91Ó°ÊÓ

The rock is a distance$4.5×{10}^{13}m$ from the center of the star.

the magnitude of its momentum $1.35×{10}^{17}kg·\mathrm{m}/\mathrm{s}$. is and the angle is $126°$

At a later time, when the rock is a distance $d_2=1.3×{10}^{13}m$from the center of the star.

The rotating analogue of linear momentum is angular momentum (also known as moment of momentum or rotational momentum). Because it is a conserved quantity—the total angular momentum of a closed system remains constant—it is a significant quantity in physics.

Identify the unknown:

The magnitude of the final momentum

List the Knowns:

initial distance:$d_1=4.5×{10}^{13}\text{m}$

Magnitude of initial momentum:$p_1=1.35×{10}^{17}\text{kg}·\mathrm{m}/\mathrm{s}$

Angel between initial distance vector and initial momentum vector:$\mathrm{Î\pm }=126°$

Final distance:$d_2=1.3×{10}^{13}\text{m}$

Angel between final distance vector and final momentum vector:$90°$

Set Up the Problem:

Angular momentum of a particle about location A:

${\stackrel{→}{L}}_A={\stackrel{→}{r}}_A×\stackrel{→}{p}$

So, the magnitude of angular momentum about location A:${\stackrel{→}{L}}_A=r_{A}p\sin \mathrm{θ}$

The position vector and the force vector are collinear, then the net torque on the rock is zero and the total angular momentum remains constant. It means that, angular momentum at given first point is same as the angular momentum at second point:

${\stackrel{→}{L}}_{A,1}={\stackrel{→}{L}}_{A,2}$$r_{A,1}{p}_1\sin {\mathrm{θ}}_1=r_{A,2}{p}_2\sin {\mathrm{θ}}_2$

$d_1{p}_1\sin \mathrm{Î\pm }=d_2{p}_2\sin \left(90\right)$

$p_2=\frac{d_1{p}_1\sin \mathrm{Î\pm }}{d_2}$

Solve the Problem:

$p_2=\frac{4.5×{10}^{13}×1.35×{10}^{17}×\sin \left(126\right)}{1.3×{10}^{13}}$