Using variation of force and time given below, final velocity of a particle of mass $2 \mathrm{~kg}$ moving…
Using variation of force and time given below, final velocity of a particle of mass $2 \mathrm{~kg}$ moving with initial velocity $6 \mathrm{~m} / \mathrm{s}$ will be
$10 \mathrm{~m} / \mathrm{s}$
$5 \mathrm{~m} / \mathrm{s}$
$12 \mathrm{~m} / \mathrm{s}$
$0 \mathrm{~m} / \mathrm{s}$
Solution
Now, using the impulse-momentum theorem to find the final velocity
$\begin{aligned}
& 12=2 \times(v-6) \\
& v-6=\frac{12}{2}=6 \\
& v=12 \mathrm{~m} / \mathrm{s}
\end{aligned}$
So, with the given force-time data $(0,6)$ and $(4,0)$, the final velocity of the particle is $12 \mathrm{~m} / \mathrm{s}$
