Question

A particle of mass m is projected with velocity v making an angle of $45^\circ$ with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be

• A. mv $\sqrt 2$
• B. zero
• C. 2 mv
• D. mv / $\sqrt 2$

Answer 1

Answer: (A)

mv $\sqrt 2$

Answer 2

The horizontal momentum does not change. The change in vertical momentum is
$mv \sin \theta - ( - mv \sin \, \theta) = 2 mv \frac{1}{ \sqrt 2 } = \sqrt 2$ mv.