Question

A force exerts an impulse $I$ on a particle changing its velocity from initial velocity $u\text{m/s}$ to final velocity $-2u\text{m/s}$. If the applied force and initial velocity are oppositely oriented along the same line, find the work done by force.

• A. $\frac{1}{2}|I|u$
• B. $\frac{3}{2}|I|u$
• C. $\frac{1}{3}|I|u$
• D. $2|I|u$

Answer 1

The correct option is A $\frac{1}{2}|I|u$


Taking rightward motion as $+ve$.
Impulse, $I=$ change in momentum $\Delta p$
$\therefore I=m(v_{final}-v_{initial})$
$=m(-2u-u)$
$=-3mu$
$\Rightarrow |I|=3mu...\left(i\right)$

Let work done by impulsive force is $W$.
From work energy theorem,
$W=\Delta K.E$
$=K.E_{final}-K.E_{initial}=\frac{1}{2}m{v}_{final}^2-\frac{1}{2}m{v}_{initial}^2=\frac{1}{2}m(v_{final}^2-v_{initial}^2)$
$=\frac{1}{2}m\left(\right(2u{)}^2-u^2)=\frac{3}{2}m{u}^2$
$\therefore W=\frac{3}{2}m{u}^2...\left(ii\right)$
From equation $\left(i\right)$ and $\left(ii\right)$
$W=\frac{1}{2}|I|u$