Conservation of momentum during mass accretion

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From Wikipedia:

Mass accretion
The following derivation is for a body that is gaining mass (accretion). A body of time-varying mass $m$ moves at a velocity $mathbf{v}$ at an initial time $t$. In the same instant, a particle of mass $dm$ moves with a velocity $mathbf{u}$. The initial momentum can be written as
$$mathbf{p}_1=mmathbf{v}+mathbf{u}dm tag{1}$$
Now at a time $t+dt$, let both the main body and the particle accrete into a body of velocity $mathbf{v}+dmathbf{v}$. Thus the new momentum of the system can be written as
$$mathbf{p}_2=(m+dm)(mathbf{v}+dmathbf{v})=mmathbf{v}+mdmathbf{v}+mathbf{v}dm+dmdmathbf{v} tag{2}$$
Since $dmdmathbf{v}$ is the product of two small values, it can be ignored, meaning during $dt$ the momentum of the system varies for
$$dmathbf{p}=mathbf{p}_2-mathbf{p}_1=(mmathbf{v}+mdmathbf{v}+mathbf{v}dm)-(mmathbf{v}+mathbf{u}dm)=mdmathbf{v}-(mathbf{u}-mathbf{v})dm tag{3}$$
Therefore, by Newton's second law
$$mathbf{F}_{ext}=frac{dmathbf{p}}{dt}=frac{mdmathbf{v}-(mathbf{u}-mathbf{v})dm}{dt}=mfrac{dmathbf{v}}{dt}-(mathbf{u}-mathbf{v})frac{dm}{dt} tag{4}$$
Noting that $mathbf{u}-mathbf{v}$ is the velocity of dm relative to $m$, symbolized as $mathbf{v}_{rel}$, this final equation can be arranged as
$$mathbf{F}_{ext}+mathbf{v}_{rel}frac{dm}{dt}=mfrac{dmathbf{v}}{dt} tag{5}$$

Shouldn't $dmathbf{p} = mathbf{p}_2-mathbf{p}_1 = 0$, by conservation of momentum of a system?

Semoi · Answer

Suppose you are standing still and I throw a basketball to you. In addition, let's suppose you are standing on a skateboard  while catching the basketball.

Before you catch the basketball your mass is $m_0=M$ and your velocity is $v_0=0m/s$. Hence, your momentum is $p_0 = m_0 cdot v_0 = 0 kg m/s$.
When you catch the basketball you experience a force. Hence, when you catch the ball, which has a mass $m_{ball}$ and a velocity $v_{ball}$, your experience  a change of momentum.
After catching the basketball you have acquired the momentum $p_1$ of the the basketball, due to conservation of momentum. However, since the momentum is now distributed to a larger mass, your skateboard does not roll with the velocity $v_{ball}$, but only with
$v_1 = frac{m_{ball}}{M+m_{ball}}v_{ball}$.

Pranav K · Answer

A body moves of mass $m$ moves with a velocity $u1$

Right behind it, is another body of mass $m$ moving with velocity $u2$

Let $u2>u1$

After some time $t$ the 2 bodies accrete to form a new body of mass $2m$ moving with velocity $v$

By Conservation of Energy we have:
$$frac{m(u1)^2}{2}+frac{m(u2)^2}{2}=frac{2m(v)^2}{2}$$
$$v = pmsqrt{frac{u1^2+u2^2}{2}}$$

momentum before accretion = $mu1+mu2$

momentum after accretion = $2m(pmsqrt{frac{u1^2+u2^2}{2}})$

Now calculating change in momentum before and after accretion:

$$triangle p = mu1 + mu2 - 2m(pmsqrt{frac{u1^2+u2^2}{2}})$$
$$triangle p = m(u1 + u2 - sqrt2(pmsqrt{u1^2+u2^2}))$$
$$triangle p ne 0$$

Hence momentum is not conserved after accretion.

Not_Einstein · Answer

If we rewrite your expression (3) as ??? + ??? −??? and set it to 0 (as it should be since there is no external force) we get d(m?) = ???. So maybe all this is saying is that to this approximation of small changes, the change in the main object's momentum is given by the momentum of the small added mass.

Answered by Not_Einstein on February 26, 2021

Aravind Emmadishetty · Answer

The above formula is derived assuming external forces are present within the system. If there are no external forces $dp$ is equal to $0$.
The mass accretion is similar to the process in a rocket system with direction of $u$ reversed. In free space there are no external forces on rocket therefore $F_{ext} = 0$. Similarly, near earth surface there exists gravitational force therefore $F_{ext} = -Mg$ (here direction of rocket is considered positive).
Refer "Introduction to mechanics by KeplerKolenkov".

Conservation of momentum during mass accretion

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