Shell gets separated in two. What's v of two parts?

[arddi2007]

Homework Statement

The shell of a shotgun, after being fired, with a velocity of v=1000 m/s gets split into two parts with equal masses. One of the two parts continues to move on the same direction as the whole (not separated) shell did, with a velocity of v=1500 m/s.

a) Find the velocity of the other part of the shell
b) Compare the E_k of the whole (not separated) shell with the two separated shells. What can we conclude?

Homework Equations

E_k=mv²/2

The Attempt at a Solution

I tried to solve this, took it to a few people but was unable to get any proper solution. I do not know the correct results, but I know that this problem CAN be solved.

Thank you very much for any help at all.

 

[tiny-tim]

welcome to pf!

hi arddi2007! welcome to pf!

show us what you've tried, and where you're stuck, and then we'll know how to help!

 

[arddi2007]

Thank you for your reply tiny-tim. I am really unsure which formulas to use, since all I know is what v is equal to. I am unsure how to find the mass of the shells, in order to find the E_k. I have tried to solve this equation for many days, I even took it to physics teachers and they were unable to give me any concrete answers. Basically, there is not much progress I've arrived on solving this equation, so if you could only show me which formulas to use, I would be most grateful! Thank you!

 

[tiny-tim]

Thank you for your reply tiny-tim. I am really unsure which formulas to use, since all I know is what v is equal to. I am unsure how to find the mass of the shells, in order to find the E_k. I have tried to solve this equation for many days, I even took it to physics teachers and they were unable to give me any concrete answers. Basically, there is not much progress I've arrived on solving this equation, so if you could only show me which formulas to use, I would be most grateful! Thank you!

Well, starting with a), you have v_i and v_f1 and m, and you want v_f2 …

so what formula do you think applies, that uses those variables?

 

[arddi2007]

Well, starting with a), you have v_i and v_f1 and m, and you want v_f2 …

so what formula do you think applies, that uses those variables?

Not sure... We also know that m gets split in two equal pieces but we've got no specific value for it?! I'm probably missing out something really big! :P

 

[tiny-tim]

Not sure... We also know that m gets split in two equal pieces but we've got no specific value for it?! I'm probably missing out something really big! :P

they're "equal masses" … call the original mass "m", then the two parts are m/2 each …

get on with it!​

 

[arddi2007]

I think I have got something, but I may be doing it all wrong.

So, we have:

v=1000 m/s
m

E_k=(mv²)/2

v₁=1500 m/s
m₁=m/2

E_k1=(m₁v₁²)/2

v₂=?
m₁=m/2

E_k2=(m₁v₂²)/2

This is the part that I have the doubts for:

"E_k=E_k1+E_k2"

Although it doesn't seem commonsense to me, I'm still going to continue solving that:

(mv²)/2=(m₁v₁²)/2+(m₁v₂²)/2

(Removed all /2)

m*1000000 m²/s² = m/2*2250000 m²/s² + m/2*v₂²

m*1000000 m²/s² = m/2 (2250000 m²/s² + v₂²)

m*1000000 m²/s² / m/2 = 2250000 m²/s² + v₂²

2 * 1000000 m²/s² = 2250000 m²/s² + v₂²

2 000 000 m²/s² - 2 250 000 m²/s² = v₂²

v₂² = - 250 000 m²/s²

I have sure missed out something important! :P

 

[tiny-tim]

Well, starting with a), you have v_i and v_f1 and m, and you want v_f2 …

so what formula do you think applies, that uses those variables?

"E_k=E_k1+E_k2"

so you're using the formula for conservation of energy …

what part of the question makes you think that energy is conserved?

 

[arddi2007]

so you're using the formula for conservation of energy …

what part of the question makes you think that energy is conserved?

Not any... That's why I said I'm really unsure about that. I'm generally good with physics but this problem has been eating me up. I cannot figure out any other formula to use in this case. Could you please at least show me which formula(s) to use?

Thank you very much for all your help!

 

[tiny-tim]

in collisions and in "reverse collisions" (explosions), energy is never conserved unless the question specifically says so

but https://www.physicsforums.com/library.php?do=view_item&itemid=53" is always conserved (because there are no external forces)

 

[arddi2007]

Totally forgot about momentum. :P

I did the same thing, and formed a system of equations using the following:

p=m*v=m*1000 m/s

p1=m1*v1=m/2 * 1500 m/s
p2=m1*v2=m/2 * v2
p=p1+p2

m*1000m/s=m/2*1500m/s + m/2*v2

...

2000m/s=1500m/s+v2
v2=500m/s

Is this any bit correct?

 

[tiny-tim]

m*1000m/s=m/2*1500m/s + m/2*v2

...

2000m/s=1500m/s+v2
v2=500m/s

Is this any bit correct?

Yes, that's fine …

how could you not get that on your own?

(btw, note that the centre of mass is still going at the original speed, 1000 )

anyway, what is your answer to b) now?​

 

[arddi2007]

Thank you very much, it now seems all so simple! :P

b) Ek=mv2\2=m*1000000m2/s2 / 2 = 500 000 m2/s2

Ek1=m1v1²\2=...=562500m2/s2*m(mass)
Ek2=m1v2²\2=...=62500m2/s2*m(mass)

Conclusion: Ek>Ek1+Ek2 which means energy cannot be conserved.

Is this correct?Thank you very very much for your help tiny-tim. I still have a few more problems I cannot solve and I hope that you will help me again. PF is lucky to have members like you! Thanks!

 

[tiny-tim]

Conclusion: Ek>Ek1+Ek2 which means energy cannot be conserved.

Is this correct?

Technically, it means that mechanical energy is not conserved …

(energy itself is … "lost" energy has become heat and sound energy)

otherwise, yes. ​

 

[AndyNewton]

The original question contains the curious phrase "The shell ... gets split into two parts ..."

The fact that the phrase sounds a little odd can alert us to the essential core of this question...
We know the shell becomes two parts - but we don't know how that happens or why it happens.
Is it compressed internally, and then simply breaks apart into two pieces under its own elastic forces? Does the shell hit something else?

To be pedantic, the question as written is actually unsolvable because it leaves too many possibilities open - including the possibility that the shell hits something else in which case all bets are off! However, we have to be a little generous to the question writer :-) So, it is reasonable to assume that the shell breaks apart due to something happening within the shell itself. In other words, there are no external forces acting on the shell. Then ... what is the only thing we know for sure is conserved? Total Momentum is conserved, as no external forces are acting on the shell.

That is the idea that tiny-tim has been gently nudging you towards in the discussion above.

Having worked out the implications, using conservation of momentum, you find that the total kinetic energy of the two parts of the shell is greater than the original kinetic energy of the original shell before splitting into two parts. We know that must be so, using the information given in the question, and the (reasonable) assumption that nothing else hits the shell, and conservation of momentum.

You can now deduce something about what must have happened to the shell to create the stated behaviour. The the final part of the question points you in that direction by "...what can we conclude?".

We must conclude that this is no ordinary shotgun "shell". It must be a special James-Bond-issue shell cooked up by "Q" ... what type of shell is it?

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See also the film "Deep Impact" and the nature and results of the first attempt made by the Messiah mission to save the earth.
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