Question

A particle having a mass $0.5kg$ is projected under gravity with a speed of $98m/s$ at an angle of ${60}^o$. The magnitude of the change in momentum in $kg-m/s$ of the particle after $10s$ is

• A. $490$
• B. $98$
• C. $49$
• D. $0.5$

Answer 1

Answer: (C)

$49$

As the particle has no acceleration in horizontal direction, thus its velocity component in horizontal direction remains the same all the time whereas its vertical component changes.

Using $v_y=u_y+gt$

Magnitude of change in vertical component of velocity $\Delta V_y=gt=9.8\times 10=98m/s$

$\therefore$ Change in momentum $\Delta p=m\Delta V_y$

$⟹$ $\Delta p=\left(0.5\right)\left(98\right)=49$ $kg.m/s$