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(C) The impulse $J$ applied to the particle is equal to the change in momentum $\Delta p = mv_f - mv_i$. Since the particle is initially at rest,$v_i

Vedclass · Accepted Answer

$\frac{2F_0T}{3m}$

Vedclass · Answer

(C) The impulse $J$ applied to the particle is equal to the change in momentum $\Delta p = mv_f - mv_i$. Since the particle is initially at rest,$v_i = 0$,so $mv_f = \int_{0}^{T} F(t) dt$.Substituting the given force expression:$mv_f = \int_{0}^{T} F_0 \left[ 1 - \left( \frac{2t - T}{T} \right)^2 \right] dt$.Let $u = \frac{2t - T}{T}$,then $du = \frac{2}{T} dt$,or $dt = \frac{T}{2} du$.When $t = 0, u = -1$. When $t = T, u = 1$.$mv_f = F_0 \int_{-1}^{1} (1 - u^2) \frac{T}{2} du = \frac{F_0 T}{2} \left[ u - \frac{u^3}{3} \right]_{-1}^{1}$.$mv_f = \frac{F_0 T}{2} \left[ (1 - 1/3) - (-1 + 1/3) \right] = \frac{F_0 T}{2} \left[ 2/3 + 2/3 \right] = \frac{F_0 T}{2} \left( \frac{4}{3} \right) = \frac{2F_0 T}{3}$.Therefore,$v_f = \frac{2F_0 T}{3m}$.

$A$ particle of mass $m$,initially at rest,is acted on by a force $F = F_0 \left\{ {1 - {{\left( {\frac{{2t - T}}{T}} \right)}^2}} \right\}$ during the interval $0 \leq t \leq T$. The velocity of the particle at the end of the interval is:

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