Question

A particle of mass $1kg$ is projected at an angle of ${45}^o$ to the horizontal with an initial velocity of $20m/s$. Change in momentum during it's time of flight is

• A. $-10\sqrt{2}kgm/s$
• B. $-20\sqrt{2}kgm/s$
• C. $-30\sqrt{2}kgm/s$
• D. $-40\sqrt{2}kgm/s$

Answer 1

The correct option is B $-20\sqrt{2}kgm/s$

The horizontal component of velocity does not change.
The vertical component of velocity initially is: $u\sin \theta$
The vertical component of velocity finally is: $-u\sin \theta$
Change in velocity: $\Delta v=-u\sin \theta -u\sin \theta =-2u\sin \theta$
Change in momentum is $m\Delta v=-2mu\sin \theta =2\times 1\times 20\times \sin {45}^0=20\sqrt{2}kgm/s$