Projectile Motion: Methodology Check

[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

Is my methodology correct or should I go about this problem in a different wayCheers guys!

 

[Mentz114]

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

 

[Doc Al]

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

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

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

And when cancelling out I got my final answer as

v=gt/2sinθ

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

 

[KiNGGeexD]

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

 

[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?

 

[Doc Al]

Ok so the object is falling at g?

So v=g*t

So p(falling)=mgt/2

M/2 as the mass has halfed?

No.

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

 

[KiNGGeexD]

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

 

[Doc Al]

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

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

 

[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

 

[Doc Al]

Ok so I had the momentum before :)

What is it? Give magnitude and direction.

 

[KiNGGeexD]

p(before)= m((vcosθ)i+(vsinθ)j))

 

[Doc Al]

p(before)= m((vcosθ)i+(vsinθ)j))

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

 

[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?

 

[Doc Al]

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?

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.)

 

[KiNGGeexD]

v=tanθ/2

 

[KiNGGeexD]

 

[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?

 

[Doc Al]

v=tanθ/2

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?

 

[KiNGGeexD]

It is zero

 

[Doc Al]

It is zero

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?

 

[KiNGGeexD]

Zero lol

 

[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?

 

[KiNGGeexD]

So it's velocity is the same?

 

[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 into 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

 

[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

 

[Doc Al]

Zero lol

Right.

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?

The momentum of the second piece must equal the total momentum.

So it's velocity is the same?

No. Realize that the mass of the second piece is only half the original.

 

[KiNGGeexD]

Ahh ok that sounds reasonable!So like I said above

P(total)=mv(after)/2

So v= 2p/m and the I just put in my total momentum as an equation?

 

[Doc Al]

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

OK.

So v after just equals v/2??

Careful with that algebra!

m(vcosθ)=m(vcosθ)/2

Which is basically the same thing!

Rewrite that as:
m(vcosθ) = m(v_after)/2

Solve for v_after.

 

[KiNGGeexD]

V(after)=2(vcosθ)Where the v in the RHS is initial?

 

[Doc Al]

V(after)=2(vcosθ)


Where the v in the RHS is initial?

Right.