Different result with kinetic energy and momentum when a bullet hits a target

[BWR]

TL;DR

kinetic energy and momentum give different result.

A bullet hits a target. Assume that all the energy is transferred to the target without destroying it.
I want to know the target’s velocity after the impact.

bullet: m₁= 123 kg, V_x= 2, V_y= 3, V_z= 4.
target: m₂= 999 kg

Calculations using kinetic energy
Ec_bullet = .5 * m₁ * V² = 1783.5 J
V_target² = Ec_bullet * 2 / m₂, V_target = 1.8896 m/s

Since I'm particularly interested in V target along the Z axis, I calculate:

Vz_target² = (.5 * m₁ * V_z²) * 2 / m₂, Vz_target = 1.40356 m/s


Calculations using momentum
p_bullet = m₁ * V = 662.375 kgm/s
V_target = p_bullet / m₂ = 0.663038 m/s which is different from 1.8896 m/s.

Where I'm wrong?

 

[Ibix]

Is the collision elastic or inelastic?

 

[haruspex]

Your first calculation doesn’t assume "all the energy is transferred to the target". It assumes the target gets all the energy the bullet originally had, but says nothing about the energy the bullet is left with. If you were to continue and calculate that you would find it still has rather a lot, so you have violated energy conservation.

The problem is that a collision with a stationary object can only transfer all of the energy if the bodies have equal mass. Instead, try specifying a coefficient of restitution parameter and finding the value which maximises the energy acquired by the target.

Also, as a matter of technique, never plug in numerical values until you have to. Keep the working entirely symbolic. It has many, many advantages.

 

[kuruman]

To begin with, this problem is absurdly unrealistic. A 123 kg bullet? The mass of a common 55 mm artillery shell is about 43 kg.

Setting that aside, we can certainly write down the energy of the bullet as
$K_b=\frac{1}{2}m_bv_0^2$. The question is, how much of that is transferred to the target? The problem wants us to assume all of it, namely $K_b.$ Once that assumption is made, there should be no energy left for the "bullet" because if there were, not all the energy is transferred. Proceeding then as OP did, we obtain $$\frac{1}{2}m_bv_0^2=\frac{1}{2}m_t v_t^2 \implies v_t=\sqrt{\frac{m_b}{m_t}}v_0.$$However, we cannot stop here because we have to ensure that momentum is also conserved. $$m_bv_0\overset{?}{=}m_t\sqrt{\frac{m_b}{m_t}}v_0 \implies m_b\overset{?}{=}\sqrt{m_tm_b}.$$The equality holds only if the masses are equal which is not the case here.

@haruspex's suggestion to find the maximum energy that can be transferred consistently with momentum conservation is a reasonable alternative interpretation, but it is not at all obvious that this is what the author of the problem had in mind.

I am also concerned about OP's interest in calculating the velocity of the target "along the Z axis." What Z axis?

To @BWR: Is the statement of the problem complete? If not, please post it exactly what was given to you.

 

[BWR]

My English is not good, but here I'm trying to understand 2 things.

1) If $\frac{1}{2}m_bv_b^2 = \frac{1}{2}m_tv_t^2$ and $m_bv_b = m_tv_t$ why the $v_t$ obtained from $K$ is different from that obtained from $p$?

2) Once I understood a reasonable method to do the calculations, I'm mainly interested in $v_t$ along the Z axis which is the axis parallel to $V_z$.

I have a basic math skill, so please explain things in a way that's easy to understand. Thank you.

 

[A.T.]

...why the vt obtained from K is different from that obtained from p?

As several posters have allready pointed out, the problem statement is self-contradictory, so it's not surprising that you get contradicing answers.

 

[Frabjous]

Do the momentum calculation. I get vx=.25m/s, vy=.37m/s, vz=.49m/s for the block. This gives the block a kinetic energy of 219J which is less than the original energy; i.e., some of the original kinetic energy is "lost" to inelastic processes.

 

[haruspex]

My English is not good, but here I'm trying to understand 2 things.

1) If $\frac{1}{2}m_bv_b^2 = \frac{1}{2}m_tv_t^2$ and $m_bv_b = m_tv_t$ why the $v_t$ obtained from $K$ is different from that obtained from $p$?

Because you are specifying an impossible combination of conditions.
In an actual one dimensional collision, there are two unknowns, the two final velocities. By requiring conservation of (kinetic) energy, conservation of momentum and specifying one of the final velocities (a stationary bullet) you have three equations but only two variables. A contradiction is the most likely result.
Certainly momentum should be conserved, so you will either have to drop conservation of energy or allow the bullet some residual KE.

 

[Frabjous]

so you will either have to drop conservation of energy or allow the bullet some residual KE.

Conservation of energy and conservation of momentum hold. This is an inelastic collision, so conservation of KINETIC energy does NOT hold. This is similar to the problem where two bodies stick together after impact that is regularly presented in textbooks.

 

[A.T.]

This is similar to the problem where two bodies stick together after impact that is regularly presented in textbooks.

From the OP:

Assume that all the energy is transferred to the target ...

 

[BWR]

Because you are specifying an impossible combination of conditions.
In an actual one dimensional collision, there are two unknowns, the two final velocities. By requiring conservation of energy, conservation of momentum and specifying one of the final velocities (a stationary bullet) you have three equations but only two variables. A contradiction is the most likely result.

I'd say that answers my question. Thank you.

Certainly momentum should be conserved, so you will either have to drop conservation of energy or allow the bullet some residual KE.

Does that mean that I should use momentum to get a reasonable result?

 

[Frabjous]

From the OP:

Kinetic energy and energy are not the same thing

 

[A.T.]

Kinetic energy and energy are not the same thing

I know. So what? There still a contradiction between 'bodies stick together after impact' and 'all the energy is transferred to the target'.

 

[A.T.]

that mean that I should use momentum to get a reasonable result?

You can do that, if you know that the bullet gets stuck in the target, or know its velocity after exit / ricochet.

 

[Frabjous]

I know. So what? There still a contradiction between 'bodies stick together after impact' and 'all the energy is transferred to the target'.

They are both inelastic collisions. I will let you figure out what “similar” means on your own.

For this problem, one uses conservation of momentum and v final = 0 for the bullet.

 

[BWR]

You can do that, if you know that the bullet gets stuck in the target, or know its velocity after exit / ricochet.

I suppose that the bullet gets stuck in the target and it transfer all its K to the target; is that a contradiction?

But neither info was in the original problem statement. Where did you get the problem from?

From my mind.

 

[A.T.]

For this problem, one uses conservation of momentum and v final = 0 for the bullet.

If energy received by the target means to include all forms of energy, then so should the energy lost by the bullet. So 'transferring all energy' would mean that the bullet is annihilated and releases its relativistic rest energy into the target. Without destroying it tough.

But seriously, since the OP is reproducing the problem statement from memory, it's quite possible that the original formulation is more precise, and matches your interpretation.

 

[A.T.]

I suppose that the bullet gets stuck in the target and it transfer all its K to the target; is that a contradiction

Yes, as allready explained by others. The bullet would still be moving (together with the target), so it would still have some kinetic energy.

 

[Gleb1964]

However, we cannot stop here because we have to ensure that momentum is also conserved. $$m_bv_0\overset{?}{=}m_t\sqrt{\frac{m_b}{m_t}}v_0 \implies m_b\overset{?}{=}\sqrt{m_tm_b}.$$The equality holds only if the masses are equal which is not the case here.

May it is better to assume that summary momentum of <bullet+target> concerved before and after collision.

 

[kuruman]

May it is better to assume that summary momentum of <bullet+target> concerved before and after collision.

Total momentum is always conserved in a collision. The point of my post was to show that all the energy of the "bullet" cannot be transferred to the target as claimed because addition of the necessary condition that momentum is conserved constrains the masses to be equal. In other words, because of momentum conservation, the energy transfer between the given masses cannot be as claimed because something has to be changed, either the masses or the amount of energy transferred.

 

[Herman Trivilino]

TL;DR: kinetic energy and momentum give different result.

You mean the conservation laws give different results. Conservation of momentum is a purely dynamical relation. Conservation of energy isn't, it includes thermal interactions.

You're getting a "different result" because you are ignoring the thermal interactions.

 

[haruspex]

I'd say that answers my question. Thank you.


Does that mean that I should use momentum to get a reasonable result?

Yes, use momentum, but that is only one equation and you need two.
Are you familiar with coefficient of restitution? That is a property of the pair of materials involved. The value is from 0 to 1, with 0 being coalescence and 1 being perfectly elastic.
(In an oblique collision, you have to decompose the motions along the direction normal to and parallel to the contacting surfaces. The parallel velocities don’t change; the coefficient of restitution applies along the normal. This also assumes the mass centres lie on the normal at the instant of collision; otherwise you have to worry about spin. Oh, and no friction.)
So you could choose a value for that. (For historical reasons, it is usually represented by the unknown "e".)
If you choose a different constraint for the second equation, such as the bullet finishing at rest, you may find that corresponds to a coefficient outside the range [0,1], which would be unphysical.

 

[haruspex]

Conservation of energy and conservation of momentum hold.

True, I was being lazy, which could mislead a student .I should have specified mechanical energy.
Ah, no, that doesn’t do it because that would allow one of the bodies to have turned some of the KE into elastic potential energy and not released it because of some latch feature. So I have to say kinetic energy.
No, still no good. Perhaps an oblique surface at the contact point on the target resulted in the target acquiring spin, so it has to be conservation of linear kinetic energy, right?

This is an inelastic collision, so conservation of KINETIC energy does NOT hold.

In post #1, the OP specified that the bullet gives all its KE to the target (implying conservation of KE, and that the bullet finishes at rest) but found this was incompatible with conservation of momentum. Those three constraints cannot all be true (unless the masses are equal). Whether that means the collision must be made inelastic or the bullet must be allowed some residual KE is a choice for the OP.
There appears to be a desire to maximise the velocity of the target, in which case the correct choice is perfectly elastic.

 

[BWR]

Are you familiar with coefficient of restitution?

No, but I found something in the web. Unfortunately I have no idea for its value, but probably I don't need it.

Suppose I fire a (small) bullet at a (big) flying target.
The bullet slows down upon impact, but the target accelerates with the bullet stuck to it.
The target brings the bullet to a complete stop relative to the target, but both are moving relative to the ground (my “inertial” reference frame).
Here's what I wrote:

$m_b v_{bi} + m_t v_{ti} = m_b v_{bf} + m_t v_{tf}$, but $v_{bf} = v_{tf} = v_f$
$m_b v_{bi} + m_t v_{ti} = (m_b + m_t) v_f$
$v_f = (m_b v_{bi} + m_t v_{ti}) / (m_b + m_t)$

Is that correct?

Since the bullet accelerates the target, is there any chance to calculate that acceleration? What is the acceleration along the z-axis?

 

[haruspex]

No, but I found something in the web. Unfortunately I have no idea for its value, but probably I don't need it.

Suppose I fire a (small) bullet at a (big) flying target.
The bullet slows down upon impact, but the target accelerates with the bullet stuck to it.
The target brings the bullet to a complete stop relative to the target, but both are moving relative to the ground (my “inertial” reference frame).

That is called coalescence. It is the case which maximises the lost KE and corresponds to coefficient of restitution = 0.

Here's what I wrote:

$m_b v_{bi} + m_t v_{ti} = m_b v_{bf} + m_t v_{tf}$, but $v_{bf} = v_{tf} = v_f$
$m_b v_{bi} + m_t v_{ti} = (m_b + m_t) v_f$
$v_f = (m_b v_{bi} + m_t v_{ti}) / (m_b + m_t)$

Is that correct?

Yes.

Since the bullet accelerates the target, is there any chance to calculate that acceleration? What is the acceleration along the z-axis?

In general, one does not know how long the collision process lasts, so the acceleration is unknown.
An elastic collision approximates SHM, so the acceleration increases then decreases.
A coalescence will tend to have a more constant acceleration. You could estimate it from the depth of penetration.

 

[BWR]

A coalescence will tend to have a more constant acceleration. You could estimate it from the depth of penetration.

Thank you. Once I estimated the time, how do I calculate the acceleration along the z-axis? should I use $F = \Delta p / \Delta t$, $F_z = \Delta p_z / \Delta t$, $a_z = F_z / (m_b+m_t)$?

EDIT
Since I know $\Delta V = v_f - v_{ti}$, I could use $\Delta V_z / \Delta t = a_z$ , right?

 

[haruspex]

Thank you. Once I estimated the time, how do I calculate the acceleration along the z-axis? should I use $F = \Delta p / \Delta t$, $F_z = \Delta p_z / \Delta t$, $a_z = F_z / (m_b+m_t)$?

EDIT
Since I know $\Delta V = v_f - v_{ti}$, I could use $\Delta V_z / \Delta t = a_z$ , right?

Yes, just be clear about which reference frame you want.

 

[BWR]

I surely want an inertial frame (the ground is good enough). Why did you tell me that?

 

[haruspex]

I surely want an inertial frame (the ground is good enough). Why did you tell me that?

Because I could not be sure which you wanted and I felt it would be an easy mistake to make. But you seem to know what you are doing.

 

[kuruman]

Here's what I wrote:

$m_b v_{bi} + m_t v_{ti} = m_b v_{bf} + m_t v_{tf}$, but $v_{bf} = v_{tf} = v_f$
$m_b v_{bi} + m_t v_{ti} = (m_b + m_t) v_f$
$v_f = (m_b v_{bi} + m_t v_{ti}) / (m_b + m_t)$

Is that correct?

I am adding this note to complete your understanding. The expression you derived for the final velocity of the coalesced object is the velocity of the center of mass (CM) which remains unchanged throughout the collision because there are no external forces acting on the combined bullet plus target system. This constancy of the CM velocity is the essence of momentum conservation.

Your expression is correct because it says that bullet and target (assumed to be point masses) right after the collision move at the same velocity as the CM. In other words when the coalescence is complete, they do not separate, the CM is where the bullet and target are and all three move together as one.