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Chileman

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I'm bored... boredofstudies wahey! But seriously, can anyone please provide any challenging questions on Year 11 Physics? SPECIFICALLY AND ONLY Module 1: Kinematics or Module 2: Dynamics please!

I'll make one for the first two modules of Chemistry tomorrow and also most of Year 11 Maths Extension 1 (apart from combinatorics)
 

MoeyNeeds90+

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I'm bored... boredofstudies wahey! But seriously, can anyone please provide any challenging questions on Year 11 Physics? SPECIFICALLY AND ONLY Module 1: Kinematics or Module 2: Dynamics please!

I'll make one for the first two modules of Chemistry tomorrow and also most of Year 11 Maths Extension 1 (apart from combinatorics)
all urs bud https://chatgpt.com/
 

Bendwhat?_over

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I'm bored... boredofstudies wahey! But seriously, can anyone please provide any challenging questions on Year 11 Physics? SPECIFICALLY AND ONLY Module 1: Kinematics or Module 2: Dynamics please!

I'll make one for the first two modules of Chemistry tomorrow and also most of Year 11 Maths Extension 1 (apart from combinatorics)
Just look at past papers.
 

Chileman

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Just look at past papers.
yea that's a good idea

but i feel like u guys making questions up on the spot will be more fun

and besides half of tshc don't have a lot of half-yearlys and my school never gives past papers that easily (until a week before the test)
 

cheesynooby

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theres no way something like this is gonna be assessed but
2 cubic masses m and M. M is on a horizontal frictionless surface, and m is against the side of M and not touching the frictionless surface. What horizontal force must be applied to m to keep the two blocks together? g, µ (static friction coefficient between m and M), m and M are known.
Screenshot 2025-02-28 at 2.59.05 pm.png
 

Chileman

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Is this right?

I got stuck trying to figure out whether the reaction force (N) is in the same direction as the applied force, but I did end up getting this.
 

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cheesynooby

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Is this right?

I got stuck trying to figure out whether the reaction force (N) is in the same direction as the applied force, but I did end up getting this.
ngl idk the answer i just saw it in a textbook somewhere but I think the force applied to M horizontally should be N (if F is only applied to m) and there should probably be M in your answer since clearly if M was super large (say, a giant basically immovable wall) F would go down
 

cheesynooby

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Screenshot 2025-03-02 at 3.38.20 pm.pngi got something like this and the equation seems to make sense? btw when i said the force applied to M horizontally should be N i meant only N
 

pl4smaa21

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I'm bored... boredofstudies wahey! But seriously, can anyone please provide any challenging questions on Year 11 Physics? SPECIFICALLY AND ONLY Module 1: Kinematics or Module 2: Dynamics please!

I'll make one for the first two modules of Chemistry tomorrow and also most of Year 11 Maths Extension 1 (apart from combinatorics)
This problem is a bit challenging, or at least I found it challenging ig (my dumbass couldn't figure it out) but the pre-req. knowledge is module 1 v-t graphs and SUVAT so no forces or anything fancy but it is a bit tricky. gl!
1740900604587.png
 

pl4smaa21

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hmm very nice, you got it correct under the interpretation that it accelerated from rest to 100 km/h and drove at constant speed until decelerating to rest again. However, this isn't the interpretation that was taken by the source of this question and honestly I don't blame you for 'misunderstanding' because it isn't worded very well. Basically, the interpretation aimed by whoever wrote this problem was that the car accelerated at a rate equal to that which requires 3.5 s to reach 100 km/h which it then snaps to decelerating at a scale of 0.7g. It wasn't made clear and but it was supposed to be understood that 100 km/h is NOT the top speed rather the opening sentence was just meant to give you the acceleration in forward motion nothing much more than that:

Here's an image of the solution so u can see what I meant:

1740910234956.png
The main goal here was to find an expression for our desired variable T i.e. minimal time taken, in terms of what variables we know , i.e. distance travelled, acceleration and deceleration. The tricky part here that I was tryna play at you with comes up in the first line involving v, sub, max . Your intuition probably would have been strong enough to deduce this but basically I was trying to play at you with this particular step. See how they used v=u+at, for the acceleration a to equate to v,sub,max but they also got this other equation involving the deceleration where they IMAGINED (srsly how tf were you meant to come up with that) that the car was instead accelerating at the rate 0.7g and from the graph it is evident that'd have taken (T-t,nought, seconds) to have achieved. so v,max, = 0.7g * (T-t,nought,) Combining these two you get an expression (THIS IS KEY) for t,nought in terms of acc. deceleration and total time. You then combine this with an expression for the area under the graph D=1 km = 1/2 * base * height = 1/2 * T * v,max, . you then sub in for v,max, (a*t,nought) and then sub in that expression for t,nought, the key expression which is in terms of acceleration, deceleration and total time. Finally you rearrange to isolate total time, T and finally you plug-in values and then ur DONE.

TL;DR for some it may have been obvious ofc some people have a very strong intuition for these kinda things but 4 yr 11 like myself i thought the trick at the beginning where you have to look at the graph in BOTH directions to make a vital eqn., relating the variables, was a hard to see and thus a challenge. Again if it wasn't a challenge sorry I tried my best.
 

Chileman

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ahh i see,

you
hmm very nice, you got it correct under the interpretation that it accelerated from rest to 100 km/h and drove at constant speed until decelerating to rest again. However, this isn't the interpretation that was taken by the source of this question and honestly I don't blame you for 'misunderstanding' because it isn't worded very well. Basically, the interpretation aimed by whoever wrote this problem was that the car accelerated at a rate equal to that which requires 3.5 s to reach 100 km/h which it then snaps to decelerating at a scale of 0.7g. It wasn't made clear and but it was supposed to be understood that 100 km/h is NOT the top speed rather the opening sentence was just meant to give you the acceleration in forward motion nothing much more than that:

Here's an image of the solution so u can see what I meant:

View attachment 46732
The main goal here was to find an expression for our desired variable T i.e. minimal time taken, in terms of what variables we know , i.e. distance travelled, acceleration and deceleration. The tricky part here that I was tryna play at you with comes up in the first line involving v, sub, max . Your intuition probably would have been strong enough to deduce this but basically I was trying to play at you with this particular step. See how they used v=u+at, for the acceleration a to equate to v,sub,max but they also got this other equation involving the deceleration where they IMAGINED (srsly how tf were you meant to come up with that) that the car was instead accelerating at the rate 0.7g and from the graph it is evident that'd have taken (T-t,nought, seconds) to have achieved. so v,max, = 0.7g * (T-t,nought,) Combining these two you get an expression (THIS IS KEY) for t,nought in terms of acc. deceleration and total time. You then combine this with an expression for the area under the graph D=1 km = 1/2 * base * height = 1/2 * T * v,max, . you then sub in for v,max, (a*t,nought) and then sub in that expression for t,nought, the key expression which is in terms of acceleration, deceleration and total time. Finally you rearrange to isolate total time, T and finally you plug-in values and then ur DONE.

TL;DR for some it may have been obvious ofc some people have a very strong intuition for these kinda things but 4 yr 11 like myself i thought the trick at the beginning where you have to look at the graph in BOTH directions to make a vital eqn., relating the variables, was a hard to see and thus a challenge. Again if it wasn't a challenge sorry I tried my best.
ohhh i see why, they added "can" to the sentence! yea now i see how to do it, it's just integration and a bit of suvat.

but in reality, this prob ain't appearing in an hsc exam, let alone a prelim exam, and let along a yr 11 school assessment
 

pl4smaa21

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ahh i see,

you


ohhh i see why, they added "can" to the sentence! yea now i see how to do it, it's just integration and a bit of suvat.

but in reality, this prob ain't appearing in an hsc exam, let alone a prelim exam, and let along a yr 11 school assessment
yea this def aint popping up in any exams or anything that's why I posted it here as a 'challenge' becuz you probably never came across it since you never really had a reason to

also I'm pretty sure you don't need to use full-blown integration when the area is a triangle its kinda overkill lol buttttt integration is the way to go when things get curvy ig
 

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