Question

A particle of mass $1kg$ is projected with velocity $20\sqrt{2}m{s}^{-1}\text{at}{45}^{\circ }$ with ground. Considering the moment when particle is at highest point, match the statements in Column -$1$ with results in Column-$2$
$(\text{Assume}g=10m/s^2)$

$\begin{array}{cc}\text{Column -1}& \text{Column -2}\\ \text{(A)Net torque on the particle}& \text{(P) 200 SI unit}\\ \text{about point of projection}& \\ \text{(B)Angular momentum of the}& \text{(Q) 400 SI unit}\\ \text{particle about point of}& \\ \text{projection}& \\ \text{(C)Angular velocity of the}& \text{(R)1.0 SI unit}\\ \text{particle about point of}& \\ \text{projection}& \\ & \text{(S)None}\end{array}$

• A. $A-P,B-Q,C-R$
• B. $A-Q,B-Q,C-S$
• C. $A-Q,B-P,C-S$
• D. $A-P,B-Q,C-S$

Answer 1

The correct option is B $A-Q,B-Q,C-S$

Given $u=20\sqrt{2}m/s$ and $\theta ={45}^{\circ },$



Range $R=\frac{u^2\sin 2\theta }{g}=\frac{\left(20\sqrt{2}{)}^2\sin \right({90}^{\circ })}{10}=\frac{800}{10}=80m$
Maximum height $H=\frac{u^2{\sin }^2\theta }{2g}=\frac{\left(20\sqrt{2}{)}^2{\sin }^2\right({45}^{\circ })}{20}=\frac{\frac{800}{2}}{20}=20m$

Torque at maximum height about point of projection $\tau =F\times r_{\perp }$
Here $r_{\perp }=\frac{R}{2}$, because $\frac{R}{2}$ is component of displacement vector perpendicular to $F=mg$ at maximum height

$\tau =\left(mg\right)\left(\frac{R}{2}\right)=\left(10\right)\left(40\right)=400N-m$

Velocity at maximum height $v=u\cos \theta =20\sqrt{2}\cos \left({45}^{\circ }\right)=20m/s$
Angular momentum of the particle at maximum height about point of projection $L=mv{r}_{\perp }=\left(m\right)\left(ucos\theta \right)\left(H\right)$

Here $r_{\perp }=H$, because $H$ is component of displacement vector perpendicular to velocity at maximum height
$L=\left(1\right)\left(20\right)\left(20\right)=400\frac{kg-m^2}{s}$

Angular velocity at maximum height $\omega =\frac{v_{\perp }}{r}$
Here $v_{\perp }=v\sin \theta$, as $v_{\perp }$ is component of velocity vector perpendicular to displacement$\left(r\right)$
$\omega =\frac{\left(vsin\theta \right)}{r}=\frac{vH}{r^2}=\frac{vH}{H^2+(\frac{R}{2}{)}^2}$
$\omega =\frac{\left(20\right)\left(20\right)}{\left[\right(40{)}^2+\left(20{)}^2\right]}=0.2rad/s$