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Conversation of energy / momentum problem weird circumstance in equations

  • Thread starter omartech
  • Start date
4
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1. Homework Statement
a projectile of mass M explodes into three fragments while in flight at velocity v. one fragment of mass M/2 travels in the original direction, one of mass 3M/10 comes to rest and the third fragment of mass M/5 travels in the opposite direction. The energy released in the explosion equals to 4.75 times the kinetic energy of the projectile before explosion. What are the velocities of the fragments?


2. Homework Equations
according to my calculations,
Mv (momentum before explosion) = M/2*v1 + M/5*v2 ... since 3M/10 comes to rest. its velocity is zero.

and .. for kinetic energy,
1/2 Mv^2 * 4.75 = 1/2 M/2 *v1^2 + 1/2 M/5 *v2^2.

from these equations we get,
v = v1/2 + v2/5 and 4.75v^2 = v1^2/4 + v2^2/10
Now.. how are these to be solved??? in terms of v?


3. The Attempt at a Solution
I hope the formation of the above two equations is correct?
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
 

Answers and Replies

kdv
336
1
1. Homework Statement
a projectile of mass M explodes into three fragments while in flight at velocity v. one fragment of mass M/2 travels in the original direction, one of mass 3M/10 comes to rest and the third fragment of mass M/5 travels in the opposite direction. The energy released in the explosion equals to 4.75 times the kinetic energy of the projectile before explosion. What are the velocities of the fragments?


2. Homework Equations
according to my calculations,
Mv (momentum before explosion) = M/2*v1 + M/5*v2 ... since 3M/10 comes to rest. its velocity is zero.
Ok but note that when you solve for v2, it must be a negative number then (from the statement of the question)
and .. for kinetic energy,
1/2 Mv^2 * 4.75 = 1/2 M/2 *v1^2 + 1/2 M/5 *v2^2.

from these equations we get,
v = v1/2 + v2/5 and 4.75v^2 = v1^2/4 + v2^2/10
Now.. how are these to be solved??? in terms of v?
Two equations for two unknowns, it should be straightforward. For example, isolate v1 from the first equation, plug in the second equation. This will leave you a single equation for the unknown v2. It's a quadratic equation so use the quadratic formula. solve but keep only the negative answer (because that fragment moves to the left). Nwo plug back in the first equation to find v1.
 
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Oh ok i didn't see that coming, but i'm still getting stuck.. i guess 3 years after you leave math u become real rusty in solving stuff ..

so finally, after i isolate v2 i get v1 = v+v2.. i put this in the main equation
to get: 47.5v^2 = 5(v+v2)^2 - 2v2^2...

solving this makes me reach.. 42.5v^2 - 3v2^2 + 10vv2 = 0. Now how do i proceed!

i'm really sorry for not being able to solve such a trivial problem
 
kdv
336
1
Oh ok i didn't see that coming, but i'm still getting stuck.. i guess 3 years after you leave math u become real rusty in solving stuff ..

so finally, after i isolate v2 i get v1 = v+v2..
How did you get this? This does not follow from your equation for conservation of momentum!

Go back to the equation you posted and isolate v1.
 
4
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yup i got the math wrong initially..

How did you get this? This does not follow from your equation for conservation of momentum!

Go back to the equation you posted and isolate v1.
i end up getting two equations.. and when i put the value of v1 in the quadratic equation to reduce it to one variable i get

27.5v^2 + 8*v*v2 - 14/5v2^2 = 0.. i'm sure about this till here..

now i'm confused as to how to solve v2 in terms of v..

how do i treat this quadratic equation.. which is the ax^2 + bx + c = 0.. i don't know which ones are what and how to proceed..

ohh btw .. this is my first ever forum experience and so far its excellent with amazing responses..thanks alot kdv!
 
kdv
336
1
i end up getting two equations.. and when i put the value of v1 in the quadratic equation to reduce it to one variable i get

27.5v^2 + 8*v*v2 - 14/5v2^2 = 0.. i'm sure about this till here..
looks good.
now i'm confused as to how to solve v2 in terms of v..

how do i treat this quadratic equation.. which is the ax^2 + bx + c = 0.. i don't know which ones are what and how to proceed..
The key idea is that you are looking for v2 in terms of v so you must think of v as if it was a number.

So your equation is

[tex] -\frac{14}{5} v_2^2 + 8 v v_2 + 27.5 v^2 =0 [/tex]
The value of "a" is then -14/5. The value of "b" is [itex] 8 v [/itex] and c i s[itex] 27.5 v^2 [/itex]. So you can solve for v2 in terms of v. And as I said before, keep only the negative solution since v2 must be negative.

At the end you should always put back your values of v1 and v2 in your two initial equations to check that you got the correct answers.

By the way, I just reread the question and I am not sure if the starting equation for energy is correct. The question states that the energy released in the explosion is 4.75 times the initial kinetic energy. This is energy produced by the explosion (chemical potential energy transformed in to kinetic energy) so, if I read the question correctly, the total kinetic energy at the end is

[tex]1/2 M v^2 + 4.75 times 1/2 M v^2 = 5.75 M v^2/2[/tex]

where the first term is the initial energy and the second term is the energy released in the explosion. I might be wrong but this is the way I interpret the question. But this is the kind of question that could be interpreted in different ways so don't take my word for it.

ohh btw .. this is my first ever forum experience and so far its excellent with amazing responses..thanks alot kdv!
I am glad it was helpful. You will find this site incredibly useful and interesting!

That's very kind. Thank you.
 
4
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great!

looks good.
The key idea is that you are looking for v2 in terms of v so you must think of v as if it was a number.

So your equation is

[tex] -\frac{14}{5} v_2^2 + 8 v v_2 + 27.5 v^2 =0 [/tex]
The value of "a" is then -14/5. The value of "b" is [itex] 8 v [/itex] and c i s[itex] 27.5 v^2 [/itex]. So you can solve for v2 in terms of v. And as I said before, keep only the negative solution since v2 must be negative.

At the end you should always put back your values of v1 and v2 in your two initial equations to check that you got the correct answers.

By the way, I just reread the question and I am not sure if the starting equation for energy is correct. The question states that the energy released in the explosion is 4.75 times the initial kinetic energy. This is energy produced by the explosion (chemical potential energy transformed in to kinetic energy) so, if I read the question correctly, the total kinetic energy at the end is

[tex]1/2 M v^2 + 4.75 times 1/2 M v^2 = 5.75 M v^2/2[/tex]

where the first term is the initial energy and the second term is the energy released in the explosion. I might be wrong but this is the way I interpret the question. But this is the kind of question that could be interpreted in different ways so don't take my word for it.


I am glad it was helpful. You will find this site incredibly useful and interesting!

That's very kind. Thank you.
Thanks a lot .. i finally see how to solve these equations.. i was really confused. it was just a long time i hadn't done this.. but yes i'll look into the question clearly to see whether we should add the 4.75 or multiply.. in either case.. we'll get the same kind of equation.. thanks alot again.. and i'll try and be helpful to others seeking help as well ..

cheers.
 
kdv
336
1
Thanks a lot .. i finally see how to solve these equations.. i was really confused. it was just a long time i hadn't done this.. but yes i'll look into the question clearly to see whether we should add the 4.75 or multiply.. in either case.. we'll get the same kind of equation.. thanks alot again.. and i'll try and be helpful to others seeking help as well ..

cheers.
You are very welcome.

Best wishes
 

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