Question

The initial speed of a body of mass $2.0\text{kg}$ is $5.0{\text{ms}}^{-1}$. A force acts for 4 seconds in the direction of motion of the body. The force-time graph is shown in figure. Calculate the impulse of the force and also the final speed of the body.

• A. $8.50\text{N s},9.25{\text{ms}}^{-1}$
• B. $4.25\text{N s},9.25{\text{ms}}^{-1}$
• C. $4.25\text{N s},4.25{\text{ms}}^{-1}$
• D. $8.50\text{N s},4.25{\text{ms}}^{-1}$

Answer 1

The correct option is A $8.50\text{N s},9.25{\text{ms}}^{-1}$

The magnitude of the impulse of a force is equal to the area between the force-time curve and the time-axis.
$\therefore$ Impulse of given force i.e., $J=$ area of $oabcde$
$J=$ area of triangle $o{a}^{\prime }a$ + area of reactangle $a{a}^{\prime }{b}^{\prime }b$ + area of trapezium $b{b}^{\prime }{c}^{\prime }c$ + area of rectangle $c{c}^{\prime }ed$
$=\frac{1}{2}\times 1.5\times 3+1\times 3+\frac{1}{2}(3+2)0.5+2\times 1$
$=2.25+3+1.25+2=8.50\text{N s}$
Impulse = change in momentum $\left(m\Delta v\right)$
Where $\Delta v$ is the increase in velocity.
$\therefore \Delta v=\frac{\text{Impulse}}{m}=\frac{8.50}{2}{\text{ms}}^{-1}=4.25{\text{ms}}^{-1}$
Final speed = Initial speed + $\Delta v$
$\text{Final speed 00000}=5.00{\text{ms}}^{-1}+4.25{\text{ms}}^{-1}=9.25{\text{ms}}^{-1}$