# Accidental explosion (Kleppner 3.4)

• geoffrey159
In summary, an instrument-carrying projectile explodes at the top of its trajectory and breaks into two pieces. The larger piece, with three times the mass of the smaller piece, lands at a horizontal distance of 8/3 times the original distance of the explosion. This can be determined using the equation of motion for the center of mass and considering the external forces, such as gravity. Neglecting air resistance and Earth curvature, the final position of the larger piece can be calculated.
geoffrey159

## Homework Statement

An instrument-carrying projectile accidentally explodes at the top of its trajectory. The horizontal distance between the launch point and the point of explosion is ##L##. The projectile breaks into two pieces which fly apart horizontally. The larger piece has three times the mass of the smaller piece. To the surprise of the scientist in charge, the smaller piece returns to Earth at the launching station. How far away does the larger piece land? Neglect air resistance and effects due to the Earth curvature.

Center of mass

## The Attempt at a Solution

The center of mass of the two pieces is ##\vec R = \frac{1}{4m} (m \vec r_1 + 3m \vec r_2)##,
where ##\vec r_1## is the position of the small piece and ##\vec r_2## the position of the large piece.

The question is to find ##r_{2,x}(t_f)##, where ##t_f## is the time both pieces are back to the ground.

At explosion time ##t_e##, the external forces are ##\vec f_{ext} = ( 0, -4mg)## as the pieces fly apart horizontally.
The equation of motion for the center of mass is :
## R_x(t) = tV_{0,x} ##
## R_y(t) = t(V_{0,y}-\frac{gt}{2})##
Now we are given that:
1. ## V_{0,x} = \frac{L}{t_e} ##
2. ## R_y(t_f) = 0 \Rightarrow t_f = \frac{2V_{0,y}}{g} ##
3. ## \dot R_y(t_e) = 0 \Rightarrow V_{0,y} = g t_e ## because ##R_y(t_e)## is a maximum of the curve ## t \rightarrow R_y(t) ##
4. ## r_{1,x}(t_f) = 0 ## as piece 1 goes back to origin

Using everything above:
## R_x(t_f) = t_f V_{0,x} = 2t_e\frac{L}{t_e} = 2L ##.

Using the definition of ##\vec R##, we get ## r_{2,x}(t_f) = \frac{8}{3}L ##

I found this one hard, is it correct?

There is no fext!
The ##R_x(t_f) = 2L## follows directly from symmetry, then using the definition of R gives r2. Looks fine to me.
That's not all that hard :)

Why do you say there are no external forces since the pieces are in motion ?

You're right, the gravitational force is present and has the same value before and after the explosion. I misinterpreted (= read too fast).

Thanks, because if I was wrong, it would mean that I get a right answer with a wrong proof. And that is worse than anything ;-)

## 1. What is an accidental explosion?

An accidental explosion is a sudden, unintended and uncontrolled release of energy, often caused by a chemical reaction or ignition of a flammable substance.

## 2. What factors can contribute to an accidental explosion?

There are several factors that can contribute to an accidental explosion, including the presence of flammable materials, inadequate safety measures, human error, and external factors such as extreme temperatures or pressure.

## 3. How can accidental explosions be prevented?

Accidental explosions can be prevented by implementing proper safety protocols, conducting regular safety inspections, and providing adequate training for handling hazardous materials. It is also important to follow proper storage and handling procedures for flammable substances.

## 4. What are the potential consequences of an accidental explosion?

The consequences of an accidental explosion can range from minor injuries and property damage to severe injuries and even fatalities. It can also have a significant impact on the surrounding environment and community.

## 5. Can accidental explosions be predicted?

While there are certain warning signs and risk factors that can indicate a potential for an accidental explosion, it is not always possible to predict when or where it may occur. Therefore, it is important to always take necessary precautions and follow safety protocols to minimize the risk of an accidental explosion.

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