question_answer

A projectile of mass m is thrown with velocity \[\upsilon \] from the ground at an angle of \[\mathbf{45}{}^\circ \] with the horizontal. What is the magnitude of change in momentum between leaving and arriving back at the ground?

Answer:

                The vertical component of velocity at the point of projection is \[\upsilon \] sin \[\theta \] and at the point of return is \[-\upsilon \] sin \[\theta \] . \[\therefore \] Magnitude of change in momentum \[=m(\upsilon \sin \theta +\upsilon \sin \theta )=2m\upsilon \sin \theta \] \[=2m\upsilon \sin {{45}^{\circ }}=\sqrt{2}m\upsilon \] (in the vertical direction).