Question

A particle having a mass of $0.5kg$ is projected with a speed of $98m{s}^{-1}$ at an angle of ${60}^o$. The magnitude of change in momentum of the particle after $10s$ in $Ns$ is

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

Answer 1

The correct option is B $49$

Change in momentum $\Delta$P = Mass $\times$ ($v_f-v_i$)
where $v_f$ = velocity of the particle at t = 10 seconds
and $v_i$ = velocity of the particle at t = 0 s

Horizontal velocity will be constant throughout the flight so we don't need to consider that. Hence we will consider only vertical velocities in $v_i$ and $v_f$.

Now, on vertical axis, we know that a = g = 9.8 m/$s^2$

By the first equation of motion,
$v_f-v_i$ = at

Hence $\Delta$P = M $\times$ at

$⟹$ $\Delta$P = 0.5 kg $\times$ 9.8 m/$s^2$ $\times$ 10 s

$⟹$ $\Delta$P = 49 Ns