# Find separation between these two particles - kinematics

1. Aug 17, 2007

### a150daysflood

1. The problem statement, all variables and given/known data
Two particles A and B are projected from the same point O with angle speed 30degree from the horizontal. If A has the speed of projection equal to (7)^0.5m/s and B has twice the projection speed,find the separation between them,when their velocities are mutually perpendicular to each other.

2. Relevant equations

3. The attempt at a solution

I can't solve it.
I need hints regarding the perpendicular part,how can i form an equation linking both of them?
Or anything link between the perpendicular part.
Thanks for helping.

Last edited: Aug 17, 2007
2. Aug 17, 2007

### Dick

You need an equation for the velocity vector of each particle as a function of time. When you have them, take the dot product and set it equal to zero and solve for t. That's when they are perpendicular.

3. Aug 17, 2007

### a150daysflood

Can you explain slightly more? I still don't quite understand.
if i set the dot product to 0,isn't it the maximum point?
thanks for helping

4. Aug 17, 2007

### Dick

The dot product between two vectors a.b=|a|*|b|*cos(theta) where theta is the angle between them. So being perpendicular means the dot product is zero. This is not like setting a derivative to zero.

5. Aug 17, 2007

### nicktacik

If you have
$$\vec{V}_{A}(t)$$
and
$$\vec{V}_{B}(t)$$

then you can find the time when their velocities are perpendicular.

If two vectors are perpendicular, then their dot product is zero. It is crucial that you understand this point.

6. Aug 17, 2007

### a150daysflood

Because cos(90) is zero thus their dot product is zero right?
by the way,any website that can help in explaining how to derive the velocity as a function of time?
Thanks a lot.

Last edited: Aug 17, 2007
7. Aug 17, 2007

### Dick

You use split the motion into x and y components and use kinematical equations. Like y(t)=y0+vy0*t+(1/2)*g*t^2. Does that look familiar? Do you know how to find the position as a function of time? Then just differentiate it.

8. Aug 17, 2007

### a150daysflood

okay thanks :)

Last edited: Aug 17, 2007
9. Aug 17, 2007

### a150daysflood

So,
B( 2 x 7^(0.5) x cos(30) , 2 x 7^(0.5) x sin(30) -9.81t )
A( 7^(0.5) x cos(30) , 7^(0.5) x sin(30) -9.81t )

When i use dot product,i get this :
21/2 + 7/2 -26.0t - 13.0t + 96.2t² = 0
96.2t² - 39.0t + 14 = 0
I can't solve this equation,where did i went wrong?
Thanks a lot

10. Aug 17, 2007

### Dick

Uh, I think your equation is pretty close to correct. That you can't find a real solution would mean there is no solution and not that you did something wrong. Which is funny, because the original phrasing of the problem seem to imply you would find one. Thinking...

11. Aug 17, 2007

### Dick

You can think of it this way. If the initial speeds were equal, the vectors would never be perpendicular. And if they were close to equal the same is true. I'm getting that you need the speed factor to be somewhere over 10 before they are ever perpendicular. Where did you get this problem?

12. Aug 18, 2007

### a150daysflood

one of my problems for my physics olympiad training,i can't seem to solve any of them on my own probably due to lack of knowledge about calculus.
i only learnt c mathematics and not further mathematics.i have to read up on dot product for this question.
anyone interesting in teaching some other questions can PM me. Thanks for your help and time.Have a good day.

The answer given to me for that question is 0.51m.

Last edited: Aug 18, 2007
13. Aug 18, 2007

### learningphysics

Did you give the question exactly as it is written? I solved the problem, and I got the same equation you did...

14. Aug 18, 2007

### a150daysflood

yea i copied it word for word.

15. Aug 18, 2007

### learningphysics

Hmmm... something fishy. I wouldn't worry about it... I think the way you solved it was right... I think there's no solution to this question as it is written. ie the velocities are never perpendicular.

Last edited: Aug 18, 2007