Question

a force ( f ) that varies with time ( t ) is exerted on an object. the force is described by the equation ( f(t)=f_{0}e^{-bt} ) where ( f_{0} ) and ( b ) are positive constants. if the object is at rest at time ( t = 0 ), the momentum of the object a very long time later is most nearly
a zero
b ( bf_{0} )
c ( \frac{f_{0}}{b} )
d ( -\frac{f_{0}}{b} )

Explanation:

Step1: Use impulse - momentum theorem

The impulse - momentum theorem states that \(J=\Delta p\). Since the object is at rest at \(t = 0\), \(p(0)=0\), so \(p(t)=J=\int_{0}^{t}F(t')dt'\). Given \(F(t)=F_{0}e^{-bt}\), then \(p(t)=\int_{0}^{t}F_{0}e^{-bt'}dt'\).

Step2: Calculate the integral

Let \(u=-bt'\), \(du=-bdt'\). When \(t' = 0\), \(u = 0\); when \(t'=t\), \(u=-bt\). Then \(\int_{0}^{t}F_{0}e^{-bt'}dt'=\frac{F_{0}}{-b}\int_{0}^{-bt}e^{u}du\).
Using the integral formula \(\int e^{u}du=e^{u}+C\), we have \(\frac{F_{0}}{-b}[e^{u}]_{0}^{-bt}=\frac{F_{0}}{b}(1 - e^{-bt})\).

Step3: Evaluate the limit as \(t

ightarrow\infty\)
As \(t
ightarrow\infty\), \(e^{-bt}
ightarrow0\) (because \(b>0\)). So \(\lim_{t
ightarrow\infty}p(t)=\frac{F_{0}}{b}\).

Answer:

C. \(\frac{F_{0}}{b}\)