Solve Exploding Projectile Problem: Find Distance & Energy

  • Thread starter Smartguy94
  • Start date
  • Tags
    Projectile
In summary: So, what energy was released?I am not going to work out the distance or the energy as I have not been following your work. You have all the equations to do this. Just check the result of the explosion. The exploded part was going a lot faster that the 3 kg so has a lot more energy!You know the speed of the 3 kg part after the explosion, You know the speed of the other bit before the explosion. So you can work out how much energy was added to the 3 kg part in the explosion. This is the energy released. Check your work carefully!Looks like you have worked out how far the 3 kg moved on after the explosion. It
  • #1
Smartguy94
41
0

Homework Statement


A 12 kg shell is launched at an angle of 64 degrees above the horizontal with an initial speed of 120 m/s. When it is at its highest point, the shell exploded into two fragments, one three times the mass of the other. The two fragments reach the ground at the same time. You can ignore air resistance. The heavier fragment lands back at the original launching point.
How far from the launch does the lighter fragment land?
How much energy is released in the explosion?


Homework Equations



p=mv
k=.5mv^2
x=vt
vf=vi+at

The Attempt at a Solution



V(horizontal)=52.604m/s
V(Vertical)=107.885m/s

KE(Vertical)= .5mv(vertical)= 69796.576
KE(Horizontal)= .5mv(horizontal)= 16603.424

KE(start) = PE(max height)
69796.576 = mgh
h = 593.5

vf=vi + at
t=11.006

x = vt
x = 578.941

If b = mass of small piece
b + 3b = 12
b = 3
small mass = 3
big mass = 9

KE(Horizontal) = .5mv^2
16603.424=.5(3)v^2
v=105.21

x = vt
x = 1157.93

total x = 1157.93 + 578.941 = 1736.87m

annnd its wrong :confused:
 
Physics news on Phys.org
  • #2
The horizontal component of the velocity of the larger mass is equal in magnitude to the horizontal component of the velocity of the entire shell at launch. Since neither of the two fragments has any vertical component of velocity and momentum is conserved you can figure out the horizontal component of the smaller fragment's velocity.
 
  • #3
Smartguy94 said:

Homework Statement


A 12 kg shell is launched at an angle of 64 degrees above the horizontal with an initial speed of 120 m/s. When it is at its highest point, the shell exploded into two fragments, one three times the mass of the other. The two fragments reach the ground at the same time. You can ignore air resistance. The heavier fragment lands back at the original launching point.
How far from the launch does the lighter fragment land?
How much energy is released in the explosion?

Homework Equations



p=mv
k=.5mv^2
x=vt
vf=vi+at

The Attempt at a Solution



V(horizontal)=52.604m/s
V(Vertical)=107.885m/s

KE(Vertical)= .5mv(vertical)= 69796.576
KE(Horizontal)= .5mv(horizontal)= 16603.424


KE(start) = PE(max height)
69796.576 = mgh
h = 593.5

vf=vi + at
t=11.006

x = vt
x = 578.941

If b = mass of small piece
b + 3b = 12
b = 3
small mass = 3
big mass = 9

KE(Horizontal) = .5mv^2
16603.424=.5(3)v^2
v=105.21

x = vt
x = 1157.93

total x = 1157.93 + 578.941 = 1736.87m

annnd its wrong :confused:
The bits I highlighted in red could be the source of your error. For a projectile, the KE at the top is not zero, that is only for things fired vertically. The KE at the top is a minimum, but not zero. KE is not a vector, so does not have vertical and horizontal components. It is a scalar, so just has magnitude

The PE at maximum height will be the difference between the KE at the start and the KE at the top.

I have not really followed your calculations exactly, just the overlying assumtions you have made.

EDIT: Note; once you have the right answer, I will show you the overview method I would have used to get it.
 
  • #4
JHamm said:
The horizontal component of the velocity of the larger mass is equal in magnitude to the horizontal component of the velocity of the entire shell at launch. Since neither of the two fragments has any vertical component of velocity and momentum is conserved you can figure out the horizontal component of the smaller fragment's velocity.

i think i got the same answer from what you are saying

you are saying that the larger mass, which is 9kg, have velocity of horizontal launch = 52.604

so since the conservation of momentum is

m1v1 + m2v2 = 0
9(52.604) +3v2 = 0
v2 = 157.812

then to get the distance

x = vt
x = 157.812 (11.006)
x = 1736.879m

and is wrong :confused:
 
  • #5
Smartguy94 said:
i think i got the same answer from what you are saying

you are saying that the larger mass, which is 9kg, have velocity of horizontal launch = 52.604

so since the conservation of momentum is

m1v1 + m2v2 = 0
9(52.604) +3v2 = 0
v2 = 157.812

then to get the distance

x = vt
x = 157.812 (11.006)
x = 1736.879m

and is wrong :confused:

The larger mass was coming back!, so its velocity will be -52.604, and the other one will be going much faster.
 
  • #6
PeterO said:
The bits I highlighted in red could be the source of your error. For a projectile, the KE at the top is not zero, that is only for things fired vertically. The KE at the top is a minimum, but not zero. KE is not a vector, so does not have vertical and horizontal components. It is a scalar, so just has magnitude

The PE at maximum height will be the difference between the KE at the start and the KE at the top.

I have not really followed your calculations exactly, just the overlying assumtions you have made.

EDIT: Note; once yu have the right answer, I will show you the overview method I would have used to get it.

second try

vf=vi + at
t=11.006

x = vit
x = 578.947

12 v(horizontal) = 3v(small) + 9v(large)
12 (52.60) = 3v(small) + 9(52.60)
v(small) = 52.6

x = vt
x = 578.9156

x total = 1157.8626

and wrong again... :confused:
 
  • #7
PeterO said:
The larger mass was coming back!, so its velocity will be -52.604, and the other one will be going much faster.

m1v1 + m2v2 = 0
9(-52.604) +3v2 = 0
v2 = 157.812

x = vt
x = 157.812 (11.006)
x = 1736.879m

it was a typo for the first one but the result is the same
 
  • #8
Smartguy94 said:
m1v1 + m2v2 = 0
9(-52.604) +3v2 = 0
v2 = 157.812

x = vt
x = 157.812 (11.006)
x = 1736.879m

Sorry, didn't notice your eroneous zero.

The momentum at the time of the explosion was not zero. the 12 kg mass was traveling at 52.504 m/s at the time!
 
  • #9
PeterO said:
Sorry, didn't notice your eroneous zero.

The momentum at the time of the explosion was not zero. the 12 kg mass was traveling at 52.504 m/s at the time!

m1v1 + m2v2 = mv
9(-52.604) +3v2 = 12(52.604)
v2 = 368.228

x = vt
x = 368.228(11.006)
x = 4050.5m

still wrong hahaha
 
  • #10
Smartguy94 said:
m1v1 + m2v2 = mv
9(-52.604) +3v2 = 12(52.604)
v2 = 368.228

x = vt
x = 368.228(11.006)
x = 4050.5m

still wrong hahaha

Looks like you have worked out how far the 3 kg moved on after the explosion. It had already traveled quite a distance before the explosion so will presumably land further away that 4050.
 
  • #11
PeterO said:
Looks like you have worked out how far the 3 kg moved on after the explosion. It had already traveled quite a distance before the explosion so will presumably land further away that 4050.

oooh yeaaa i forgot to add it to the original x -__-

now how do i find the energy??

.5mv^2 + mgh??
.5(12)(52.604)^2 + 12(9.8)(578.947)
84687.25J

guess not
 
Last edited:
  • #12
Smartguy94 said:
oooh yeaaa i forgot to add it to the original x -__-

now how do i find the energy??

.5mv^2 + mgh??

The explosion took place at a specific height - so potential doesn't come into it - pure kinetic.

You know how fast the 12 kg was going before explosion 0.5 mv2 for that

You know how fast each of the pieces were going after explosion. 0.5 mv2 a couple more times.

Presumable the energy after is greater than the energy before → the energy from the explosion.
 
  • #13
PeterO said:
The explosion took place at a specific height - so potential doesn't come into it - pure kinetic.

You know how fast the 12 kg was going before explosion 0.5 mv2 for that

You know how fast each of the pieces were going after explosion. 0.5 mv2 a couple more times.

Presumable the energy after is greater than the energy before → the energy from the explosion.

you are the best man, thanks so much
 
  • #14
Smartguy94 said:
you are the best man, thanks so much

Fine: Now my method.

As you carefully calculated, this projectile takes 11 seconds to reach maximum height, covering a horizontal distance of 580m
[these are rounded off figures for explanation purposes]

In the explosion, the two parts [3/4 and 1/4 of the original] head off horizontally.

the large part lands back where the original launch took place, so came back at the same speed as it reached the point.
The velocity of that part changed by 2V [from V to -V] so the other mass, being 1/3 the mass will undergo a 6V change, reaching +7V.

In the 11 seconds taken to get back to the ground, the 9kg travel the 580m [back to where it started]
In the same time, the 3kg will travel 7 times as far [7 times as fast], so will finish a total of 8 times the original 580m → 4640m

Energy wise.

The 12 kg mass arrives at maximum height at speed V so has 1/2 x 12 x V2 of KE. [call that quantity X for the moment]

3/4 of that mass travels at the same speed after explosion, so has 3X/4 units of energy.

1/4 of that mass is traveling at 7 times that speed, so from a mass point of view will have 1/4 the enrgy, but from a speed point of view will have 49 times the energy, so a total of 49X/4.

total energy is thus 13X

Thus the explosion contributes 12X joules of energy

X = 0.5 x 12 x 52.62

You do the numbers - best done with the actual values with all their decimal places.

NOTE: that if you can keep this, what I call global, view of the situation you can keep focussed on what is happening.

I refer to it as living the problem to see the answer, and resorting to the calculator to get the second decimal place of the answer.

Good Luck
 

What is the Exploding Projectile Problem?

The Exploding Projectile Problem is a physics problem that involves calculating the distance and energy of a projectile that explodes into multiple pieces while in motion.

Why is it important to solve the Exploding Projectile Problem?

Solving the Exploding Projectile Problem can provide valuable insights into the behavior of objects in motion and can have practical applications in fields such as engineering and ballistics.

What information is needed to solve the Exploding Projectile Problem?

To solve the Exploding Projectile Problem, you will need to know the initial velocity of the projectile, the mass and velocity of each piece after the explosion, and the angle at which the projectile was launched.

What are the steps to solve the Exploding Projectile Problem?

The steps to solve the Exploding Projectile Problem include breaking down the initial velocity into horizontal and vertical components, using the conservation of momentum to determine the velocity of each piece after the explosion, and using trigonometry and kinematic equations to calculate the distance and energy of each piece.

What are some real-world examples of the Exploding Projectile Problem?

The Exploding Projectile Problem can be seen in various scenarios such as a bullet fragmenting after hitting a target, a fireworks display where projectiles explode into different patterns, or a rock exploding into pieces after being hit by a hammer.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
14
Views
4K
  • Introductory Physics Homework Help
Replies
1
Views
7K
  • Introductory Physics Homework Help
Replies
15
Views
7K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
Back
Top