A) Impulse
B) Change in kinetic energy per unit mass
C) Change in momentum per unit mass
D) Total change in energy
Correct Answer: C
Solution :
Area under acceleration displacement curve \[=\int{a.dx}\] \[=\int{\frac{dv}{dt}.dx}\] \[=\int\limits_{u}^{v}{v\,\,dv}\left[ v=\frac{dx}{dv} \right]\] \[={{\left[ \frac{{{v}^{2}}}{2} \right]}^{v}}_{u}\] \[=\frac{1}{2}\left[ {{\operatorname{v}}^{2}}-{{\operatorname{u}}^{2}} \right]\] = change in K.E. energy per unit mass
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