Mechanics

EMPhillips1989 · 2008-02-25 02:22:36

hi
im a bit confused on the following question

when two particals a and b are projected vertically upwards from the same point one after another, with initial speed U.
if

is the time interval between the two projectiles where

find the time before the particals collide

does anyone know how to attempt this sort of question please help!!!!!

EMPhillips1989 · 2008-02-25 04:28:08

PLEASE HELP!!!!!!!!!!!
im really stuck

Last edited by EMPhillips1989 (2008-02-25 07:20:23)

EMPhillips1989 · 2008-02-25 07:58:40

well i figure that when the two particles collide the postion vectors will equall one another

the position vector of particle 1 is given by r=xi+zk
as the only force acting on it is gravity, from newtons second law mr''=-mgk
integrating with respect to t gives: r'=-gkt+c (where c is a constant)
at t=0, r'=0 so c=0 so
r'=-gkt
integrating again gives:

the position vector of particle two is given by:

when i set these equal to one another i get

as
does anyone know how i can get an answer please help!!!!

mathsyperson · 2008-02-25 08:36:55

EMPhillips1989 wrote:

...r'=-gkt+c (where c is a constant)
at t=0, r'=0 so c=0...

r' is velocity, and you're told that it's initially launched at U m/s. So then c=U and then after integration you get:

(I removed the k vector since the problem is only in one dimension anyway)

Similarly:

Equate these:

The restriction of λ <2u/g is there to make sure that the 2nd particle is launched before the first one lands. Otherwise this formula wouldn't work.

EMPhillips1989 · 2008-02-25 08:45:30

oh yes i see where my integration went wrong but is there a way of getting an exact result for t in terms of seconds?

luca-deltodesco · 2008-02-25 09:01:32

his workings ended with an equation to do that.

EMPhillips1989 · 2008-02-25 09:18:45

yes but im slightly confused as i'm not given u or g i might be having a blonde moment!

luca-deltodesco · 2008-02-25 10:08:47

yes you are

u is the initial velocity. g is the acceleration due to gravity.

there is no numerical value, because they are left as variables - but its perfectly find to quote an answer using them.

EMPhillips1989 · 2008-02-25 10:25:29

im now trying to work out the speed immediately before the impact
which equation do i plug t into??

luca-deltodesco · 2008-02-25 10:44:27

for the velocity of a and b you have
u - gt for a
and
u - g(t-λ) for b

EMPhillips1989 · 2008-02-25 22:46:03

would i not do it this way?

so i set r=0 at time of collision

simplifying:

adding t for particle one gives:

so
adding t for particle two give:

using

to get:

so

Last edited by EMPhillips1989 (2008-02-25 23:33:56)

luca-deltodesco · 2008-02-26 00:02:05

the way i would do it is to differentiate the equations for displacement and substitute the equation for t

the first equation i gave

similary for particle b you have

the other equation i gave

ofcourse, doing this isn't exactly necessary, since in the process of finding the two equations, you had to get these two anyways

if you plug in the equation for t from before, you get

Last edited by luca-deltodesco (2008-02-26 00:02:35)

JaneFairfax · 2008-02-26 03:24:58

EMPhillips1989 wrote:

so i set r=0 at time of collision

r is NOT 0 at the time of collision!!

However, if you want r to be the distance between the two particles, then the equation is

and you can set that to be 0 at the time of collision.

Last edited by JaneFairfax (2008-02-26 03:30:52)

Math Is Fun Forum

#1 2008-02-25 02:22:36

Mechanics

#2 2008-02-25 04:28:08

Re: Mechanics

#3 2008-02-25 07:58:40

Re: Mechanics

#4 2008-02-25 08:36:55

Re: Mechanics

#5 2008-02-25 08:45:30

Re: Mechanics

#6 2008-02-25 09:01:32

Re: Mechanics

#7 2008-02-25 09:18:45

Re: Mechanics

#8 2008-02-25 10:08:47

Re: Mechanics

#9 2008-02-25 10:25:29

Re: Mechanics

#10 2008-02-25 10:44:27

Re: Mechanics

#11 2008-02-25 22:46:03

Re: Mechanics

#12 2008-02-26 00:02:05

Re: Mechanics

#13 2008-02-26 03:24:58

Re: Mechanics

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