Question
Question: How is \(F = v\dfrac{{dm}}{{dt}}\)?...
How is F=vdtdm?
Solution
We are asked to state a situation for which F=vdtdm. We can use the concept of differentiation. We can use the product rule for differentiation. We can then use Newton’s second law of motion and then finally evaluate the situation for which this is true.
Formulae used:
F=dtdp
Where, F is the force on an object, p is the linear momentum of the object and t is the time.
p=mv
Where, m is the mass of the object and v is the velocity of the object.
dtd(uv)=dtdu+dtdv
Complete step by step answer:
We know,
p=mv
Where, p is the linear momentum of the object, m is the mass of the object and v is the velocity of the object.
Also, we know,
F=dtdp
Now, use the formula of momentum in this equation, we get
F=dtd(mv)
Now, we use the product rule for the differentiation,
F=mdtdv+vdtdm−−−−(i)
Now, for real life situations, we consider mass (m) to be constant as the mass of objects in real life situations, its mass remains constant. As the differentiation of constant is zero. Thus,
vdtdm=0
Hence, the equation (i) transforms to
F=mdtdv
Now, the requirement of our question is transforming the equation (i) to
F=vdtdm
Which means the term mdtdv to be zero.
That means, velocity (v) should be constant and mass (m) should be a variable.
Now, this case is only possible only for the situation when the object moves at relativistic speeds.In this situation, the mass varies as the object moves and the velocity of the object remains nearly equal to the speed of light in vacuum throughout its motion.Thus, for this situation, the equation (i) transforms to
F=vdtdm
As the velocity remains constant and thus the parameter mdtdv=0.
Hence, F=vdtdm when an object moves at relativistic speeds.
Note: Students often get confused with the equation as they think the parameter mdtdv will remain but they should always keep in mind that this parameter will approach zero as the velocity of the object approaches relativistic speeds.