Momentum is a property of a moving object with mass `m ` kg and velocity `v(t) ` m/s (possibly time-dependent) and is measured in kg m/s. Letting momentum be represented by `p(t) ` (so that the momentum can be possibly changing with time) it can be defined by the equation
`p(t) = mv(t) ` where `p(t) ` is in units kg.m/s
Note that although momentum and velocity may depend on time `t ` here, mass is constant with time (unless the object is gaining or losing mass as it proceeds).
Now, Newton's 1st Law of Motion (also Galileo's Law of Inertia) states that
Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it. (see first link below)
and this contains within it the Law of the Conservation of Momentum, which is that momentum of an object moving at a constant velocity is preserved unless an external event (such as collision with another object, or friction due to contact with a surface or air resistance) changes the motion of the object. If an object in motion is accelerating due to force acting upon it, the momentum increases with the increase in velocity, but in each split second the momentum is preserved unless lost to events external to the system in question.
Further, Newton's 2nd Law of Motion states that force acting on a moving body is equal to its mass `m ` in kg multiplied by its acceleration `a(t) ` in m/s^2 (which can depend on time `t ` ), that is
`F(t) = ma(t) ` where `F(t) ` is measured in kg.m/s^2 ( = Newtons, N)
So if, as in the question, an object has momentum equal `p(t)= 5` kg.m/s then we know that
`mv(t) = p(t) = 5 `
Since acceleration (measured in m/s^2) is the rate of change of velocity with respect to time then ` a(t) ` satisfies
`a(t) = (del v(t))/(del t) `
that is, the derivative of `v(t) ` with respect to time `t ` . Since `a(t) ` can change with time in a system, this implies that `F(t) ` may change with time also.
We have then that
`F(t) = m a(t) = m ((del v(t))/(del t)) `
Since the mass `m ` is constant in time we have equivalently that
`F(t) = (del ( m v(t)))/(del t) `
Now, we have been given the momentum at time say `t = t_0 ` so we can write that
`p(t) = m v(t) = f(t)/f(t_0) ` where we know that `f(t)/f(t_0) = 5 `
We now see that, since `t_0 ` is a constant with respect to time,
`F(t) = (del (m v(t)))/(del(t)) = ( del p(t))/(del t) ` ` = 1/f(t_0). (del f(t))/(del t) = (f(t)/f(t_0)) (1/f(t) .(del f(t))/(del t)) `
` = 5 (1/f(t) . (del f(t))/(del t)) `
The expression in braces in this last term is in fact ` (del log(f(t)))/(del t)` so that we have finally that an object with momentum `p(t) = 5 ` kg.m/s has a force acting upon it of
`F(t) = (5. del log(f(t)))/(del t) ` kg.m/s^2 (or N)
Only if `f(t) ` is constant with time is the force acting on the object constant and equal to 5 kg.m/s^2, otherwise it is time-dependent.