Q 11-03-129JEE MainJEE Main 2021 (18 Mar, Shift 2)Easy
The projectile motion of a particle of mass 5 g is shown in the figure.
The initial velocity of the particle is $5\sqrt2\ \text{m s}^{-1}$ and the air resistance is assumed to be negligible. The magnitude of the change in momentum between the points $A$ and $B$ is $x\times10^{-2}\ \text{kg m s}^{-1}$. The value of $x$, to the nearest integer, is ______.
Numerical value type. Enter your answer.
Answer: 5
The horizontal component of velocity is unchanged; the vertical component reverses from $u\sin45^\circ$ up to $u\sin45^\circ$ down.
$$|\Delta p| = 2mu\sin45^\circ = 2\times0.005\times5\sqrt2\times\frac{1}{\sqrt2} = 0.05 = 5\times10^{-2}\ \text{kg m s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics