A force exerts an impulse I on a particle changing its speed from…

MCQ

$A$ force exerts an impulse $I$ on a particle changing its speed from $u$ to $2u$. The applied force and the initial velocity are oppositely directed along the same line. The work done by the force is

A
$\frac{3}{2}{\text{I}}\,{\text{u}}$
✓
$\frac{1}{2}{\text{I}}\,{\text{u}}$
C
$I \,u$
D
$2\, I\, u$

✓

Answer

Correct option: B.

$\frac{1}{2}{\text{I}}\,{\text{u}}$

b
$\mathrm{I}=\Delta \mathrm{P}$

$\mathrm{I}=2 \mathrm{mu}-(-\mathrm{mu})$

$\mathrm{I}=3 \mathrm{mu} \ldots(\mathrm{i})$

According to work energy theorem

$\mathrm{W}=\Delta \mathrm{T}$

$\mathrm{W}=\mathrm{T}_{\mathrm{f}}-\mathrm{T}_{\mathrm{i}}$

$\mathrm{W}=\frac{1}{2} \mathrm{m}(2 \mathrm{u})^{2}-\frac{1}{2} \mathrm{m}(-\mathrm{u})^{2}$

$\mathrm{W}=\frac{1}{2} \mathrm{m}\left(4 \mathrm{u}^{2}\right)-\frac{1}{2} \mathrm{mu}^{2}$

$\mathrm{w}=\frac{3 \mathrm{mu}^{2}}{2}$ $...(ii)$

From equation $(i)$ and $(ii)$

$\mathrm{W}=\frac{1}{2} \mathrm{Iu}$

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