Question

A particle of mass $3\text{kg}$ has been acted upon by a force $F\left(\text{N}\right)$ whose variation with time $\left(\text{s}\right)$ is shown in the figure.


If the particle is moving along a straight line with initial velocity of $4\text{m/s}$, then find the final velocity of the particle.

• A. $5.33\text{m/s}$
• B. $2\text{m/s}$
• C. $14.5\text{m/s}$
• D. $8.83\text{m/s}$

Answer 1

The correct option is D $8.83\text{m/s}$

From the impulse-momentum theorem:
Impulse $J=\int Fdt=\Delta P....\left(i\right)$
where $\Delta P=P_f-P_i$
$\int Fdt=\text{Area under}F-t\text{graph}$
Area under graph
$=(4\times 3)+(\frac{1}{2}\times 3\times 3)-(\frac{1}{2}\times 2\times 2)$
$\therefore \text{Area}=14.5\text{Ns}$
Now, $P_f=m{v}_f=3\times v_f$
$P_i=m{v}_i=3\times 4=12\text{Ns}$
Hence, from Eq $\left(i\right)$:
$14.5=(3\times v_f)-12$
$\therefore v_f=\frac{26.5}{3}=8.83\text{m/s}$