Question

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 is

• A. $9.25\,ms^{-1}$
• B. $5\,ms^{-1}$
• C. $14.25 \,ms^{-1}$
• D. $4.25\,ms^{-1}$

Answer 1

Answer: (C)

Impulse = Change in momentum $=m(v_{2}-v_{1}) \dots(i)$
Again impulse = Area between the graph and time axis
$=\frac{1}{2}\times 2 \times 4+2\times 4 +\frac{1}{2}(4+2.5)\times 0.5+2 \times 2.5$
$=4+8+1.625+5=18.625 \dots(ii)$
From (i) and (ii), $m(v_{2}-v_{1})=18.625$
$\Rightarrow v_{2}=\frac{18.625}{m}+v_{1}=\frac{18.625}{2}+5$
$=14.25 \,m/s$