Some Physics questions. [Kinematics O.O]
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1. A stone of mass 200g falls vertically for 2s and embeds itself 2cm in soft mud.
(A) Assuming that the stone was released from rest, from what height was it released?
(B) What is the average deceleration in the mud? (this is the hard one)
2. If a hovercraft was travelling at 20m/s and its max uniform acceleration is -1m/s, how long distance can it travel before it comes to a halt?
Thanks these are the questions i have after doing 2 Physics paper 2s. I think they look kinda the same. There are acceleration distance velocity and time 4 variables in the calculations.
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1. A stone of mass 200g falls vertically for 2s and embeds itself 2cm in soft mud.
(A) Assuming that the stone was released from rest, from what height was it released?
(B) What is the average deceleration in the mud? (this is the hard one)
(A) Using the kinematics equations,
s = ut + ½at²,
where u = 0, t = 2, and a = g = 10 (for O levels)s = 0 + ½ (10)(2)² = 20m
(B) v = u + at
v = 0 + 10 * 2 = 20 m/s just before hitting the mud.Using v² = u² - 2as, where a is the deceleration
s = 0.02m, v = 0, u = 20
0 = 20² - 2a(0.02)
a = 10000 m/s => please help confirm2. If a hovercraft was travelling at 20m/s and its max uniform acceleration is -1m/s, how long distance can it travel before it comes to a halt?
Using v² = u² + 2as, where
a = -1
v = 0
u = 200 = 20² + 2(-1)s
s = 200mSome more kinematics questions can be found here:
ExamWorld Kinematics -
a = 10000 m/s => please help confirm
This is correct. But arent those formulas not in syllabus for this year?
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erm.. to do kinematics, i think this are the only formulas.. which u are supposed to be using till university..
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the formulae is not in the syllabus anymore. So you need to use another way to do it.
anyway, 1A) velocity = acceleration x time
= 2 x 10
= 20m/s
1B), KE = half mass velocity square
= 40J
Work done = F x Distance
40J = F x 0.02m
F = 2000N
F = ma
2000N = 0.2kg x a
a = 10 000m /s2
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but it is still good to know those formulas :D
Anyway, those formulas originated from the graphs themselves. For example, for part 1 (A), you can actually sketch the graph out, and calculate the area under graph
You can do the same thing for question 2 as well
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nothing wrong with formula, i learned them too.
but i don't like to teach the kids extra stuff when they can't even handle the basic ones.
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Originally posted by eagle:
(A) Using the kinematics equations,
s = ut + ½at²,
where u = 0, t = 2, and a = g = 10 (for O levels)s = 0 + ½ (10)(2)² = 20m
(B) v = u + at
v = 0 + 10 * 2 = 20 m/s just before hitting the mud.Using v² = u² - 2as, where a is the deceleration
s = 0.02m, v = 0, u = 20
0 = 20² - 2a(0.02)
a = 10000 m/s => please help confirmUsing v² = u² + 2as, where
a = -1
v = 0
u = 200 = 20² + 2(-1)s
s = 200mSome more kinematics questions can be found here:
ExamWorld KinematicsEagle would you be so kind as to list down the different variations of a=v-u / t please?
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Originally posted by crimsontactics:
Eagle would you be so kind as to list down the different variations of a=v-u / t please?
There are actually only 3 main kinds
Derived from graph:
s = ut + ½at²
v = u + at (same as a =(v-u) / t)from the above, if u substitute t off (maths), you will get
v² = u² + 2as -
Originally posted by skythewood:
the formulae is not in the syllabus anymore. So you need to use another way to do it.
anyway, 1A) velocity = acceleration x time
= 2 x 10
= 20m/s
1B), KE = half mass velocity square
= 40J
Work done = F x Distance
40J = F x 0.02m
F = 2000N
F = ma
2000N = 0.2kg x a
a = 10 000m /s2
They want the HEIGHT leh. :D
So maybe just use GPE = mgh for the second part?
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Originally posted by eagle:
There are actually only 3 main kinds
Derived from graph:
s = ut + ½at²
v = u + at (same as a =(v-u) / t)from the above, if u substitute t off (maths), you will get
v² = u² + 2asthanks.
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Just a question.
When I see a deceleration of 10000ms-2, I almost think it's impossible, but then it's calculated from the kinematics equation.
Is it really that high in practical?
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FYI, forumlaes are included in Physics (Pure) Syllabus
You could use them to solve questions.......
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Originally posted by secretliker:
Just a question.
When I see a deceleration of 10000ms-2, I almost think it's impossible, but then it's calculated from the kinematics equation.
Is it really that high in practical?
why not..? u have something falling from a height of 20m high and stopping within a distance of 0.02m..
20m/s converts to 72km/h.. slightly less den a car's speed on the expressway..
to stop within 2cm.. thats quite a deceleration..
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Originally posted by SBS n SMRT:
FYI, forumlaes are included in Physics (Pure) Syllabus
You could use them to solve questions.......
you mean this v = u + at?
Aint in the textbook. But indeed you could use it.
The 2 methods are actually quite different approaches; one involving energy. XD
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Originally posted by purpledragon84:
why not..? u have something falling from a height of 20m high and stopping within a distance of 0.02m..
20m/s converts to 72km/h.. slightly less den a car's speed on the expressway..
to stop within 2cm.. thats quite a deceleration..
True~
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Originally posted by Garrick_3658:
They want the HEIGHT leh. :D
So maybe just use GPE = mgh for the second part?
True~ (too)
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So the school setting this paper did intend students to use the energy method since they asked for the height in part (a)
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Originally posted by eagle:
There are actually only 3 main kinds
Derived from graph:
s = ut + ½at²
v = u + at (same as a =(v-u) / t)from the above, if u substitute t off (maths), you will get
v² = u² + 2asSry im missing some link XD. With what formula do you subsitute your t with?
If it is using s = ut + ½at², how do you get this formula?
Thanks
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Originally posted by Garrick_3658:
They want the HEIGHT leh. :D
So maybe just use GPE = mgh for the second part?
yup, you are right.
was just concentrating on part B since TS says this is the hard part.
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Originally posted by davidche:
Sry im missing some link XD. With what formula do you subsitute your t with?
If it is using s = ut + ½at², how do you get this formula?
Thanks
assume acceleration is constant.
let's say from A to B. s is the area under the graph.
s = (A +B)/2 x t
= At/2 + Bt/2
=At/2 + (A + at)t/2 (v = u + at)
= At/2 + At/2 + (at^2)/2
=At + (at^2)/2
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Originally posted by davidche:
Sry im missing some link XD. With what formula do you subsitute your t with?
If it is using s = ut + ½at², how do you get this formula?
Thanks
for s = ut + ½at²
Draw your velocity time graph, starting from u at t=0 and ending at v at t =t (all draw positive for simplicity)
Area under graph is a rectangle plus a triangle.
Area of rectangle will give ut
Are of triangle will give ½(at)(t), where at is v - uSum up, you will get the formula ;)
Finally, from a = (v-u)/t,
t = (v-u)/a
Sub into s = ut + ½at²Resolve and simplify, you will get v² = u² + 2as
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I understand liao. Thanks guys