A particle of mass m is projected with velocity v making an angle of 45^o&nbs

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3-2.Motion in Plane

medium

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

A$\sqrt 2 mv$

B$0$

C$2mv$

D$\frac{{mv}}{{\sqrt 2 }}$

(AIPMT-2008)

Solution

$\begin{array}{l}
The\,horizontal\,momentum\,does\,not\\
change.\,The\,change\,in\,vertical\,\\
momentum\,is\\
\,\,\,\,mv\sin \,\theta \, – \left( { – mv\sin \theta } \right) = 2mv\frac{1}{{\sqrt 2 }}\\
\,\,\,\,\,\, = \sqrt {2\,} mv
\end{array}$

Standard 11

Physics

The angle of projection for a projectile to have same horizontal range and maximum height is

hard

(AIPMT-2012)

For a given angle of the projectile if the initial velocity is doubled the range of the projectile becomes

medium

(AIIMS-2011)

The velocity at the maximum height of a projectile is $\frac{\sqrt{3}}{2}$ times its initial velocity of projection $(u)$. Its range on the horizontal plane is ………….

medium

A particle is projected with an angle of projection $\theta$ to the horizontal line passing through the points $( P , Q )$ and $( Q , P )$ referred to horizontal and vertical axes (can be treated as $x$-axis and $y$-axis respectively).

The angle of projection can be given by

hard

(AIIMS-2015)

At the top of the trajectory of a projectile, the directions of its velocity and acceleration are

easy

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

Solution

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A particle is projected with an angle of projection $\theta$ to the horizontal line passing through the points $( P , Q )$ and $( Q , P )$ referred to horizontal and vertical axes (can be treated as $x$-axis and $y$-axis respectively).

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