Question

A particle of mass $1\text{kg}$ is initially at rest. A force starts acting on it in one direction whose magnitude changes with time. The (F - t) graph is shown in figure. The velocity of the particle at the end of $10\text{s}$ is

• A. $150\text{m/s}$
• B. $120\text{m/s}$
• C. $100\text{m/s}$
• D. $112.5\text{m/s}$

Answer 1

The correct option is D $112.5\text{m/s}$

Given:
Velocity of particle at $(t=0)$; $v_i=0$
Let velocity of particle at $(t=10\text{s})$ is $v_f$.
Using Impulse$\left(J\right)$ = change in momentum $\Delta P$

As we know area under (F -t) graph gives change in momentum.

$\therefore$ we have, $m(v_f-v_i)=$ Area under (F - t) graph from $t=0\text{to}t=10\text{s}$

$\Rightarrow$ $m(v_f-v_i)$= Area under (F - t) graph from $(t=0\text{to}t=5\text{s})+$Area under (F - t) graph from $(t=5\text{to}t=10\text{s})$
$\Rightarrow m{v}_f=\frac{1}{2}\times 5\times 15+5\times 15$
$\Rightarrow 1.v_f=\frac{1}{2}\times 5\times 15+5\times 15$
$\therefore v_f=112.5\text{m/s}$