# Center of mass project exploding at max height

• J-dizzal
In summary, J-dizzal has solved the original problem. The question now is how to solve it if the masses are not equal.
J-dizzal

## Homework Statement

A shell is shot with an initial velocity http://edugen.wileyplus.com/edugen/shared/assignment/test/session.quest2529559entrance1_N10030.mml?size=14&ver=1436312588112 of 17 m/s, at an angle of θ0 = 63° with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass (see the figure). One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and that air drag is negligible?

http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c09/fig09_46new.gif

m2/M d = xcom

## The Attempt at a Solution

I have found the correct answer for this one its 35.77m which i solve by inspection and symmetry. My problem is how do i find the distance of the other particle if it was not symmetric? I tried using,
xcom = m1x1 + m2x2 / M​
where M = sum of all masses. And solving this eq for x2 But i keep getting 23.86m which is the distance of xcom.

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The problem states that it falls straight down from the top of its trajectory. Can you find the x coordinate of the maximum height?

In the middle of your working you have a line that ends with xcom=17..., but I have no idea how you came to that. What relative masses are you assuming?

brainpushups said:
The problem states that it falls straight down from the top of its trajectory. Can you find the x coordinate of the maximum height?
J-dizzal has solved the original problem. The question now is how to solve it if the masses are not equal?

haruspex said:
J-dizzal has solved the original problem. The question now is how to solve it if the masses are not equal?

Ah. Thanks. I misread his question.

J-dizzal said:
My problem is how do i find the distance of the other particle if it was not symmetric?

To be clear: you mean that the shell still explodes at apex in the horizontal direction and that that one of the pieces comes to rest and falls straight down?

brainpushups said:
Ah. Thanks. I misread his question.

To be clear: you mean that the shell still explodes at apex in the horizontal direction and that that one of the pieces comes to rest and falls straight down?

Yes, that's my interpretation. The difference is that the mass fragments' masses might not necessarily be the same this time.

The thing is, whether or not the mass fragments are of equal mass or not, it doesn't change where the "center of mass" lands. In other words, it won't have a effect on $x_{COM}$.

Given the diagram in the attachment, the OP has already figured out that $x_{COM}$ = 23.84 m. But as @haruspex points out, the OP later treats the $x_{COM}$ as 17.89 meters later in the calculations. 'Not sure where this 17.89 m value is coming from.

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In that case my suggestion would be to work symbolically and perhaps write the total mass M as the sum of two masses m and nm where n is the factor that relates the two masses.

collinsmark said:
Yes, that's my interpretation. The difference is that the mass fragments' masses might not necessarily be the same this time.

The thing is, whether or not the mass fragments are of equal mass or not, it doesn't change where the "center of mass" lands. In other words, it won't have a effect on $x_{COM}$.

Given the diagram in the attachment, the OP has already figured out that $x_{COM}$ = 23.84 m. But as @haruspex points out, the OP later treats the $x_{COM}$ as 17.89 meters later in the calculations. 'Not sure where this 17.89 m value is coming from.
brainpushups said:
In that case my suggestion would be to work symbolically and perhaps write the total mass M as the sum of two masses m and nm where n is the factor that relates the two masses.

ok i get it now, my mistake was setting m2= the mass as if it were equal to m1 and using this eq;
x2= (M(xcom) - m1x2) / (m2).​
I believe this eq will work if i had the mass of m2.

haruspex said:
In the middle of your working you have a line that ends with xcom=17..., but I have no idea how you came to that. What relative masses are you assuming?
sorry i should of erased that equation, but 17.895m was the xcom when setting the coordinate system to x=0 at the center of the parabola along the x-axis. it was my initial attempt to find xcom using the eq xcom= (m2/M)d
so then 17.895 + 5.965 = 23.84

## 1. What is the "center of mass project exploding at max height"?

The "center of mass project exploding at max height" is a hypothetical scenario used in physics experiments to study the behavior of objects at the peak of their vertical motion. It involves a projectile exploding at the highest point of its trajectory, resulting in the separation of its center of mass into two distinct components.

## 2. Why is this project important in the field of science?

This project is important because it helps us understand the concept of center of mass and its role in the motion of objects. It also allows us to study the effects of explosions and their impact on the overall motion of a projectile.

## 3. How is the center of mass calculated in this project?

The center of mass is calculated by taking into account the mass and position of each component of the projectile after the explosion. The center of mass is then calculated using the formula: xcm = (m1x1 + m2x2) / (m1 + m2), where xcm is the position of the center of mass and m1 and m2 are the masses of the two components.

## 4. What factors can affect the outcome of this project?

The outcome of this project can be affected by various factors such as the initial velocity and direction of the projectile, the strength and location of the explosion, and the mass and distribution of the projectile's components. Other external factors such as air resistance and gravitational pull may also impact the results.

## 5. How can the results of this project be applied in real-world scenarios?

The results of this project can be applied in real-world scenarios such as understanding the behavior of objects in explosions, designing safer and more efficient projectile systems, and analyzing the motion of objects in space. It can also be used to study the impact of explosions on structures and the environment.

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