Question

A particle of mass $2\text{kg}$ is projected on horizontal ground with an initial velocity $u=20\text{m/s}$ making an angle ${60}^{\circ }$ with the vertical. Find out the angular momentum of the particle about the point of projection when it just strikes the ground. [in ${\text{kg m}}^2\text{/s}$]

• A. $200\sqrt{3}$
• B. $100\sqrt{3}$
• C. $\text{zero}$
• D. $400\sqrt{3}$

Answer 1

The correct option is D $400\sqrt{3}$

Given,
Mass of particle, $m=2\text{kg}$
Initial velocity of particle, $u=20\text{m/s}$


Assume particle hits the ground after time $t$ at point P.
Vertical component of initial velocity $=u_y$
$u_y=u\sin \alpha =u\sin {30}^{\circ }=20\times \frac{1}{2}=10\text{m/s}$
As we know, vertical component of particle's speed, at the time of projection and when it hits the ground will be same. Hence, $v_y=10\text{m/s}$
As we can see, at point P, horizontal component of velocity passes through the origin i.e $r_{\perp }=0,$
Hence, there is no angular momentum due to horizontal component of particle's velocity)

Angular momentum because of vertical component of velocity :
$L=m{v}_y{r}_{\perp }$
Here $r_{\perp }$ = Range of projectile = $R$
$R=\frac{u^2\sin 2\alpha }{g}=\frac{{20}^2\times \sin \left({60}^{\circ }\right)}{10}=20\sqrt{3}\text{m}$
Hence $L=(m{v}_y\times R)=2\times 10\times 20\sqrt{3}$
$L=400\sqrt{3}$ ${\text{kg m}}^2\text{/s}$