# A particle of mass m is projected with velocity v making an angle of 45 with the horizontal from level ground. when the particle lands on the level ground the magnitude of the change in its momentum will be :

Change in momentum =

$$\begin{array}{l} \Delta \vec{P} \end{array}$$

=

$$\begin{array}{l} \Delta P_{x}\hat{i} + \Delta P_{x}\hat{j} \end{array}$$

Since

$$\begin{array}{l} \Delta P_{x} = 0 \end{array}$$
$$\begin{array}{l}\Delta P_{y}\hat{j} = \vec{P}_{fy} – \vec{P}_{iy} \end{array}$$
$$\begin{array}{l}P_{iy} = mvsin45 and P_{fy} = mvsin45 \end{array}$$

=

$$\begin{array}{l} – 2mvsin45 = -\sqrt{2}mv_{j}\left | \Delta = \vec{P} \right | \end{array}$$

=

$$\begin{array}{l} \sqrt{2}mv \end{array}$$

Explore more such questions and answers at BYJU’S.