Force without acceleration

My doubt is very basic and fundamental, by Newton's second law we can say that $F=\frac{dp}{dt}$. Hence, there can also be possible cases when $F=\frac{dm}{dt}v$, when the body is moving with constant velocity in the presence of a force! Then what is the effect of that force as a whole, what is it doing? We have always thought of force as an agent of acceleration, something that provides acceleration, but here the body is under the influence of a net force and still possess a constant velocity!! This whole idea seems to be absurd and can anyone help me in absorbing this concept.

Yes such a situation is possible, but you are no longer considering point mechanics (where $m$ is by definition constant), but the mechanics of a system consisting of multiple point particles. In other words: to arrive at such an equation with changing mass, you have to analyse a system of point masses, for each of which $F = m\dot v$ (in other words, it all depends on how the mass is gained).

A simple model leading to an equation such as the above is the following. Consider an object, let's say an asteroid, of mass $M$ that moves through space filled with small objects at rest of mass $m$, let's say dust. The small objects are at rest. We assume that if the large object hits a dust particle there will be a completely inelastic collision (idealized to occure instantaneously). In other words we can compute the velocity afterwards by momentum conservation (energy is not conserved, since the non-elastic deformation of the two colliding objects creates heat): $$p = Mv = (M+m)v'$$ so the velocity after such an event will be $$v' = \frac{M}{M+m} v.$$ Now we can say that $M$ depends on $t$ since the asteroid gains mass $m$ each time it hits a dust particle. Each of these events can be handled as above, the momentum is conserved but the mass of the asteroid changes, in other words, we arrive at the equation $$F = \dot p = \partial_t (M(t) v(t)) = \dot M(t) v(t) + M(t) \dot v(t).$$ The force $F$ is assumed to only apply to the asteroid, not the dust. So if there is a dust trail which the asteroid sweeps up the mass will rise, and it will slow down, unless an external force is applied.

• Point mechanics do not require constant mass. Point mechanics is an abstraction of non-rotating bodies. Mass can still vary, as can be seen in this question physics.stackexchange.com/q/216895 – Alejandro Menaya Aug 23 '18 at 16:53
• Yes you can do that, but to understand the physical meaning of that construction, you have to do what this answer is doing. If the mass changes due to other mechanisms (e.g. dust particles with non-zero momentum) just using a changing mass will give wrong results. – Sebastian Riese Aug 23 '18 at 16:56
• I can agree with you in this specific example, however the dynamics of a point particle with varying mass is still point particle mechanics, which was what I wanted to notice. – Alejandro Menaya Aug 23 '18 at 16:58
• Your last equation is missing something. The right side is a momentum, but the left and middle have momenutm per time. – Aaron Stevens Aug 23 '18 at 17:22
• yes, indeed it is wrong, I'll fix it. – Sebastian Riese Aug 24 '18 at 16:44

This is the idea behind a rocket. Very simplified, while the rocket looses fuel mass, the exhaust produces thrust

The answer of your question itself lies in it. You have written F to be equal to $F=\frac{dm}{dt}v$. It becomes a variable mass system just like a rocket!

A Special Relativistic view :

In the rest system $\:\mathcal{S}_{o}\:$ of a particle, see ($\alpha$), by a mechanism power is transferred to the particle with rate $\:\overset{\boldsymbol{\cdot}}{\mathrm{q}}_{o}\:$. This rate is with respect to the proper time $\:\tau\:$ and this power changes the rest mass $\:m_{o}\:$ of the particle: $$\overset{\boldsymbol{\cdot}}{\mathrm{q}}_{o}=\dfrac{\mathrm{d}\left(m_{o}c^{2}\right)}{\mathrm{d}\tau}=c^{2}\dfrac{\mathrm{d}m_{o}}{\mathrm{d}\tau} \tag{B-01}$$ In an other inertial system $\:\mathcal{S}\:$ moving with constant 3-velocity $\:\boldsymbol{-}\mathbf{w}\:$ with respect to $\:\mathcal{S}_{o}\:$, the particle is moving with constant velocity $\:\mathbf{w}\:$, see ($\beta$), under the influence of a 'force' $$\boldsymbol{\mathcal{h}}=\dfrac{\overset{\boldsymbol{\cdot}}{\mathrm{q}}_{o}}{c^{2}}\mathbf{w}=\dfrac{\mathrm{d}m_{o}}{\mathrm{d}\tau}\mathbf{w}=\gamma(w)\dfrac{\mathrm{d}m_{o}}{\mathrm{d} t}\mathbf{w} \tag{B-02}$$ This 'force' $\:\boldsymbol{\mathcal{h}}\:$, although acting on the particle, keeps its velocity $\:\mathbf{w}\:$ constant. So, its 3-acceleration is $\:\mathbf{a}=\mathrm{d}\mathbf{w}/\mathrm{d}t =\boldsymbol{0}\:$ and consequently its 4-acceleration $\:\mathbf{A}=\boldsymbol{0}$. This 'force' is defined as heatlike.