question_answer

A projectile of mass m is thrown with a velocity u making an angle \[60{}^\circ \] with the horizontal, neglecting air resistance, the change in momentum of the departure from A to its arrival at B, to along the vertical directions

A)  \[2\,mu\]                          

B)  \[\sqrt{3}\,mu\]

C)  \[\,mu\]                                             

D)  \[\frac{\,mu}{\sqrt{3}}\]

Correct Answer: B

Solution :

                 Let body be projected with an initial velocity u, at an angle \[{{60}^{o}}\] with the horizontal, then change in momentum \[(\Delta p)\] is given by                                 \[\Delta p=mu\,\sin {{60}^{o}}-(-mu\,\sin {{60}^{o}})\]                 \[\Delta p=2\,mu\,\sin {{60}^{o}}\] \[\Rightarrow \]               \[\Delta p=2\,\,mu\,\frac{\sqrt{3}}{2}\] \[\Rightarrow \]               \[\Delta p=\,\sqrt{3}\,mu\]