Question

The magnitude of force acting on a particle moving along x-axis varies with time (t) as shown in figure. If at t=o the velocity of [particle is $v_0$, then its velocity at $t=T_0$ will be

• A. $v_0+\frac{\pi F_0{T}_0}{4m}$
• B. $v_0+\frac{\pi F_0}{2m}$
• C. $v_0+\frac{\pi T_0^2}{4m}$
• D. $v_0+\frac{\pi F_0{T}_0}{m}$

Answer 1

Answer: (A)

$v_0+\frac{\pi F_0{T}_0}{4m}$

From the graph, radius of semi-circle $=F_0=\frac{T_0}{2}$

As we know that,

Area under $F-t$ graph gives change in momentum

i.e ., $m\Delta v$

Therefore, Area $=m\Delta v=\frac{\pi {F^2}_0}{2}=\frac{\pi {F^2}_0}{2}\times \frac{T_0}{2}$

( substituting the value of $F_0$)

$\begin{array}{c}m\Delta v=\frac{\pi F_0{T}_0}{4}\\ \Delta v=\frac{\pi F_0{T}_0}{4m}\end{array}$

Thus, if final velocity is $v$ and initial velocity is $V_0$. then

$V-V_0=\Delta V$

and, $V=V_0+\Delta V=V_0+\frac{\pi F_0{T}_0}{4m}$

$\therefore$ Option $A$ is correct.