Unable to solve Mechanics problems

Original post by DylanJ42

in the other thread they have stated time = 1.6, does this mean 1.6 secs after the second ball starts moving or 1.6 secs from the very start? the former and i can help you, the latter and i cannot

1.6 seconds after the second ball starts moving.

Reply 3

stochasym

Is it 23/7 sec from the very start?

Reply 4

DylanJ42

Original post by fablereader

1.6 seconds after the second ball starts moving.

Reply 5

Original post by DylanJ42

Thank you very much! This has helped clear things up tremendously in my head. I think one of the main problems was that I got t = 2.6, and since it wasn't the number of the answer I ignored the remaining context (after the release of the second ball), but reading this helped me keep that in mind. That got me t = 1.6 s and s = 17.28 m (from the bottom).

Reply 6

DylanJ42

Original post by fablereader

thats exactly what i got too :biggrin:

nice work

Reply 7

stochasym

Funny enough, I thought that t₁=t and t₂=t-2. I don't know where I found that :colondollar:

. Thanks.

Original post by fablereader

Reply 8

OK, I'm having trouble with another problem. This one just seems confusing to me, and I'm not entirely sure what they want. Could anyone help with this?

'A body falls from rest from the top of a tower. During the last second of its motion it falls 7/16 of the whole distance.

a) Show that the time of descent is independent of the value of g.

b) Find the height of the tower in terms of g.'

Note that g is gravity, an acceleration that is approximately 9.8 m s^-2.

Reply 9

Original post by fablereader

You know it is dropped from rest:

h = \frac{1}{2}gt^2

where

t

is the total time and

h

is the height.

You know that in

t-1

seconds, it falls a distance of

\frac{9h}{16}

, so we also have our second equation:

\frac{9h}{10} = \frac{1}{2}g(t-1)^2

Now divide the second expression by the first:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\frac{\frac{9h}{16}}{h} = \frac{\frac{1}{2}g(t-1)^2}{\frac{1}{2}gt^2} = \frac{(t-1)^2}{t^2} \iff \left(\frac{3}{4}\right)^2 = \left(\frac{t-1}{t}\right)^2 \end{equation*}

Reply 10

Original post by fablereader

This is probably a typo - it's obviously false as written (consider the motion with g = 0 m/s^2 for example) - I guess they meant independent of m, the mass, which is the import of Galileo's famous experiment.

Reply 11

Original post by atsruser

I think the case for

g=0

is the only exception since you get an indeterminate form of

\frac{0}{0}

in my answer, but otherwise, I think my approach was correct(?).

Reply 12

Original post by Zacken

I think the case for

g=0

is the only exception since you get an indeterminate form of

\frac{0}{0}

in my answer, but otherwise, I think my approach was correct(?).

I'm not sure what you're claiming to show - that the time of descent of an object falling a distance h in a gravitational field g is independent of g? Or something else?

If it's the former, then it's false - we know that

h = 0.5 gt^2 \Rightarrow t = \sqrt{2h/g}

which depends on g. Am I confused?

Reply 13

OK, to be quite frank I think your discussion was a little bit above my level. I didn't (and still don't) understand it all (I'm no Maths genius). However, doing this division thing got me to:

9/16 = (t² - 2t + 1)/t²

Multiplying it out and putting everything on one side I got:

0 = 7t² - 32t + 16.

Using the quadratic formula I got:

t = 4 or t = 0.5714285714

(edited 8 years ago)

Reply 14

Original post by Zacken

I think the case for

g=0

is the only exception since you get an indeterminate form of

\frac{0}{0}

in my answer, but otherwise, I think my approach was correct(?).

OK, I think I understand. The question is very poorly written. What it should say is:

"Given the information provided, show that we can calculate the time of descent without knowing the value of g".

It is true that we can make that calculation, but essentially the "value of g" has been encoded in the ratio of the distances fallen.

Reply 15

Original post by fablereader

Just square root both sides first - that's much easier.

Reply 16

Original post by atsruser

Just square root both sides first - that's much easier.

OK, that get's me:

3/4 = (t-1)/t

Multiplying it out and putting it all on one side I get:

t = 4

Looking at the answers, that's basically what they wanted, I think (their answer: Time is 4 s).

Thanks for helping!

Reply 17

Original post by atsruser

That makes much more sense, the way I originally thought of it was "Here's a situation that happens in a really weird field where the gravitation strength is irrelevant, but we're not going to tell you that, we'll only tell you that the distance fallen in the last second is this much" but your explanation makes much more sense, I originally did a double take when I first read the question; that'll teach me to trust my instincts more. Thanks!

Reply 18

OK, having another difficulty. Again, g is gravity (i.e. acceleration).

'A stone falls past a window of height 2.5 m in 0.5 s. Taking g = 10 m s^-2, find the height from which the stone fell.'

Please help!

Reply 19