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
- mv $ \sqrt 2 $
- zero
- 2 mv
- mv / $ \sqrt 2 $
The Correct Option is A
Solution and Explanation
$mv \sin \theta - ( - mv \sin \, \theta) = 2 mv \frac{1}{ \sqrt 2 } = \sqrt 2 $ mv.
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Concepts Used:
Motion in a Plane
It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions.
Equations of Plane Motion
The equations of motion in a straight line are:
v=u+at
s=ut+½ at2
v2-u2=2as
Where,
- v = final velocity of the particle
- u = initial velocity of the particle
- s = displacement of the particle
- a = acceleration of the particle
- t = the time interval in which the particle is in consideration