Projectile Motion: Methodology Check

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In summary, the conversation involves a discussion of a problem involving a ball being ejected from a mortar at an angle and exploding into two pieces, with one piece dropping vertically. The participants discuss the methodology for finding the velocity of the other piece after the explosion, considering momentum and the change in mass. Ultimately, they determine that the velocity of the other piece is equal to twice the initial horizontal velocity.
  • #1
KiNGGeexD
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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 way?:)Cheers guys!
 
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  • #2
There's not enough information. What does 'drop vertically' mean ? Also your answer should be a vector.
 
  • #3
KiNGGeexD said:
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?
 
  • #4
Hold on I've noticed a flaw in my work! I shall get back with an update ASAP:)
 
  • #5
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
KiNGGeexD said:
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.
 
  • #7
Ok so where would I go from the fact that momentum is conserved?
 
  • #8
KiNGGeexD said:
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.
 
  • #9
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
KiNGGeexD said:
Ok so I had the momentum before :)
What is it? Give magnitude and direction.
 
  • #11
p(before)= m((vcosθ)i+(vsinθ)j))
 
  • #12
KiNGGeexD said:
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.
 
  • #13
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
KiNGGeexD said:
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.)
 
  • #15
v=tanθ/2
 
  • #16
ImageUploadedByPhysics Forums1386716315.091743.jpg
 
  • #17
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
KiNGGeexD said:
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?
 
  • #19
It is zero
 
  • #20
KiNGGeexD said:
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?
 
  • #21
Zero lol
 
  • #22
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
So it's velocity is the same?
 
  • #24
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
 
  • #25
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
 
  • #26
KiNGGeexD said:
Zero lol
Right.

KiNGGeexD said:
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.

KiNGGeexD said:
So it's velocity is the same?
No. Realize that the mass of the second piece is only half the original.
 
  • #27
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?
 
  • #28
KiNGGeexD said:
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(vafter)/2

Solve for vafter.
 
  • #29
V(after)=2(vcosθ)Where the v in the RHS is initial?
 
  • #30
KiNGGeexD said:
V(after)=2(vcosθ)


Where the v in the RHS is initial?
Right.
 
  • #31
Ok! Should I subscript the velocity before and after though? :)

Thanks for your help by the way
 

Related to Projectile Motion: Methodology Check

1. What is projectile motion?

Projectile motion is the motion of an object through the air or other medium under the influence of gravity. It is a type of motion that is characterized by a constant acceleration due to gravity, and a curved path known as a parabola.

2. What is the methodology used to study projectile motion?

The methodology used to study projectile motion involves breaking down the motion into two components: horizontal and vertical. The horizontal component is constant and is not affected by gravity, while the vertical component is affected by gravity and follows a parabolic path. This allows for the use of equations and principles from kinematics and dynamics to analyze the motion.

3. What factors affect the trajectory of a projectile?

The trajectory of a projectile is affected by several factors, including initial velocity, angle of launch, air resistance, and the acceleration due to gravity. The weight and shape of the projectile can also have an impact on its trajectory.

4. How do you calculate the range of a projectile?

The range of a projectile is the horizontal distance it travels before hitting the ground. It can be calculated using the equation: R = (v02sin(2θ))/g, where R is the range, v0 is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity.

5. How is projectile motion used in real life?

Projectile motion is used in many real-life applications, such as sports, engineering, and physics. For example, in sports, the trajectory of a ball is crucial in determining its path and where it will land. In engineering, the principles of projectile motion are used to design and test projectiles, such as rockets and missiles. In physics, projectile motion is used to understand the motion of objects in free fall, as well as to study the effects of air resistance on objects in motion.

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