Mixing
Why does momentum appear to be not conserved in this elastic collision?
$begingroup$
Here's a bit of a fun physics paradox, which I will pose, and then answer below.
(These ideas were inspired by Grant Sanderson's fascinating video on how the digits of $pi$ are hidden within elastic collisions.)
An object is travelling to the right, encounters an immovable barrier, undergoes a perfectly elastic collision, and returns to the left at an equal but opposite velocity.
Since the collision is perfectly elastic, kinetic energy and momentum are both conserved.
But the momentum after the collision is not equal to the momentum after the collision, because $p ne -p$. The net change, $Delta p$ should be zero, but instead:
$$Delta p = p_f - p_i = -p - p = -2p ne 0$$
What's up with that?
physics recreational-mathematics
asked Jan 23 at 4:21
Adam HrankowskiAdam Hrankowski
2,094930
$endgroup$
|
show 2 more comments
$begingroup$
Here's a bit of a fun physics paradox, which I will pose, and then answer below.
(These ideas were inspired by Grant Sanderson's fascinating video on how the digits of $pi$ are hidden within elastic collisions.)
An object is travelling to the right, encounters an immovable barrier, undergoes a perfectly elastic collision, and returns to the left at an equal but opposite velocity.
Since the collision is perfectly elastic, kinetic energy and momentum are both conserved.
But the momentum after the collision is not equal to the momentum after the collision, because $p ne -p$. The net change, $Delta p$ should be zero, but instead:
$$Delta p = p_f - p_i = -p - p = -2p ne 0$$
What's up with that?
physics recreational-mathematics
asked Jan 23 at 4:21
Adam HrankowskiAdam Hrankowski
2,094930
$endgroup$
1
$begingroup$
Actually I feel that this is more like a physical problem (which is more proper for the physics SE) instead of a mathematical one, because once we defined what it means to be "immovable" then perhaps there's no difficulty on verifying that the mathematical identity for conversation of momentum must hold...
$endgroup$
– Macrophage
Jan 23 at 4:24
$begingroup$
I'm inclined to agree. The mathematics is in manipulating all those symbols, but the paradoxy part is indeed in the physics. :D
$endgroup$
– Adam Hrankowski
Jan 23 at 4:26
$begingroup$
@Macrophage Is it proper to cross post it to Physics SE if it's already on MSE?
$endgroup$
– Adam Hrankowski
Jan 23 at 4:28
1
$begingroup$
I'm not sure about cross-posting. But I think if any moderator happens to think this is off-topic and decides to close this one, then you may safely post the question on Physics SE.
$endgroup$
– Macrophage
Jan 23 at 4:31
1
$begingroup$
However, I'm not sure what's the convention for answering one's own question on Physics SE. Here at MSE, I find it not very common.
$endgroup$
– Macrophage
Jan 23 at 4:32
|
show 2 more comments
$begingroup$
Here's a bit of a fun physics paradox, which I will pose, and then answer below.
(These ideas were inspired by Grant Sanderson's fascinating video on how the digits of $pi$ are hidden within elastic collisions.)
An object is travelling to the right, encounters an immovable barrier, undergoes a perfectly elastic collision, and returns to the left at an equal but opposite velocity.
Since the collision is perfectly elastic, kinetic energy and momentum are both conserved.
But the momentum after the collision is not equal to the momentum after the collision, because $p ne -p$. The net change, $Delta p$ should be zero, but instead:
$$Delta p = p_f - p_i = -p - p = -2p ne 0$$
What's up with that?
physics recreational-mathematics
asked Jan 23 at 4:21
Adam HrankowskiAdam Hrankowski
2,094930
$endgroup$
Here's a bit of a fun physics paradox, which I will pose, and then answer below.
(These ideas were inspired by Grant Sanderson's fascinating video on how the digits of $pi$ are hidden within elastic collisions.)
An object is travelling to the right, encounters an immovable barrier, undergoes a perfectly elastic collision, and returns to the left at an equal but opposite velocity.
Since the collision is perfectly elastic, kinetic energy and momentum are both conserved.
But the momentum after the collision is not equal to the momentum after the collision, because $p ne -p$. The net change, $Delta p$ should be zero, but instead:
$$Delta p = p_f - p_i = -p - p = -2p ne 0$$
What's up with that?
physics recreational-mathematics
physics recreational-mathematics
asked Jan 23 at 4:21
Adam HrankowskiAdam Hrankowski
2,094930
asked Jan 23 at 4:21
Adam HrankowskiAdam Hrankowski
2,094930
edited Jan 25 at 16:43
Adam Hrankowski
asked Jan 23 at 4:21
Adam HrankowskiAdam Hrankowski
2,094930
asked Jan 23 at 4:21
Adam HrankowskiAdam Hrankowski
2,094930
asked Jan 23 at 4:21
Adam HrankowskiAdam Hrankowski
2,094930
2,094930
1
$begingroup$
Actually I feel that this is more like a physical problem (which is more proper for the physics SE) instead of a mathematical one, because once we defined what it means to be "immovable" then perhaps there's no difficulty on verifying that the mathematical identity for conversation of momentum must hold...
$endgroup$
– Macrophage
Jan 23 at 4:24
$begingroup$
I'm inclined to agree. The mathematics is in manipulating all those symbols, but the paradoxy part is indeed in the physics. :D
$endgroup$
– Adam Hrankowski
Jan 23 at 4:26
$begingroup$
@Macrophage Is it proper to cross post it to Physics SE if it's already on MSE?
$endgroup$
– Adam Hrankowski
Jan 23 at 4:28
1
$begingroup$
I'm not sure about cross-posting. But I think if any moderator happens to think this is off-topic and decides to close this one, then you may safely post the question on Physics SE.
$endgroup$
– Macrophage
Jan 23 at 4:31
1
$begingroup$
However, I'm not sure what's the convention for answering one's own question on Physics SE. Here at MSE, I find it not very common.
$endgroup$
– Macrophage
Jan 23 at 4:32
|
show 2 more comments
1
$begingroup$
Actually I feel that this is more like a physical problem (which is more proper for the physics SE) instead of a mathematical one, because once we defined what it means to be "immovable" then perhaps there's no difficulty on verifying that the mathematical identity for conversation of momentum must hold...
$endgroup$
– Macrophage
Jan 23 at 4:24
$begingroup$
I'm inclined to agree. The mathematics is in manipulating all those symbols, but the paradoxy part is indeed in the physics. :D
$endgroup$
– Adam Hrankowski
Jan 23 at 4:26
$begingroup$
@Macrophage Is it proper to cross post it to Physics SE if it's already on MSE?
$endgroup$
– Adam Hrankowski
Jan 23 at 4:28
1
$begingroup$
I'm not sure about cross-posting. But I think if any moderator happens to think this is off-topic and decides to close this one, then you may safely post the question on Physics SE.
$endgroup$
– Macrophage
Jan 23 at 4:31
1
$begingroup$
However, I'm not sure what's the convention for answering one's own question on Physics SE. Here at MSE, I find it not very common.
$endgroup$
– Macrophage
Jan 23 at 4:32
1
1
$begingroup$
Actually I feel that this is more like a physical problem (which is more proper for the physics SE) instead of a mathematical one, because once we defined what it means to be "immovable" then perhaps there's no difficulty on verifying that the mathematical identity for conversation of momentum must hold...
$endgroup$
– Macrophage
Jan 23 at 4:24
$begingroup$
Actually I feel that this is more like a physical problem (which is more proper for the physics SE) instead of a mathematical one, because once we defined what it means to be "immovable" then perhaps there's no difficulty on verifying that the mathematical identity for conversation of momentum must hold...
$endgroup$
– Macrophage
Jan 23 at 4:24
$begingroup$
I'm inclined to agree. The mathematics is in manipulating all those symbols, but the paradoxy part is indeed in the physics. :D
$endgroup$
– Adam Hrankowski
Jan 23 at 4:26
$begingroup$
I'm inclined to agree. The mathematics is in manipulating all those symbols, but the paradoxy part is indeed in the physics. :D
$endgroup$
– Adam Hrankowski
Jan 23 at 4:26
$begingroup$
@Macrophage Is it proper to cross post it to Physics SE if it's already on MSE?
$endgroup$
– Adam Hrankowski
Jan 23 at 4:28
$begingroup$
@Macrophage Is it proper to cross post it to Physics SE if it's already on MSE?
$endgroup$
– Adam Hrankowski
Jan 23 at 4:28
1
1
$begingroup$
I'm not sure about cross-posting. But I think if any moderator happens to think this is off-topic and decides to close this one, then you may safely post the question on Physics SE.
$endgroup$
– Macrophage
Jan 23 at 4:31
$begingroup$
I'm not sure about cross-posting. But I think if any moderator happens to think this is off-topic and decides to close this one, then you may safely post the question on Physics SE.
$endgroup$
– Macrophage
Jan 23 at 4:31
1
1
$begingroup$
However, I'm not sure what's the convention for answering one's own question on Physics SE. Here at MSE, I find it not very common.
$endgroup$
– Macrophage
Jan 23 at 4:32
$begingroup$
However, I'm not sure what's the convention for answering one's own question on Physics SE. Here at MSE, I find it not very common.
$endgroup$
– Macrophage
Jan 23 at 4:32
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The paradox is created by assuming the "immovable" absorbs no momentum at all.
Let $m$ and $M$ be the masses of the moving object and the barrier, respectively.
Let $v_i$ be the velocity of the moving object prior to the collision. The initial conditions are:
$$p_{total}= mv_i$$
$$E_k{_{total}}= frac{1}{2}m{v_i}^2$$
Let $v_f$ be the velocity of the moving object after the collision. And let $V$ be the velocity of the "immovable" barrier after the collision.
$$p_{total}= mv_f + MV$$
$$E_k{_{total}}= frac{1}{2}m{v_f}^2+frac{1}{2}M{V}^2$$
Assign values as follows:
$m = 1$, $v_i = 1$ and $M=10^{10}$
Units are irrelevant. The purpose of this exercis is to show comparative values. The mass, $M$ is immensely larger than the mass, $m$. Eventually, it will be shown that the velocity, $V$ of the "immovable" barrier is obscenely small compared to $v_i$ and $v_f$.
Now solve for the two unknowns, $v_f$ and $V$, using substitution:
$$p_{total}= mv_i = (1)(1) = 1 = mv_f + MV = (1)v_f + (10^{10})V$$
$$v_f = -(10^{10})V+1 approx -(10^{10})V$$
Note that if we choose larger values for $M$, the ratio $frac {v_f}{V}$ becomes larger and negative. The "immovable" barrier recoils from the collision at an infinitesimal rate, in the opposite direction of $v_f$.
$$E_k{_{total}}= frac{1}{2}m{v_i}^2 = frac{1}{2}(1)(1)^2 = frac{1}{2}$$
$$frac{1}{2}= frac{1}{2}m{v_f}^2+frac{1}{2}M{V}^2 = frac{1}{2}(1)(-(10^{10})V+1)^2+frac{1}{2}(10^{10})V^2$$
Double everything to get rid of the fractions, then expand:
$$1=(10^{20})V^2−2(10^{10})V+1+(10^{10})V^2 $$
$$0=(10^{20}+10^{10})V^2−2(10^{10})V$$
$$0=V((10^{20}+10^{10})V−2(10^{10}))$$
Solving for $V$ yields two solutions:
$$V=0$$ $$ text{(The velocity before the collision)}$$
$$V=frac{2(10^{10})}{10^{20}+10^{10}}$$ $$ text{(The itty-bitty velocity after the collision)}$$
Applying what was calculated earlier:
$$v_f = -(10^{10})V+1 = frac{-2(10^{20})}{10^{20}+10^{10}}+1[[approx -1]]$$
The total momentum after the collision:
$$p_{total}= mv_f + MV$$
$$= (1)(frac{-2(10^{20})}{10^{20}+10^{10}}+1) + (10^{10})(frac{2(10^{10})}{10^{20}+10^{10}})=1$$
$$p_f=p_i=1 \ therefore Delta p=0$$
To see more intuitively the distribution of momentum after the collision, approximate each of the figures, by eliminating the (relatively) insignificant $10^{10}$ from each denominator.
$$mv_f=(1)(frac{-2(10^{20})}{10^{20}+10^{10}}+1) approx frac{-2(10^{20})}{10^{20}+10^{10}}+1 =frac{-2(10^{20})}{10^{20}}+1 = -2 + 1 = 1$$
$$MV = (10^{10})(frac{2(10^{10})}{10^{20}+10^{10}})=frac{2(10^{20})}{10^{20}+10^{10}}approx frac{2(10^{20})}{10^{20}} = 2$$
The "immovable" barrier actually carries a momentum that has twice the magnitude of the moving object. However, its velocity is mindbogglingly small, and approaches zero as the mass of the barrier approaches infinity. We could show this more rigorously with painful limits and epsilons and such, but the intuition shown here should suffice.
Finally, to correct the diagram in the question:
answered Jan 23 at 4:21
Adam HrankowskiAdam Hrankowski
2,094930
$endgroup$
$begingroup$
Of course, the "immovable" barrier could be rigidly attached to the Earth. Every time you move, you (temporarily) change the momentum of the Earth (if you don't consider yourself as part of the Earth).
$endgroup$
– Robert Israel
Jan 23 at 4:35
$begingroup$
Sure. We could treat the mass units as kilograms, then add $5.972 x 10^{24}$ to M. :D
$endgroup$
– Adam Hrankowski
Jan 23 at 4:37
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The paradox is created by assuming the "immovable" absorbs no momentum at all.
Let $m$ and $M$ be the masses of the moving object and the barrier, respectively.
Let $v_i$ be the velocity of the moving object prior to the collision. The initial conditions are:
$$p_{total}= mv_i$$
$$E_k{_{total}}= frac{1}{2}m{v_i}^2$$
Let $v_f$ be the velocity of the moving object after the collision. And let $V$ be the velocity of the "immovable" barrier after the collision.
$$p_{total}= mv_f + MV$$
$$E_k{_{total}}= frac{1}{2}m{v_f}^2+frac{1}{2}M{V}^2$$
Assign values as follows:
$m = 1$, $v_i = 1$ and $M=10^{10}$
Units are irrelevant. The purpose of this exercis is to show comparative values. The mass, $M$ is immensely larger than the mass, $m$. Eventually, it will be shown that the velocity, $V$ of the "immovable" barrier is obscenely small compared to $v_i$ and $v_f$.
Now solve for the two unknowns, $v_f$ and $V$, using substitution:
$$p_{total}= mv_i = (1)(1) = 1 = mv_f + MV = (1)v_f + (10^{10})V$$
$$v_f = -(10^{10})V+1 approx -(10^{10})V$$
Note that if we choose larger values for $M$, the ratio $frac {v_f}{V}$ becomes larger and negative. The "immovable" barrier recoils from the collision at an infinitesimal rate, in the opposite direction of $v_f$.
$$E_k{_{total}}= frac{1}{2}m{v_i}^2 = frac{1}{2}(1)(1)^2 = frac{1}{2}$$
$$frac{1}{2}= frac{1}{2}m{v_f}^2+frac{1}{2}M{V}^2 = frac{1}{2}(1)(-(10^{10})V+1)^2+frac{1}{2}(10^{10})V^2$$
Double everything to get rid of the fractions, then expand:
$$1=(10^{20})V^2−2(10^{10})V+1+(10^{10})V^2 $$
$$0=(10^{20}+10^{10})V^2−2(10^{10})V$$
$$0=V((10^{20}+10^{10})V−2(10^{10}))$$
Solving for $V$ yields two solutions:
$$V=0$$ $$ text{(The velocity before the collision)}$$
$$V=frac{2(10^{10})}{10^{20}+10^{10}}$$ $$ text{(The itty-bitty velocity after the collision)}$$
Applying what was calculated earlier:
$$v_f = -(10^{10})V+1 = frac{-2(10^{20})}{10^{20}+10^{10}}+1[[approx -1]]$$
The total momentum after the collision:
$$p_{total}= mv_f + MV$$
$$= (1)(frac{-2(10^{20})}{10^{20}+10^{10}}+1) + (10^{10})(frac{2(10^{10})}{10^{20}+10^{10}})=1$$
$$p_f=p_i=1 \ therefore Delta p=0$$
To see more intuitively the distribution of momentum after the collision, approximate each of the figures, by eliminating the (relatively) insignificant $10^{10}$ from each denominator.
$$mv_f=(1)(frac{-2(10^{20})}{10^{20}+10^{10}}+1) approx frac{-2(10^{20})}{10^{20}+10^{10}}+1 =frac{-2(10^{20})}{10^{20}}+1 = -2 + 1 = 1$$
$$MV = (10^{10})(frac{2(10^{10})}{10^{20}+10^{10}})=frac{2(10^{20})}{10^{20}+10^{10}}approx frac{2(10^{20})}{10^{20}} = 2$$
The "immovable" barrier actually carries a momentum that has twice the magnitude of the moving object. However, its velocity is mindbogglingly small, and approaches zero as the mass of the barrier approaches infinity. We could show this more rigorously with painful limits and epsilons and such, but the intuition shown here should suffice.
Finally, to correct the diagram in the question:
answered Jan 23 at 4:21
Adam HrankowskiAdam Hrankowski
2,094930
$endgroup$
$begingroup$
Of course, the "immovable" barrier could be rigidly attached to the Earth. Every time you move, you (temporarily) change the momentum of the Earth (if you don't consider yourself as part of the Earth).
$endgroup$
– Robert Israel
Jan 23 at 4:35
$begingroup$
Sure. We could treat the mass units as kilograms, then add $5.972 x 10^{24}$ to M. :D
$endgroup$
– Adam Hrankowski
Jan 23 at 4:37
add a comment |
$begingroup$
The paradox is created by assuming the "immovable" absorbs no momentum at all.
Let $m$ and $M$ be the masses of the moving object and the barrier, respectively.
Let $v_i$ be the velocity of the moving object prior to the collision. The initial conditions are:
$$p_{total}= mv_i$$
$$E_k{_{total}}= frac{1}{2}m{v_i}^2$$
Let $v_f$ be the velocity of the moving object after the collision. And let $V$ be the velocity of the "immovable" barrier after the collision.
$$p_{total}= mv_f + MV$$
$$E_k{_{total}}= frac{1}{2}m{v_f}^2+frac{1}{2}M{V}^2$$
Assign values as follows:
$m = 1$, $v_i = 1$ and $M=10^{10}$
Units are irrelevant. The purpose of this exercis is to show comparative values. The mass, $M$ is immensely larger than the mass, $m$. Eventually, it will be shown that the velocity, $V$ of the "immovable" barrier is obscenely small compared to $v_i$ and $v_f$.
Now solve for the two unknowns, $v_f$ and $V$, using substitution:
$$p_{total}= mv_i = (1)(1) = 1 = mv_f + MV = (1)v_f + (10^{10})V$$
$$v_f = -(10^{10})V+1 approx -(10^{10})V$$
Note that if we choose larger values for $M$, the ratio $frac {v_f}{V}$ becomes larger and negative. The "immovable" barrier recoils from the collision at an infinitesimal rate, in the opposite direction of $v_f$.
$$E_k{_{total}}= frac{1}{2}m{v_i}^2 = frac{1}{2}(1)(1)^2 = frac{1}{2}$$
$$frac{1}{2}= frac{1}{2}m{v_f}^2+frac{1}{2}M{V}^2 = frac{1}{2}(1)(-(10^{10})V+1)^2+frac{1}{2}(10^{10})V^2$$
Double everything to get rid of the fractions, then expand:
$$1=(10^{20})V^2−2(10^{10})V+1+(10^{10})V^2 $$
$$0=(10^{20}+10^{10})V^2−2(10^{10})V$$
$$0=V((10^{20}+10^{10})V−2(10^{10}))$$
Solving for $V$ yields two solutions:
$$V=0$$ $$ text{(The velocity before the collision)}$$
$$V=frac{2(10^{10})}{10^{20}+10^{10}}$$ $$ text{(The itty-bitty velocity after the collision)}$$
Applying what was calculated earlier:
$$v_f = -(10^{10})V+1 = frac{-2(10^{20})}{10^{20}+10^{10}}+1[[approx -1]]$$
The total momentum after the collision:
$$p_{total}= mv_f + MV$$
$$= (1)(frac{-2(10^{20})}{10^{20}+10^{10}}+1) + (10^{10})(frac{2(10^{10})}{10^{20}+10^{10}})=1$$
$$p_f=p_i=1 \ therefore Delta p=0$$
To see more intuitively the distribution of momentum after the collision, approximate each of the figures, by eliminating the (relatively) insignificant $10^{10}$ from each denominator.
$$mv_f=(1)(frac{-2(10^{20})}{10^{20}+10^{10}}+1) approx frac{-2(10^{20})}{10^{20}+10^{10}}+1 =frac{-2(10^{20})}{10^{20}}+1 = -2 + 1 = 1$$
$$MV = (10^{10})(frac{2(10^{10})}{10^{20}+10^{10}})=frac{2(10^{20})}{10^{20}+10^{10}}approx frac{2(10^{20})}{10^{20}} = 2$$
The "immovable" barrier actually carries a momentum that has twice the magnitude of the moving object. However, its velocity is mindbogglingly small, and approaches zero as the mass of the barrier approaches infinity. We could show this more rigorously with painful limits and epsilons and such, but the intuition shown here should suffice.
Finally, to correct the diagram in the question:
answered Jan 23 at 4:21
Adam HrankowskiAdam Hrankowski
2,094930
$endgroup$
$begingroup$
Of course, the "immovable" barrier could be rigidly attached to the Earth. Every time you move, you (temporarily) change the momentum of the Earth (if you don't consider yourself as part of the Earth).
$endgroup$
– Robert Israel
Jan 23 at 4:35
$begingroup$
Sure. We could treat the mass units as kilograms, then add $5.972 x 10^{24}$ to M. :D
$endgroup$
– Adam Hrankowski
Jan 23 at 4:37
add a comment |
$begingroup$
The paradox is created by assuming the "immovable" absorbs no momentum at all.
Let $m$ and $M$ be the masses of the moving object and the barrier, respectively.
Let $v_i$ be the velocity of the moving object prior to the collision. The initial conditions are:
$$p_{total}= mv_i$$
$$E_k{_{total}}= frac{1}{2}m{v_i}^2$$
Let $v_f$ be the velocity of the moving object after the collision. And let $V$ be the velocity of the "immovable" barrier after the collision.
$$p_{total}= mv_f + MV$$
$$E_k{_{total}}= frac{1}{2}m{v_f}^2+frac{1}{2}M{V}^2$$
Assign values as follows:
$m = 1$, $v_i = 1$ and $M=10^{10}$
Units are irrelevant. The purpose of this exercis is to show comparative values. The mass, $M$ is immensely larger than the mass, $m$. Eventually, it will be shown that the velocity, $V$ of the "immovable" barrier is obscenely small compared to $v_i$ and $v_f$.
Now solve for the two unknowns, $v_f$ and $V$, using substitution:
$$p_{total}= mv_i = (1)(1) = 1 = mv_f + MV = (1)v_f + (10^{10})V$$
$$v_f = -(10^{10})V+1 approx -(10^{10})V$$
Note that if we choose larger values for $M$, the ratio $frac {v_f}{V}$ becomes larger and negative. The "immovable" barrier recoils from the collision at an infinitesimal rate, in the opposite direction of $v_f$.
$$E_k{_{total}}= frac{1}{2}m{v_i}^2 = frac{1}{2}(1)(1)^2 = frac{1}{2}$$
$$frac{1}{2}= frac{1}{2}m{v_f}^2+frac{1}{2}M{V}^2 = frac{1}{2}(1)(-(10^{10})V+1)^2+frac{1}{2}(10^{10})V^2$$
Double everything to get rid of the fractions, then expand:
$$1=(10^{20})V^2−2(10^{10})V+1+(10^{10})V^2 $$
$$0=(10^{20}+10^{10})V^2−2(10^{10})V$$
$$0=V((10^{20}+10^{10})V−2(10^{10}))$$
Solving for $V$ yields two solutions:
$$V=0$$ $$ text{(The velocity before the collision)}$$
$$V=frac{2(10^{10})}{10^{20}+10^{10}}$$ $$ text{(The itty-bitty velocity after the collision)}$$
Applying what was calculated earlier:
$$v_f = -(10^{10})V+1 = frac{-2(10^{20})}{10^{20}+10^{10}}+1[[approx -1]]$$
The total momentum after the collision:
$$p_{total}= mv_f + MV$$
$$= (1)(frac{-2(10^{20})}{10^{20}+10^{10}}+1) + (10^{10})(frac{2(10^{10})}{10^{20}+10^{10}})=1$$
$$p_f=p_i=1 \ therefore Delta p=0$$
To see more intuitively the distribution of momentum after the collision, approximate each of the figures, by eliminating the (relatively) insignificant $10^{10}$ from each denominator.
$$mv_f=(1)(frac{-2(10^{20})}{10^{20}+10^{10}}+1) approx frac{-2(10^{20})}{10^{20}+10^{10}}+1 =frac{-2(10^{20})}{10^{20}}+1 = -2 + 1 = 1$$
$$MV = (10^{10})(frac{2(10^{10})}{10^{20}+10^{10}})=frac{2(10^{20})}{10^{20}+10^{10}}approx frac{2(10^{20})}{10^{20}} = 2$$
The "immovable" barrier actually carries a momentum that has twice the magnitude of the moving object. However, its velocity is mindbogglingly small, and approaches zero as the mass of the barrier approaches infinity. We could show this more rigorously with painful limits and epsilons and such, but the intuition shown here should suffice.
Finally, to correct the diagram in the question:
answered Jan 23 at 4:21
Adam HrankowskiAdam Hrankowski
2,094930
$endgroup$
The paradox is created by assuming the "immovable" absorbs no momentum at all.
Let $m$ and $M$ be the masses of the moving object and the barrier, respectively.
Let $v_i$ be the velocity of the moving object prior to the collision. The initial conditions are:
$$p_{total}= mv_i$$
$$E_k{_{total}}= frac{1}{2}m{v_i}^2$$
Let $v_f$ be the velocity of the moving object after the collision. And let $V$ be the velocity of the "immovable" barrier after the collision.
$$p_{total}= mv_f + MV$$
$$E_k{_{total}}= frac{1}{2}m{v_f}^2+frac{1}{2}M{V}^2$$
Assign values as follows:
$m = 1$, $v_i = 1$ and $M=10^{10}$
Units are irrelevant. The purpose of this exercis is to show comparative values. The mass, $M$ is immensely larger than the mass, $m$. Eventually, it will be shown that the velocity, $V$ of the "immovable" barrier is obscenely small compared to $v_i$ and $v_f$.
Now solve for the two unknowns, $v_f$ and $V$, using substitution:
$$p_{total}= mv_i = (1)(1) = 1 = mv_f + MV = (1)v_f + (10^{10})V$$
$$v_f = -(10^{10})V+1 approx -(10^{10})V$$
Note that if we choose larger values for $M$, the ratio $frac {v_f}{V}$ becomes larger and negative. The "immovable" barrier recoils from the collision at an infinitesimal rate, in the opposite direction of $v_f$.
$$E_k{_{total}}= frac{1}{2}m{v_i}^2 = frac{1}{2}(1)(1)^2 = frac{1}{2}$$
$$frac{1}{2}= frac{1}{2}m{v_f}^2+frac{1}{2}M{V}^2 = frac{1}{2}(1)(-(10^{10})V+1)^2+frac{1}{2}(10^{10})V^2$$
Double everything to get rid of the fractions, then expand:
$$1=(10^{20})V^2−2(10^{10})V+1+(10^{10})V^2 $$
$$0=(10^{20}+10^{10})V^2−2(10^{10})V$$
$$0=V((10^{20}+10^{10})V−2(10^{10}))$$
Solving for $V$ yields two solutions:
$$V=0$$ $$ text{(The velocity before the collision)}$$
$$V=frac{2(10^{10})}{10^{20}+10^{10}}$$ $$ text{(The itty-bitty velocity after the collision)}$$
Applying what was calculated earlier:
$$v_f = -(10^{10})V+1 = frac{-2(10^{20})}{10^{20}+10^{10}}+1[[approx -1]]$$
The total momentum after the collision:
$$p_{total}= mv_f + MV$$
$$= (1)(frac{-2(10^{20})}{10^{20}+10^{10}}+1) + (10^{10})(frac{2(10^{10})}{10^{20}+10^{10}})=1$$
$$p_f=p_i=1 \ therefore Delta p=0$$
To see more intuitively the distribution of momentum after the collision, approximate each of the figures, by eliminating the (relatively) insignificant $10^{10}$ from each denominator.
$$mv_f=(1)(frac{-2(10^{20})}{10^{20}+10^{10}}+1) approx frac{-2(10^{20})}{10^{20}+10^{10}}+1 =frac{-2(10^{20})}{10^{20}}+1 = -2 + 1 = 1$$
$$MV = (10^{10})(frac{2(10^{10})}{10^{20}+10^{10}})=frac{2(10^{20})}{10^{20}+10^{10}}approx frac{2(10^{20})}{10^{20}} = 2$$
The "immovable" barrier actually carries a momentum that has twice the magnitude of the moving object. However, its velocity is mindbogglingly small, and approaches zero as the mass of the barrier approaches infinity. We could show this more rigorously with painful limits and epsilons and such, but the intuition shown here should suffice.
Finally, to correct the diagram in the question:
answered Jan 23 at 4:21
Adam HrankowskiAdam Hrankowski
2,094930
edited Jan 23 at 4:43
answered Jan 23 at 4:21
Adam HrankowskiAdam Hrankowski
2,094930
answered Jan 23 at 4:21
Adam HrankowskiAdam Hrankowski
2,094930
answered Jan 23 at 4:21
Adam HrankowskiAdam Hrankowski
2,094930
2,094930
$begingroup$
Of course, the "immovable" barrier could be rigidly attached to the Earth. Every time you move, you (temporarily) change the momentum of the Earth (if you don't consider yourself as part of the Earth).
$endgroup$
– Robert Israel
Jan 23 at 4:35
$begingroup$
Sure. We could treat the mass units as kilograms, then add $5.972 x 10^{24}$ to M. :D
$endgroup$
– Adam Hrankowski
Jan 23 at 4:37
add a comment |
$begingroup$
Of course, the "immovable" barrier could be rigidly attached to the Earth. Every time you move, you (temporarily) change the momentum of the Earth (if you don't consider yourself as part of the Earth).
$endgroup$
– Robert Israel
Jan 23 at 4:35
$begingroup$
Sure. We could treat the mass units as kilograms, then add $5.972 x 10^{24}$ to M. :D
$endgroup$
– Adam Hrankowski
Jan 23 at 4:37
$begingroup$
Of course, the "immovable" barrier could be rigidly attached to the Earth. Every time you move, you (temporarily) change the momentum of the Earth (if you don't consider yourself as part of the Earth).
$endgroup$
– Robert Israel
Jan 23 at 4:35
$begingroup$
Of course, the "immovable" barrier could be rigidly attached to the Earth. Every time you move, you (temporarily) change the momentum of the Earth (if you don't consider yourself as part of the Earth).
$endgroup$
– Robert Israel
Jan 23 at 4:35
$begingroup$
Sure. We could treat the mass units as kilograms, then add $5.972 x 10^{24}$ to M. :D
$endgroup$
– Adam Hrankowski
Jan 23 at 4:37
$begingroup$
Sure. We could treat the mass units as kilograms, then add $5.972 x 10^{24}$ to M. :D
$endgroup$
– Adam Hrankowski
Jan 23 at 4:37
add a comment |
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$begingroup$
Actually I feel that this is more like a physical problem (which is more proper for the physics SE) instead of a mathematical one, because once we defined what it means to be "immovable" then perhaps there's no difficulty on verifying that the mathematical identity for conversation of momentum must hold...
$endgroup$
– Macrophage
Jan 23 at 4:24
$begingroup$
I'm inclined to agree. The mathematics is in manipulating all those symbols, but the paradoxy part is indeed in the physics. :D
$endgroup$
– Adam Hrankowski
Jan 23 at 4:26
$begingroup$
@Macrophage Is it proper to cross post it to Physics SE if it's already on MSE?
$endgroup$
– Adam Hrankowski
Jan 23 at 4:28
1
$begingroup$
I'm not sure about cross-posting. But I think if any moderator happens to think this is off-topic and decides to close this one, then you may safely post the question on Physics SE.
$endgroup$
– Macrophage
Jan 23 at 4:31
1
$begingroup$
However, I'm not sure what's the convention for answering one's own question on Physics SE. Here at MSE, I find it not very common.
$endgroup$
– Macrophage
Jan 23 at 4:32