Question

A particle of mass $M$ orginally at rest is subjected to a force whose direction is constant but magnitude varies with time according to the relation $F=F_0[1-{\left(\frac{t-T}{T}\right)}^2]$
where $F_0$ and $T$ are constants. The force acts only for the time interval $2T$. The velocity $v$ of the particle after time $2T$ is :

• A. $\frac{2F_{0}T}{M}$
• B. $\frac{F_{0}T}{2M}$
• C. $\frac{4F_{0}T}{3M}$
• D. $\frac{F_{0}T}{3M}$

Answer 1

The correct option is C $\frac{4F_{0}T}{3M}$

From the given problem we can infer that
at $t=0,u=0$
also, $F=F_0[1-{\left(\frac{t-T}{T}\right)}^2]$
or, $F=F_0-\frac{F_0(t-T{)}^2}{T^2}$
Therefore, we have
acceleration, $a=\frac{F_0}{M}-\frac{F_0{(t-T)}^2}{M{T}^2}=\frac{dv}{dt}$

or, ${\int }_0^{v}dv={\int }_{t=0}^{2T}(\frac{F_0}{M}-\frac{F_0{(t-T)}^2}{M{T}^2})dt$
$\Rightarrow v={\left[\frac{F_0}{M}t\right]}_0^{2T}-\frac{F_0}{M{T}^2}{[\frac{t^3}{3}-t^{2}T+T^{2}t]}_0^{2T}\Rightarrow v=\frac{4F_{0}T}{3M}$
The particle will have constant velocity, as force will not act after $t=2T$, which will be equal to $v$.
Hence, option (c) is correct.