Question

Hi, folks.

I need help with a whole question here:

http://www.studyjapan.go.jp/pdf/questions/09/ga-phy.pdf

It is the question 3. I solved the first, but am having trouble with the others. Would you help me, please?

Answer 1

You mean the question on page 8 right!

Answer 2

pages 9-10

Answer 3

Question (1) is just a matter of making centripetal force equals the component of weigth in the line (since there is no force from the string). But in the second one I think I need to consider a tension force...

Answer 4

I think I got (2) by conservation of energy.

(1) is m*v²/L = m*g*sin(θ),
so v=sqrt(g*L*sin(θ))

(2) is m*v_o²/2 = mv²/2 + m*g*(L+Lsin(θ)),
so v_o=sqrt(2*g*L + 3*g*Lsin(θ))

Answer 5

yeah I think that's right for (1) and (2)
for (3)
h = v^2 cos\[h=v ^{2} \cos ^{2}\theta/2g\](θ)/2g

Answer 6

got this one too, by conservation of energy (Thanks!)

Answer 7

number (4), you think I could solve by making the y (from y=y_o+v_0*t+a*t²/2) equals to the y in C? Or something like that...

Answer 8

yes, it will be solved by using this equation.
where y=l sin60, v=sqrt(gl sin 60), and t=x/vcos60

Answer 9

got it?

Answer 10

i think I got it... but what is x?

Answer 11

It is against OpenStudy policy to help with tests.

Answer 12

it is a question from a previous test (2009). Does it count?

Answer 13

Is it going towards your grade?

Answer 14

No, I will do another test, so it is a preparatory study. The test is avaiable publicly.

Answer 15

Be careful when you say you need help with a test. Generally you can get into a lot of trouble by putting "test" in the question :)

Answer 16

I see! Sorry about that, I didn't realize it.

Answer 17

edited my question

Answer 18

x=l*(cos(60) + cos(phi)

it gives me a big equation if i substitute it in y=y_o+v_o*t+a*t²/2
(a = -g)

Answer 19

When the force becomes zero, the particle starts moving in a parabolic path since it has a horizontal velocity. and x is the range of the projectile.

Answer 20

yeah, i understood that. V was calculated, and I have the angle theta. I'm trying to solve by energy conservation again.

Answer 21

y=v*t+a*t²/2
l sin60= v sin30 *(x/vcos30) -(g*x^2)/(2*v^2*cos^230)

solve for x and you'll get x= 3*l/2

Answer 22

just a minute..

Answer 23

but the value for y is (l*sin(theta)-sin(phi)), isn't it? (in your equation)

You used point B as ground zero?

Did you get any answer yet?

Answer 24

I think I got it!

Answer 25

y is the displacement of particle from B to the diameter on x-axis.

Answer 26

I think i'm started to get confused uhh

Answer 27

hmmm, didn't work. You gave a value to g?
I was using
x=l*3/2 and
x=l(cos(theta)-cos(phi)), assuming phi > PI/2

I don't know...

Answer 28

I'm very confused! haha

Answer 29

I will think more about this question. Thank you ghaidaa.
If I solve it, I will post here.

Answer 30

By the way! Answers for this question:

(1)c
(2)b
(3)e
(4)d
(5)b

(sorry about that.....)

Answer 31

you're welcome. nvm i'll try to get also

Answer 32

Still up? I've taken monbusho tests before. Guess I can help a bit.