How far from the starting point do the pieces fall on ground

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1. Oct 31, 2016

annalian

1. The problem statement, all variables and given/known data
An object is thrown with speed 20 m/Sunder the angle pi/3 rad with the horisontal direction. In the highest point, the object is divised into two pieces with same weights. One of them, after the division has the speed 0.
How far from the starting point do the pieces fall on ground?

2. Relevant equations
t=2v0sina/g

3. The attempt at a solution
The piece that has the speed 0:
x1=v0^2sin2a/2g=17.32 m
x2=v0^2sin2a/g=34.64 m
The answer in the book is 10, 20 meters.

2. Oct 31, 2016

PeroK

This problem is in two parts. Can you describe what happens in each part?

Because of this, you will need to be careful about what you mean by $v_0$ and $t$. You will have to define all your variables carefully before you use any equations to relate them.

PS In fact, it's in three parts!

3. Oct 31, 2016

annalian

The first piece follows the movement as if nothing happened (as if it was the object without being devised.)
That's why for it I used L=v0^2sin2a/g
v0=20 m/S is the initial speed of the object
a-the angle the object forms with the horizon.
The second piece just does half of the complete movement, as when it reaches the top, the speed becomes 0. So the road is half of the one of the first piece.

4. Oct 31, 2016

PeroK

I think your answer is better than the one in the book. Do you think conservation of momentum might be involved?

The book answer can't possibly be correct for the problem as stated.

5. Oct 31, 2016

annalian

No, we haven't repeated momentum yet. Do you think the answer will be right, by using the momentum?

6. Oct 31, 2016

PeroK

I would perhaps leave this question. The book answer is clearly not right. But, you might ask yourself why it said the two pieces were of equal weight? Why is the size of the two pieces important?

7. Oct 31, 2016

annalian

to use teh expression mv=m1v1+m2v2
v=v1+v2, v1=0
v2=v

8. Oct 31, 2016

PeroK

Shouldn't that be $m_1 = m_2 = \frac{m}{2}, \ \ v_2 = 2v$?

9. Oct 31, 2016

annalian

Yes, you're right

10. Oct 31, 2016

annalian

So in this case, the road for the first one would be the same, but for the second, I would have to add to the first one v0^2sin2a/2g, where v0=40 m/s?

11. Oct 31, 2016

PeroK

No, you need to think more carefully about the horizontal speed of the second piece. It is $v$ until the highest point, and $2v$ after that.

12. Oct 31, 2016

annalian

Yes, I said it is 20^2sin2a/2g+40^2sin2a/g. Is this correct?

13. Oct 31, 2016

PeroK

Yes, of course, that's correct.

I was thinking that simply it goes 3 times as far as the first piece! I wasn't thinking in terms of $g$ and angles and things.