The magnitude of force acting on a particle m | vedclass

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(A) The impulse-momentum theorem states that the impulse applied to an object is equal to the change in its momentum.$J = \int F dt = \Delta p = m(v -

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$v_0+\frac{\pi F_0 T_0}{4 m}$

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(A) The impulse-momentum theorem states that the impulse applied to an object is equal to the change in its momentum.$J = \int F dt = \Delta p = m(v - v_0)$The impulse is equal to the area under the $F-t$ curve.The given curve is a semi-ellipse with semi-major axis $a = T_0/2$ and semi-minor axis $b = F_0$. The area of a semi-ellipse is given by $\frac{1}{2} \pi a b$.Area $= \frac{1}{2} \pi \left(\frac{T_0}{2}\right) F_0 = \frac{\pi F_0 T_0}{4}$Equating impulse to the change in momentum:$m(v - v_0) = \frac{\pi F_0 T_0}{4}$$v - v_0 = \frac{\pi F_0 T_0}{4m}$$v = v_0 + \frac{\pi F_0 T_0}{4m}$Thus,the correct option is $A$.

The magnitude of force acting on a particle moving along the $x$-axis varies with time $(t)$ as shown in the figure. If at $t=0$ the velocity of the particle is $v_0$,then its velocity at $t=T_0$ will be

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