal component of motion.
- The horizontal component of motion(its velocity) would decrease in value as it decelerates due to the horizontal net force becoming more and more non zero.

(ii)
-  Air resistance of an object increases when the velocity of the object increases as well(Air Resistance ∝ Velocity).
- Unlike horizontal motion, the velocity in the vertical component would increase in value as initially, the velocity in the vertical component would be 0m/s due to it being dropped from rest.
- As the objects descends closer to the ground, the velocity in the vertical component of the object gets bigger due to the non zero net force. However, even though the velocity is getting greater, air resistance slows down the acceleration of the object meaning the vertical component of the velocity as it hits the ground won't be as high of a value as it would if air resistance was not involved at all(was negligible).

(b) Which of the horizontal or vertical components of the motion of the parcel is likely to experience the greater air resistance during:
(i) the first 2 seconds of its fall
(ii) the final 2 seconds of its fall?

(i)
My Answer: Horizontal Component.
- This is due to air resistance in the horizontal direction causing the horizontal velocity to decrease.
- The velocity in the vertical direction would still increase, even with the presence of air resistance involved.

(ii) My Answer: Vertical Component.
- This is due to air resistance causing the vertical velocity to accelerate slowly(not as much as it would if air resistance was not involved).
- Since in the final 2 seconds where the horizontal velocity would already be decelerating closer and closer to 0m/s, the effect in which air resistance would be applied to the horizontal component decreases as Air Resistance ∝ Velocity.

All replies would be much appreciated. Thanks  dream chaser

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« Reply #2226 on: January 04, 2019, 11:44:07 pm »
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Hi Guys,

Really need help to do this question. The question is in regards to uniform horizontal circular motion.

When travelling around a roundabout, John notices that the fluffy dice suspended from his rear-vision mirror swing out. If John is travelling at 8.0 m/s and the roundabout has a radius of 5.0 m, what angle will the string connected to the fluffy dice (mass 100 g) make with the vertical?

One of the things I don't get is why the radius of the car in circular motion(i.e the roundabout) would be the same as the radius of the circular motion in which the dice makes. Also, will the dice be experiencing circular motion as the car?

Also, could someone please explain to me what happens to the smaller object when an smaller object within a larger object(eg. a dice in a car) goes around a roundabout if the larger object is going at a constant speed(uniform circular motion)

All help will be greatly appreciated. Thanks « Last Edit: January 04, 2019, 11:59:31 pm by dream chaser »

dream chaser

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« Reply #2227 on: January 05, 2019, 10:20:33 am »
+1
Would someone be able to help me with the question in my previous post for physics? Thanks

studyingg Re: VCE Physics Question Thread!
« Reply #2228 on: January 05, 2019, 11:27:50 am »
0
Would someone be able to help me with the question in my previous post for physics? Thanks

I'm not sure if I fully grasped this concept, but I thinkkkk the reason the radius is the same can be explained by the second law of motion
I'm sure you know that when two bodies are connected, they experience the same acceleration. You probably also know that in the cases of centripetal acceleration, the object (in this case -- the connected body) is deviating towards the centre C. The distance from the centre to the object is the radius, and because both the dice and car are experiencing the same centripetal acceleration, the radius is 5m for both. With response to your other question, (again Im not too sure), but I would say, yes, the two objects are experiencing the same circular motion. The forces acting on the car are gravity and (perhaps) friction, while the forces acting on the dice are tension and gravity. Which means that the dice is a conical pendulum, but this is still a form of circular motion. Which leads me to say that, the smaller object and the larger object, when connected, will experience the same motion (and act as one system I guess). This is because acceleration is the same. I'm not 100% sure about this; however.

lzxnl

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« Reply #2229 on: January 05, 2019, 11:25:32 pm »
+2
Hi Guys,

Are my reasoning's for these questions okay? The question is related to projectile motion.

Question: A parcel is dropped from a height of 500m from a helicopter travelling at a speed of 20m/s.

(a) Describe the effects of air resistance on:
(i) the horizontal component of the motion of the parcel.
(ii) the vertical component of motion of the parcel.

(i)
- Air resistance of an object increases when the velocity of the object increases as well(Air Resistance ∝ Velocity). this relationship requires a few complicated assumptions but I'll let it slide
- As the object descends more and more, the air resistance would be applied to a larger effect on the horizontal component of motion. just say 'air resistance increases'
- The horizontal component of motion(its velocity) would decrease in value as it decelerates due to the horizontal net force becoming more and more non zero. the horizontal component would decay to zero; becoming more and more non-zero makes no sense here

(ii)
-  Air resistance of an object increases when the velocity of the object increases as well(Air Resistance ∝ Velocity).
- Unlike horizontal motion, the velocity in the vertical component would increase in value as initially, the velocity in the vertical component would be 0m/s due to it being dropped from rest.
- As the objects descends closer to the ground, the velocity in the vertical component of the object gets bigger due to the non zero net force. However, even though the velocity is getting greater, air resistance slows down the acceleration of the object meaning the vertical component of the velocity as it hits the ground won't be as high of a value as it would if air resistance was not involved at all(was negligible).
Assuming your air resistance is proportional to the velocity, you can write a = g - kv. Initially, when v=0, the acceleration is positive and the object speeds up. Eventually, as v increases, the acceleration decreases and asymptotes towards zero; this limiting velocity is called 'terminal velocity'
(b) Which of the horizontal or vertical components of the motion of the parcel is likely to experience the greater air resistance during:
(i) the first 2 seconds of its fall
(ii) the final 2 seconds of its fall?

(i)
My Answer: Horizontal Component.
- This is due to air resistance in the horizontal direction causing the horizontal velocity to decrease.
- The velocity in the vertical direction would still increase, even with the presence of air resistance involved.
Answer is correct, reasoning isn't great. Rather, mention that the initial horizontal velocity is 20 m/s and in 2 seconds, the vertical velocity cannot accelerate to 20 m/s, so the horizontal air resistance will always be greater

(ii) My Answer: Vertical Component.
- This is due to air resistance causing the vertical velocity to accelerate slowly(not as much as it would if air resistance was not involved).
- Since in the final 2 seconds where the horizontal velocity would already be decelerating closer and closer to 0m/s, the effect in which air resistance would be applied to the horizontal component decreases as Air Resistance ∝ Velocity.
Your second point is good, but the point is really that the horizontal velocity will be small due to air resistance and no horizontal driving force, while the vertical velocity will be much larger as gravity will then approximately balance air resistance
All replies would be much appreciated. Thanks  Hi Guys,

Really need help to do this question. The question is in regards to uniform horizontal circular motion.

When travelling around a roundabout, John notices that the fluffy dice suspended from his rear-vision mirror swing out. If John is travelling at 8.0 m/s and the roundabout has a radius of 5.0 m, what angle will the string connected to the fluffy dice (mass 100 g) make with the vertical?

One of the things I don't get is why the radius of the car in circular motion(i.e the roundabout) would be the same as the radius of the circular motion in which the dice makes. Also, will the dice be experiencing circular motion as the car?

Also, could someone please explain to me what happens to the smaller object when an smaller object within a larger object(eg. a dice in a car) goes around a roundabout if the larger object is going at a constant speed(uniform circular motion)

All help will be greatly appreciated. Thanks The dice is in the car. It essentially moves with the car, hence the radii and speeds are the same. It's like saying, if you're holding onto your phone as you run, your phone moves as fast as you do, because it's moving with you.

For your last question, it depends on the sizes of the objects involved. In this particular case, the dice is so small compared to the car that you can consider it part of the car.
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dream chaser

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« Reply #2230 on: January 07, 2019, 11:43:39 am »
+1
Thank you lzxnl and studyingg for the help. Much appreciated. Ansaki

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« Reply #2231 on: January 26, 2019, 01:02:49 pm »
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Ishtar is riding a motorised scooter along a level bike path. The combined mass of Ishtar and her scooter is 80 kg. The friction and drag forces that are acting total to 45 N. What is the magnitude of the driving force being provided by the motor if she is:
(a) moving with constant speed of 10 m/s
(b) accelerating at 1.5 m/s^2?

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« Reply #2232 on: January 26, 2019, 01:16:32 pm »
+3
Ishtar is riding a motorised scooter along a level bike path. The combined mass of Ishtar and her scooter is 80 kg. The friction and drag forces that are acting total to 45 N. What is the magnitude of the driving force being provided by the motor if she is:
(a) moving with constant speed of 10 m/s
(b) accelerating at 1.5 m/s^2?

Hey!

a) We want the scooter to NOT accelerate, ie a = 0. If we want the the scooter to not accelerate, the forces on the scooter should cancel out, so the motor should provide a force of 45N

b) F = ma -> If we want the scooter to accelerate 1.5m/s^2 we need a net force in the direction of motion to be 120N meaning including the frictional forces, we need 120+45 N which is 165N

Hope this helps!
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HolHen Re: VCE Physics Question Thread!
« Reply #2233 on: January 28, 2019, 04:57:24 pm »
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Hi!

I am going into year 12 this year and I'm not really sure how to approach the SACs that are going to be practical experiment based.

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« Reply #2234 on: February 02, 2019, 12:50:28 pm »
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What are some topics to watch out for in Physics 1/2?
',:{D