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March 29, 2024, 10:01:00 am

Author Topic: VCE Physics Question Thread!  (Read 603393 times)  Share 

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S200

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Re: VCE Physics Question Thread!
« Reply #2205 on: December 03, 2018, 07:23:35 am »
+3
Okay, thanks for letting me know. Also, how did you get 50 degrees?
I believe jirachi just approximated that number by looking at the symbol he used in his post. There is a limited number of angle symbols available... :)
The 50 is an example and is not related to the final answer....
« Last Edit: December 03, 2018, 07:25:10 am by S200 »
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dream chaser

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Re: VCE Physics Question Thread!
« Reply #2206 on: December 03, 2018, 08:02:40 am »
0
Okay, thanks for letting me know.

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Re: VCE Physics Question Thread!
« Reply #2207 on: December 03, 2018, 08:54:28 am »
+4
Hi Guys,

Can someone please explain to me why we need to calculate the centripetal force in this question. Woudn't the centripetal force always equal the tension force anyway. Or is that only when there is no angle involved? Also, what do they mean "with the horizontal"? And is the centripetal force always equal to the net force acting on the object?
Question is in the attachment. The question is related to Uniform circular motion.

Thanks


To build on Jirachi's explanation, the centripetal force is the net force which points directly towards the center of the circle. However, because the string is at an angle it has a vertical component and a horizontal component (we only want the horizontal one since only the horizontal component points to the center of the circle).
To find this horizontal component (the centripetal force) we draw up a triangle. The hypotenuse of the triangle is the tension (12 N) and we know that the angle between the tension and the horizontal is 10 degrees.
cos(angle) = adjacent/hypotenuse
cos(10) = centripetal force/12
12cos(10) = centripetal force

studyingg

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Re: VCE Physics Question Thread!
« Reply #2208 on: December 17, 2018, 06:20:12 pm »
0
is the concept of true'weightlessness' still aplicable to gravity in the new SD?

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Re: VCE Physics Question Thread!
« Reply #2209 on: December 17, 2018, 06:38:25 pm »
+4
is the concept of true'weightlessness' still aplicable to gravity in the new SD?

Yep.
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dream chaser

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Re: VCE Physics Question Thread!
« Reply #2210 on: December 20, 2018, 02:05:45 pm »
+1
Need help with this question regarding projectile motion. All help will be appreciated.  :)

Gavin is at the golf driving range. He selects a club that will launch the ball with an initial vertical velocity of 43.3m/s to a maximum height of 93.75m. The effects of air resistance can be ignored. Find the time it takes to reach the maximum height.

When I answered it, I knew in terms of vertical motion(given that I take the initial direction of motion i.e when the ball is going upwards as the positive direction):
a=-9.8m/s^2
x=93.75m
u=43.3m/s
v=0m/s
t=?

So technically, you can use either x=ut+1/2at^2 or v=u+at to find the 't' value, as we have 4 pieces of information. However, when I tried both approaches, I got separate answers. What am I doing wrong?

studyingg

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Re: VCE Physics Question Thread!
« Reply #2211 on: December 20, 2018, 02:40:20 pm »
+1
Need help with this question regarding projectile motion. All help will be appreciated.  :)

Gavin is at the golf driving range. He selects a club that will launch the ball with an initial vertical velocity of 43.3m/s to a maximum height of 93.75m. The effects of air resistance can be ignored. Find the time it takes to reach the maximum height.

When I answered it, I knew in terms of vertical motion(given that I take the initial direction of motion i.e when the ball is going upwards as the positive direction):
a=-9.8m/s^2
x=93.75m
u=43.3m/s
v=0m/s
t=?

So technically, you can use either x=ut+1/2at^2 or v=u+at to find the 't' value, as we have 4 pieces of information. However, when I tried both approaches, I got separate answers. What am I doing wrong?

I also got two seperate answers using your approach. However,  I think it's because you are not using the vertical component of the initial velocity. Was there a diagram in this question which contained an angle? I belive u should actually be u=sin(angle)*43.3

edit: or, if there isn't an angle  you should actually first use v^2= u^2 +2as to find u and then proceed to find t
« Last Edit: December 20, 2018, 02:42:54 pm by studyingg »

dream chaser

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Re: VCE Physics Question Thread!
« Reply #2212 on: December 20, 2018, 02:51:47 pm »
+1
I also got two seperate answers using your approach. However,  I think it's because you are not using the vertical component of the initial velocity. Was there a diagram in this question which contained an angle? I belive u should actually be u=sin(angle)*43.3

edit: or, if there isn't an angle  you should actually first use v^2= u^2 +2as to find u and then proceed to find t

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?

studyingg

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Re: VCE Physics Question Thread!
« Reply #2213 on: December 20, 2018, 03:04:11 pm »
+1
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?

i got 4.42 s using v=u+at , which I think is correct. I think you got -10.30 because you might have accidently used 9.8m/s^2 and not -9.8 m/s^2. (That's what I got with +9.8m/s^2). But even using the negative sign I got 5.04 s. It's a weird question tbh...

Are you sure 43.3  is the initial vertical component? Because I tried the way I described in my edit and I got 4.37s using both methods. If 43.3 is the initial vertical component then I think that there is an error in the question.

dream chaser

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Re: VCE Physics Question Thread!
« Reply #2214 on: December 20, 2018, 03:19:03 pm »
+1
i got 4.42 s using v=u+at , which I think is correct. I think you got -10.30 because you might have accidently used 9.8m/s^2 and not -9.8 m/s^2. (That's what I got with +9.8m/s^2). But even using the negative sign I got 5.04 s. It's a weird question tbh...

Are you sure 43.3  is the initial vertical component? Because I tried the way I described in my edit and I got 4.37s using both methods. If 43.3 is the initial vertical component then I think that there is an error in the question.

There is probably an error then. It says in the question the initial vertical velocity is 43.3m/s.

In regards to the negative answer, I made a mistake in calculations.

Also, can you show me your working out if you use x=ut+1/2at^2 and you take a=-9.8m/s^2, how you get your answer?
« Last Edit: December 20, 2018, 03:21:11 pm by dream chaser »

studyingg

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Re: VCE Physics Question Thread!
« Reply #2215 on: December 20, 2018, 03:39:45 pm »
+1
There is probably an error then. It says in the question the initial vertical velocity is 43.3m/s.

In regards to the negative answer, I made a mistake in calculations.

Also, can you show me your working out if you use x=ut+1/2at^2 and you take a=-9.8m/s^2, how you get your answer?

Well I rearranged the equation to form a quadratic, factorised out 4.9, and completed the square to get 5.04 and 3.79, I could take a photo if you want. I hate to be annoying but I really think that 43.3 is just the initial velocity and not the vertical component though.

dream chaser

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Re: VCE Physics Question Thread!
« Reply #2216 on: December 20, 2018, 04:01:40 pm »
+1
Well I rearranged the equation to form a quadratic, factorised out 4.9, and completed the square to get 5.04 and 3.79, I could take a photo if you want. I hate to be annoying but I really think that 43.3 is just the initial velocity and not the vertical component though.

Thanks for showing me your working out. No need for the photo. I am also feeling the same way as to the vertical initial velocity but that is what it says on the question unless it is a typo error or something, Thanks for the help by the way  :D

studyingg

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Re: VCE Physics Question Thread!
« Reply #2217 on: December 20, 2018, 04:08:23 pm »
+1
Thanks for showing me your working out. No need for the photo. I am also feeling the same way as to the vertical initial velocity but that is what it says on the question unless it is a typo error or something, Thanks for the help by the way  :D

All good!

dream chaser

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Re: VCE Physics Question Thread!
« Reply #2218 on: December 20, 2018, 04:25:35 pm »
0
All good!

Just a quick question. When you get t=3.79sec and t=5.04sec, which one would you choose as the answer?

studyingg

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Re: VCE Physics Question Thread!
« Reply #2219 on: December 20, 2018, 04:31:52 pm »
+1
Just a quick question. When you get t=3.79sec and t=5.04sec, which one would you choose as the answer?
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.