Homework Help: Projectile question

1. Dec 10, 2013

KiNGGeexD

I have another question for you guys:)! I have solved something and I just want to know if my method is ok!

A ball of mass m is ejected from a mortar at speed v and angle θ to the horizontal, at the top of its trajectory it explodes into two equal pieces of mass m/2 and one drops vertically what is the velocity of the other right after the explosion?

The first thing I done was components of velocity

v(x)= (vcosθ)
v(y)=(vsinθ)

Then calculated momentum before and made it equal to momentum after as there was no external force so momentum is conserved

Although I used (vcosθ)i+(vsinθ)j

And when cancelling out I got my final answer as

v=gt/2sinθ

Cheers guys!

2. Dec 10, 2013

Mentz114

There's not enough information. What does 'drop vertically' mean ? Also your answer should be a vector.

3. Dec 10, 2013

Staff: Mentor

OK, so what was the momentum just before the explosion?

And what was the initial momentum of the piece that "drops" vertically?

How did you get this? Why does your answer depend on time?

4. Dec 10, 2013

KiNGGeexD

Hold on I've noticed a flaw in my work! I shall get back with an update ASAP:)

5. Dec 10, 2013

KiNGGeexD

Ok so the object is falling at g?

So v=g*t

So p(falling)=mgt/2

M/2 as the mass has halfed?

6. Dec 10, 2013

Staff: Mentor

No.

They want the speed immediately after the explosion, not after some time later when it has fallen.

7. Dec 10, 2013

KiNGGeexD

Ok so where would I go from the fact that momentum is conserved?

8. Dec 10, 2013

Staff: Mentor

Start by answering the two questions I asked in my first post.

9. Dec 10, 2013

KiNGGeexD

Ok so I had the momentum before :)

So momentum after would be....

p=mv/2 as would the momentum of the other piece!

So

p(falling)=m(vsinθ) as it has no horizontal component

p(other)=m(vsinθ)+(vcosθ)/2

I missed the unit vectors but I did mean to put them in

10. Dec 10, 2013

Staff: Mentor

What is it? Give magnitude and direction.

11. Dec 10, 2013

KiNGGeexD

p(before)= m((vcosθ)i+(vsinθ)j))

12. Dec 10, 2013

Staff: Mentor

No. Note that the moment in question is when the ball is at the top of its trajectory, not when it is first launched.

13. Dec 10, 2013

KiNGGeexD

Ahh ok so then we would just have

p(before)=m(vcosθ)i

As at the top of the trajectory y component of velocity is zero?

14. Dec 10, 2013

Staff: Mentor

Right!

And after the explosion, what is the momentum of the piece that "drops"? (I would assume the "usual" meaning of "drop", though it is somewhat ambiguous.)

15. Dec 10, 2013

KiNGGeexD

v=tanθ/2

16. Dec 10, 2013

KiNGGeexD

17. Dec 10, 2013

KiNGGeexD

So as a vector

v= (vcosθ)i-tanθ/2 j

Although I have not considered the loss of mass for the x component in the above! I know horizontal velocity is constant with constant mass but mass here changes?

18. Dec 10, 2013

Staff: Mentor

Not sure where this comes from. For one thing, the right hand side has no units! So it cannot equal a velocity.

What's the initial velocity of something that is dropped vertically?

19. Dec 10, 2013

KiNGGeexD

It is zero

20. Dec 10, 2013

Staff: Mentor

Right! (That's how I interpret the given information.) So what is its momentum?

And what must be the momentum of the second piece? And thus its velocity?

21. Dec 10, 2013

KiNGGeexD

Zero lol

22. Dec 10, 2013

KiNGGeexD

I disregarded momentum for the falling part so if that is the case then the momentum is just the same which doesn't make sense?

23. Dec 10, 2013

KiNGGeexD

So it's velocity is the same?

24. Dec 10, 2013

KiNGGeexD

A ball of mass M is ejected from a mortar firework at speed v at an angle θ to the horizontal. At the top of its trajectory, it explodes in to two equal pieces of mass M/2. One piece drops vertically.

What is the velocity of the other piece right after the explosion?

That's the exact words

25. Dec 10, 2013

KiNGGeexD

Ok

v=v/2

Which agrees dimensionally and if I think about it if we disregard momentum in the falling piece I should say call it zero rather than disregard it lol!

So
mv(before)=mv/2(after)

So v after just equals v/2??

m(vcosθ)=m(vcosθ)/2

Which is basically the same thing!

This seems far to simple in the context of the problem