1. ## particle

a particle falls from rest from point A under gravity. after its fallen a distance a another particle is given a downward speed root(8ga) from the same starting point. (i)show the 2 particles collide (ii) find the time and distance from a they collide.

im using s=ut+.5at^2 s=.5gt^2 and i cant find the equation for the second one though as i dont know what its time should be

2. ## Re: particle

Originally Posted by markosheehan
a particle falls from rest from point A under gravity. after its fallen a distance a another particle is given a downward speed root(8ga) from the same starting point. (i)show the 2 particles collide (ii) find the time and distance from a they collide.

im using s=ut+.5at^2 s=.5gt^2 and i cant find the equation for the second one though as i dont know what its time should be
it's not given so it must be a parameter

Set point $A$ as $0$

the first particle moves as

$s_1(t) = -\dfrac 1 2 g t^2$

and assuming we drop the second particle $\delta$ seconds after the first, the second particle moves as

$s_2(t) = -\dfrac 1 2 g (t-\delta)^2 - \sqrt{8 g a}(t-\delta)$

set these equal to one another and solve for $t$

3. ## Re: particle

First, you should state some assumptions. I take it there is no air resistance or other resistance so that the only force is gravity so the acceleration is -g. Yes, after time, t seconds, the first object has fallen a distance (-g/2)t^2. It will have fallen distance a when (-g/2)t^2= -a or t= sqrt(2a/g). But your formula for the second object is incorrect. The distance fallen, t seconds after the first object was dropped (so we can use the same t) is (-g/2)(t-sqrt(2a/g))^2- sqrt(8ag)(t- 2a/g). The "-2a/g" is the difference in time the first object was dropped and the second object was dropped and "-sqrt(8ag)" is the initial speed of the second object. Set those two equal and solve for t.

4. ## Re: particle

Just a point about the variable names. I know you probably copied the problem from a book but it is better not to use the variable "a" in any problem dealing with an acceleration which is also called "a" in the equations. There's too much chance of confusion.

-Dan

5. ## Re: particle

Your equation has 3 variables t,a and the other one and can't type the symbol

6. ## Re: particle

Originally Posted by markosheehan
Your equation has 3 variables t,a and the other one and can't type the symbol
$t$ is the only variable

$a$ and $\delta$ are called parameters. They have a fixed value, we just don't know it.

Your final solution will include these parameters.

7. ## Re: particle

I get to t. Equal to root8ga P-.5gp^2/root8ga -gp
Do you know if that is right

8. ## Re: particle

let $d$ be the distance (particle 1) falls before (particle 2) leaves point A. Since both particles are moving only in the downward direction, let that direction be positive.

after falling a distance $d$, (particle 1) has velocity $v_1 = \sqrt{2gd}$

(particle 2) starts at $v_2 = \sqrt{8gd} = 2v_1$

$\Delta y_1 = \Delta y_2$

$d + v_1 \cdot t + \dfrac{1}{2}gt^2 = 0 + 2v_1 \cdot t + \dfrac{1}{2}gt^2 \implies t = \dfrac{d}{v_1} = \dfrac{d}{\sqrt{2gd}} = \sqrt{\dfrac{d}{2g}}$

$\Delta y_1 = d + \sqrt{2gd} \cdot \sqrt{\dfrac{d}{2g}} + \dfrac{1}{2} g \cdot \dfrac{d}{2g} = d + d + \dfrac{d}{4} = \dfrac{9d}{4}$

$\Delta y_2 = 2 \sqrt{2gd} \cdot \sqrt{\dfrac{d}{2g}} + \dfrac{1}{2} g \cdot \dfrac{d}{2g} = 2d + \dfrac{d}{4} = \dfrac{9d}{4}$