Question

A ball of mass $m$ is initially at rest. A variable force starts acting on it in a fixed direction, whose magnitude changes with time. The force - time graph is shown in the figure.


Find the speed of the ball at the end of time $t=T$ seconds.

• A. $\frac{3}{5}v\text{m/s}$
• B. $\frac{3}{4}v\text{m/s}$
• C. $\frac{4}{3}v\text{m/s}$
• D. $\frac{2}{3}v\text{m/s}$

Answer 1

The correct option is B $\frac{3}{4}v\text{m/s}$

Impulse is given by
$J=\int Fdt...\left(i\right)$
$\Rightarrow \text{Impulse}=\text{Area under F-t curve}$
$=$ Area of given trapezium
$=\frac{1}{2}(T+\frac{T}{2})\times \frac{mv}{T}=\frac{3mv}{4}$

From Impulse-momentum theorem:
$J=m\Delta v=m(v_f-v_i)...\left(ii\right)$
Equating Eq. $\left(i\right)$ and $\left(ii\right)$
$\frac{3mv}{4}=m(v_f-v_i)=m(v_f-0)$
(ball is initially at rest)
$\therefore v_f=\frac{3}{4}v\text{m/s}$

Hence, $v_f$ is the required speed of the ball at time $t=T\text{s}$.