question_answer

A body having mass M is projected with the velocity u in a direction making an angle \[45{}^\circ \] with the horizontal. When body strikes the ground, the magnitude of change of momentum will be

A)  \[\sqrt{3}\,Mu\]                             

B)  \[\frac{Mu}{\sqrt{2}}\]

C)  \[\frac{Mu}{\sqrt{3}}\]                                

D)  \[\sqrt{2}\,Mu\]

E)  \[Mu\]

Correct Answer: D

Solution :

                See the diagram The change in momentum \[\Delta P={{P}_{f}}-{{P}_{i}}=\]momentum at B - momentum at A \[=m\,({{u}_{j}}-{{u}_{i}})\] \[=M\,[(u\,\,\cos 45{}^\circ \hat{i}-u\sin 45{}^\circ \hat{j})\]\[-\,(u\,\cos 45{}^\circ \hat{i}+u\sin 45{}^\circ \hat{j})]\] \[=M\left[ \frac{u}{\sqrt{2}}\hat{i}-\frac{u}{\sqrt{2}}\hat{j} \right]-\left[ \frac{u}{2}\hat{i}+\frac{u}{\sqrt{2}}\hat{j} \right]\] \[=-\sqrt{2}Mu\,\hat{j}\] \[\therefore \]  \[|\Delta P|=\sqrt{2}Mu\]