Question

Change in momentum of a projectile in half the time of flight if momentum of projection is mu at angle $\theta$?

• A. $mu\sin \theta$
• B. $mu\cos \theta$
• C. $2mu\sin \theta$
• D. $2mu\cos \theta$

Answer 1

The correct option is A $mu\sin \theta$

The object is projected at an angle $\theta$

So, initial velocity be, $u_i$= $ucos\theta \hat{i}+usin\theta \hat{j}$

The vertical component that lifting the body upward is $usin\theta \hat{j}$

In projectile motion, horizontal component of velocity remains unchanged.

Therefore in total time of flight, ie; at the end of the motion,

velocity is, $u_f=cos\theta \widehat{i-}sin\theta \hat{j}$

Thus change in velocity in the motion, $\Delta v=u_f-u_i$

$\Delta u=-2usin\theta$

So, change in momentum, $\Delta P=-2musin\theta$

This is the change in momentum in total time of flight.

In half of time of flight, change in momentum is= -$musin\theta$

So the magnitude of change in momentum is in half of time of flight is $musin\theta$