Q. A body of $2\ kg$ has an initial speed $5\ ms^{- 1}$. A force acts on it for some time in the direction of motion. The force time graph is shown in figure. The final speed of the body
A) $9.25\ ms^{- 1}$
B) $5\ ms^{- 1}$
C) $14.25\ ms^{- 1}$
D) $4.25\ ms^{- 1}$
Explanation
(c) Impulse = Change in momentum = $m(v_{2} - v_{1})$ …(i)
Again impulse = Area between the graph and time axis
$= \frac{1}{2} \times 2 \times 4 + 2 \times 4 + \frac{1}{2}(4 + 2.5) \times 0.5 + 2 \times 2.5$
$= 4 + 8 + 1.625 + 5 = 18.625$ …(ii)
From (i) and (ii), $m\left( v_{2} - v_{1} \right) = 18.625$
$\Rightarrow v_{2} = \frac{18.625}{m} + v_{1} = \frac{18.625}{2} + 5 = 14.25\ m/s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.