Maximum possible speed of explosion fragments

[songoku]

Homework Statement

An object moves with speed of 500 m/s then explodes to three equal parts causing the kinetic energy to increase by a factor of 1.5. What is the maximum possible speed obtained by one of the three part?

Relevant Equations

conservation of momentum

Let say the object explodes to three identical parts A, B, and C and initially the mass moves to the right before explosion.

To obtain maximum possible speed (I assume C is the one with maximum speed), I imagine C should move to the right and A and B both should move to the left. Is this correct?

Thanks

 

[Delta2]

I think you have to take into account that the total kinetic energy of the three parts with mass m, will be equal to 1.5 times the initial kinetic energy of $\frac{1}{2}M(500\frac{m}{s})^2$ of the mass $M=3m$.

So I am not sure that your configuration will yield the maximum possible speed for C because in this configuration the kinetic energy will be split in three (not necessarily equal) parts. This problem looks a bit hard, are you sure you have given us the complete statement of the problem?

 

[haruspex]

I think you have to take into account that the total kinetic energy of the three parts with mass m, will be equal to 1.5 times the initial kinetic energy of 12M(500ms)212M(500ms)2 of the mass M=3mM=3m.

So I am not sure that your configuration will yield the maximum possible speed for C because in this configuration the kinetic energy will be split in three (not necessarily equal) parts. This problem looks a bit hard, are you sure you have given us the complete statement of the problem?

I don't think there's any difficulty with @songoku's plan.
If C gains velocity v then each of the others loses velocity v/2. No other way of adjusting the velocities of A and B will give C as much boost.

 

[Adesh]

If no external force causes the explosion then kinetic energy of the system cannot increase, it can remain same or it can decrease (when the explosion causes the environment to heat up). To increase the energy of the system we must do some work on it but in your case nothing external is given.

 

[haruspex]

If no external force causes the explosion then kinetic energy of the system cannot increase, it can remain same or it can decrease (when the explosion causes the environment to heat up). To increase the energy of the system we must do some work on it but in your case nothing external is given.

I think we are supposed to assume there was chemical energy available..

 

[Adesh]

I think we are supposed to assume there was chemical energy available..

Yes, that may be the possibility.

 

[Delta2]

If no external force causes the explosion then kinetic energy of the system cannot increase, it can remain same or it can decrease (when the explosion causes the environment to heat up). To increase the energy of the system we must do some work on it but in your case nothing external is given.

You are wrong here, kinetic energy of a system can increase (or decrease) due to internal forces, the momentum cannot change due to internal forces.

Example: Two charges that attract (or repulse ) each other. If we don't stabilize them the kinetic energy of the system will increase due to internal coulomb forces.

 

[Delta2]

@Adesh your post #4 would be correct if instead of "kinetic" energy at the first sentence you had said just energy or total energy.

 

[Adesh]

Two charges that attract (or repulse ) each other. If we don't stabilize them the kinetic energy of the system will increase due to internal coulomb forces.

But when we hold them in an unstable position its we who have provided them the potential energy.

 

[Adesh]

@Adesh your post #4 would be correct if instead of "kinetic" energy at the first sentence you had said just energy or total energy.

Yes I accept that I was not very clear, but as the original post involved only Kinetic Energy, I meant that total energy of the original post which was kinetic.

I would like to be sorry for an unclear post.

 

[Delta2]

Yes I accept that I was not very clear, but as the original post involved only Kinetic Energy, I meant that total energy of the original post which was kinetic.

I would like to be sorry for an unclear post.

There is hidden chemical potential energy in this problem, so the total energy is Kinetic+chemical.

 

[Delta2]

I don't think there's any difficulty with @songoku's plan.
If C gains velocity v then each of the others loses velocity v/2. No other way of adjusting the velocities of A and B will give C as much boost.

You are a bit unclear here, I was having in mind that the configuration that will yield max speed for C is $v_A=v_B=0$ why this is not the case?

 

[haruspex]

You are a bit unclear here, I was having in mind that the configuration that will yield max speed for C is $v_A=v_B=0$ why this is not the case?

Momentum conservation means you might be able to get more. As against that, you might find that stopping A and B dead means too much net KE is gained.

 

[Delta2]

Momentum conservation means you might be able to get more. As against that, you might find that stopping A and B dead means too much net KE is gained.

Yes there seems to be a tradeoff, if A and B gain/lose speed then C will gain/lose some speed because of conservation of momentum, but it will lose/gain some speed because of conservation of kinetic(+chemical) energy.

 

[songoku]

I think you have to take into account that the total kinetic energy of the three parts with mass m, will be equal to 1.5 times the initial kinetic energy of $\frac{1}{2}M(500\frac{m}{s})^2$ of the mass $M=3m$.

So I am not sure that your configuration will yield the maximum possible speed for C because in this configuration the kinetic energy will be split in three (not necessarily equal) parts. This problem looks a bit hard, are you sure you have given us the complete statement of the problem?

Yes that is the complete statement of the question.

Momentum conservation means you might be able to get more. As against that, you might find that stopping A and B dead means too much net KE is gained.

Yes there seems to be a tradeoff, if A and B gain/lose speed then C will gain/lose some speed because of conservation of momentum, but it will lose/gain some speed because of conservation of kinetic(+chemical) energy.

I have checked the case where the final speed of A and B = 0. It means that by conservation of momentum, C will gain three times the speed of M (initial object before explosion) and the final kinetic energy will be three times of previous so as both of you suspected, the gain in KE is too much.

So based on that case, I think my initial thought (A and B move to left and C to the right) is not possible because C will have speed higher than when the case where A and B stop, meaning that the ratio of KE will be higher than 3My other thought is to consider 2-D momentum, where C moves to the right, A moves to upper left making angle θ with respect to horizontal and B moves to lower left making angle β with respect to horizontal.

I am not sure this is the case because there are too many unknowns (final speed of A,B,C and also θ and β), unless I take θ = β but can I do it? I also do not know whether this will result in C gaining highest speed possible.

Thanks

 

[Delta2]

Yes that is the complete statement of the question.My other thought is to consider 2-D momentum, where C moves to the right, A moves to upper left making angle θ with respect to horizontal and B moves to lower left making angle β with respect to horizontal.

I am not sure this is the case because there are too many unknowns (final speed of A,B,C and also θ and β), unless I take θ = β but can I do it? I also do not know whether this will result in C gaining highest speed possible.

Thanks

That's why I said that this problem seems to be hard. All the possible configurations for the velocities of A,B,C involve 2D or 3D momentum.
However I think that your original scenario might be the one that gives the maximum speed for C(that is A,B,C all moving along the same line, let's say x-axis, with C towards the positive x-axis and A,B towards the negative x-axis). You just have to take into consideration conservation of kinetic energy and not only conservation of momentum when solving the equations for the final speeds of A,B,C.

 

[haruspex]

my initial thought (A and B move to left and C to the right) is not possible because C will have speed higher than when the case where A and B stop

They only need to move to the left relative to their prior motion. If you take the plunge and solve the equations according to your original thought you will get the answer; it just won't be quite what you expected.

 

[Delta2]

@haruspex it seems to me that we will have 3 unknowns (the velocities of A,B,C) and two equations (one from each conservation principle). Do we further assume that $v_A=v_B$?

 

[haruspex]

@haruspex it seems to me that we will have 3 unknowns (the velocities of A,B,C) and two equations (one from each conservation principle). Do we further assume that $v_A=v_B$?

In principle, it is not necessary to assume anything; there must be a maximum achievable speed subject to the given constraints. But a rigorous solution might be rather painful.
It is intuitive that the three will stay in a straight line, which gives us just three unknowns for the velocities. We have the final KE per unit mass and final velocity sum. Maximising the final velocity of C is the third equation.
But I think it is also fairly clear that A and B will turn out equal.

 

[songoku]

They only need to move to the left relative to their prior motion. If you take the plunge and solve the equations according to your original thought you will get the answer; it just won't be quite what you expected.

Ok so, let:
M = mass of object before explosion
m = mass of each part after explosion = 1/3 M
v_A = speed of A after explosion (move to the left)
v_B = speed of B after explosion (move to the left)
v_C = speed of C after explosion (move to the right)

Conservation of momentum:
M . 500 = 1/3 M . v_C - 1/3 M . v_A - 1/3 M . v_B
1500 = v_C - v_A - v_B
v_A = v_C - v_B - 1500 ... (i)KE final = 1.5 KE initial
1/2 . 1/3 M . v_A² + 1/2 . 1/3 M . v_B² + 1/2 . 1/3 M . v_C² = 3/2 . 1/2 . M . 500²

2 v_A² + 2 v_B² + 2 v_C² = 9 . 500²

2 (v_C - v_B - 1500)² + 2 v_B² + 2 v_C² = 9 . 500²

2 v_C² + 2 v_B² + 2 . 1500² - 4 v_C . v_B - 6000 v_C + 6000 v_B + 2 v_B² + 2 v_C² = 9 . 500²

4 v_C² + 4 v_B² + 2 . 1500² - 4 v_C . v_B - 6000 v_C + 6000 v_B = 9 . 500²

How to proceed after this?

Thanks

 

[haruspex]

Ok so, let:
M = mass of object before explosion
m = mass of each part after explosion = 1/3 M
v_A = speed of A after explosion (move to the left)
v_B = speed of B after explosion (move to the left)
v_C = speed of C after explosion (move to the right)

Conservation of momentum:
M . 500 = 1/3 M . v_C - 1/3 M . v_A - 1/3 M . v_B
1500 = v_C - v_A - v_B
v_A = v_C - v_B - 1500 ... (i)KE final = 1.5 KE initial
1/2 . 1/3 M . v_A² + 1/2 . 1/3 M . v_B² + 1/2 . 1/3 M . v_C² = 3/2 . 1/2 . M . 500²

2 v_A² + 2 v_B² + 2 v_C² = 9 . 500²

2 (v_C - v_B - 1500)² + 2 v_B² + 2 v_C² = 9 . 500²

2 v_C² + 2 v_B² + 2 . 1500² - 4 v_C . v_B - 6000 v_C + 6000 v_B + 2 v_B² + 2 v_C² = 9 . 500²

4 v_C² + 4 v_B² + 2 . 1500² - 4 v_C . v_B - 6000 v_C + 6000 v_B = 9 . 500²

How to proceed after this?

Thanks

How do you find the max of one variable as you vary another?

 

[songoku]

How do you find the max of one variable as you vary another?

1st derivative = 0? You mean I take derivative of the equation with respect to v_B?

Thanks

 

[Delta2]

Yes but you got to be carefull of what thing's 1st derivative you going to take. I think you first have to transform the equation to $v_c=f(v_b)$ where f some function that will have square roots (I am afraid) and then solve $f'(v_b)=0$

 

[Delta2]

I think there is a way to show that the maximum of $v_c$ will happen when $$v_a=v_b$$. So plug this at your two equations $$1500=v_c-v_a-v_b$$ and $$KE_{final}=1.5KE_{initial}$$ and see what you get.

 

[songoku]

Yes but you got to be carefull of what thing's 1st derivative you going to take. I think you first have to transform the equation to $v_c=f(v_b)$ where f some function that will have square roots (I am afraid) and then solve $f'(v_b)=0$

I think there is a way to show that the maximum of $v_c$ will happen when $$v_a=v_b$$. So plug this at your two equations $$1500=v_c-v_a-v_b$$ and $$KE_{final}=1.5KE_{initial}$$ and see what you get.

After doing the derivative, I got v_A = v_B then after substituting back, my answer for maximum possible speed of C is 1000 m/s

How can I be sure that this is the maximum possible speed?

Thanks

 

[haruspex]

I think you first have to transform the equation to $v_c=f(v_b)$

Unnecessary. Just go ahead and differentiate wrt v_b and substitute 0 for $\frac{dv_c}{dv_b}$.

 

[haruspex]

After doing the derivative, I got v_A = v_B then after substituting back, my answer for maximum possible speed of C is 1000 m/s

How can I be sure that this is the maximum possible speed?

Thanks

You have completed the proof on the assumption that the three velocities are in the same line.
If not, say C flies off along the y axis. A and B must have equal and opposite velocity components in the x direction. If these are nonzero you can reduce them and spend the saved energy on the y components, keeping momentum conserved and boosting C.

 

[songoku]

Sorry I don't understand

If not, say C flies off along the y axis. A and B must have equal and opposite velocity components in the x direction.

Would not this make final momentum in x - direction equals to zero? Hence the momentum is not conserved?

Thanks

 

[haruspex]

Sorry I don't understandWould not this make final momentum in x - direction equals to zero? Hence the momentum is not conserved?

Thanks

sorry, I was assuming that C would be in the original direction and defining that as the Y axis.

 

[songoku]

Oh ok

Thank you very much haruspex, delta2 and adesh