Solving a Physics Shell Explosion Question

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The physics problem involves a shell shot at 20 m/s at a 60-degree angle, which explodes into two equal fragments at the top of its trajectory. One fragment falls vertically with zero horizontal velocity, while the other continues moving horizontally. The conservation of momentum in the x-direction is key to solving the problem, as the horizontal component of the velocity must be calculated for the moving fragment. The trajectory of the shell before the explosion can be used to determine the position at the time of the explosion. Ultimately, the horizontal distance traveled by the moving fragment after the explosion can be computed based on these principles.
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Hey everyone! I'm a newbie here, how are you all today?
Anywho, I was doing my physics homework and I came across this question, and I'm stuck :( Can anyone tell me just how to start it?

A shell is shot with an initial velocity of 20 m/s, at an angle of 60 degrees with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. 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?

Totally stuck! Don't even know where to start :(
Thanks for your help!
 
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edit: scratch my idea. I have to think about this more.

2nd edit: ok. Assuming the the explosion means "breaks into two parts without making too much of a fuss" and air resistance is neglected the system's momentum's x-component is conserved and you can solve the x component of velocity of the half that keeps moving in the x-direction. Can you do it from here on?

If approached like this the problem is so unrealistic that I'm not sure if I'm giving you good advice. Maybe some of the official homework helpers could confirm this? It feels odd to think about a system that breaks up like this but since you can break up the equations of motions into components the system's momentum's x-component should be conserved.
 
Last edited:
clucky said:
Hey everyone! I'm a newbie here, how are you all today?
Anywho, I was doing my physics homework and I came across this question, and I'm stuck :( Can anyone tell me just how to start it?

A shell is shot with an initial velocity of 20 m/s, at an angle of 60 degrees with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. 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?

Totally stuck! Don't even know where to start :(
Thanks for your help!

I will assume you can find the position of the shell at explosion. At that time the velocity is horizontal. Momentum will be conserved. Immediately after explosion one half of the shell will be at rest, then it falls. How fast and in what direction will the other half be moving immediately after the explosion? How does the horizontal distance traveled after explosion compare to the horizontal distance traveled before the explosion?
 
OlderDan said:
I will assume you can find the position of the shell at explosion. At that time the velocity is horizontal. Momentum will be conserved. Immediately after explosion one half of the shell will be at rest, then it falls. How fast and in what direction will the other half be moving immediately after the explosion? How does the horizontal distance traveled after explosion compare to the horizontal distance traveled before the explosion?

wahh! I doubted myself in vain.
 

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