How Does an Explosion Affect the Center of Mass in Projectile Motion?

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SUMMARY

The discussion centers on a physics problem involving a mortar shell that explodes at the apex of its trajectory, breaking into two pieces of mass (1/5)m and (4/5)m. The key conclusion is that the center of mass (x_cm) of the exploded pieces remains at the original landing distance (r) from the mortar due to the conservation of momentum. The explosion exerts only internal forces, allowing the center of mass to follow the same trajectory as the shell, confirming that x_cm = r.

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  • Understanding of projectile motion principles
  • Knowledge of conservation of momentum
  • Familiarity with center of mass calculations
  • Basic grasp of internal vs. external forces in physics
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  • Explore advanced projectile motion problems involving multiple objects
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Hi, I was hoping to gather some feedback for one of my homework problems. Any help would be much appreciated. The problem is as follows:

A mortar fires a shell of mass m at speed v_0. The shell explodes at the top of its trajectory as designed. However, rather than creating a shower of colored flares, it breaks into just two pieces, a smaller piece of mass (1/5)m and a larger piece of mass (4/5)m. Both pieces land at exactly the same time. The smaller piece lands perilously close to the mortar (at a distance of zero from the mortar). The larger piece lands a distance d from the mortar. If there had been no explosion, the shell would have landed a distance r from the mortar. Assume that air resistance and the mass of the shell's explosive charge are negligible.

"The two exploded pieces of the shell land at the same time. At the moment of landing, what is the distance x_cm from the mortar to the center of mass of the exploded pieces?"


We're supposed to express it in terms of r. I had no problem w/ the second part of the problem, which wants us to express x_cm in terms of D. That was merely (4/5*m*d)/m, since the other component of the center of mass had a position of 0. I'm stumped on how I can approach the problem expressing it in terms of r, however. The hint given was that "The explosion only exerts internal forces on the particles. The only external force acting on the two-piece system is gravity, so the center of mass will continue along the original trajectory of the shell," which I didn't really find helpful. The position of the first piece would still be 0, wouldn't it? So wouldn't that just leave us with (4/5)*r/m ? This, however, is incorrect.

Edit: After thinking about it some more, would x_cm merely be r?
Edit2: Sweet, it is. Thanks for looking.
 
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Hi there,

It seems like you have already arrived at the correct answer, which is great! The key concept to understand in this problem is the conservation of momentum. When the shell explodes, the total momentum of the system (shell + explosive charge) remains the same, but it is now distributed between the two pieces.

The hint given about the internal and external forces is trying to convey that the explosion does not change the total momentum of the system, but only redistributes it within the system. This means that the center of mass will still follow the same trajectory as the original shell, which is why the position of the first piece is still 0.

Using this understanding, we can see that the center of mass will be located at a distance r from the mortar, since the total mass of the system is still m and the position of the first piece is 0. Therefore, x_cm = r.

I hope this helps clarify any confusion and good job on solving the problem! Keep up the good work.
 

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