91Ó°ÊÓ

Before starting, let's list the given values: 1. Mass of the particle (m) = 0.5 kg 2. Position of the particle in the x-direction (x) = 5.0 m 3. Velocity of the particle in the y-direction (v_y) = 2.0 m/s

First, we will calculate the position vector \(\boldsymbol{r}\) and linear momentum \(\boldsymbol{p}\). The position vector is given as \(\boldsymbol{r} = x\boldsymbol{\hat{i}} + y\boldsymbol{\hat{j}}\). Since the particle's position only has an x-component, the position vector is \(\boldsymbol{r} = (5.0 \,\mathrm{m}) \boldsymbol{\hat{i}}\). Next, we calculate the linear momentum as \(\boldsymbol{p} = m \boldsymbol{v}\). Since the particle is only moving in the positive y-direction, the momentum vector is \(\boldsymbol{p} = (0.5 \,\mathrm{kg})(2.0 \,\mathrm{m}/\mathrm{s}) \boldsymbol{\hat{j}} = 1.0 \,\mathrm{kg} \cdot \mathrm{m}/\mathrm{s} \boldsymbol{\hat{j}}\).

Now, we will compute the angular momentum by finding the cross product of the position and momentum vectors: \[\boldsymbol{L} = \boldsymbol{r} \times \boldsymbol{p}\] The cross product of two vectors \(\boldsymbol{A} = A_x\boldsymbol{\hat{i}} + A_y\boldsymbol{\hat{j}} + A_z\boldsymbol{\hat{k}}\) and \(\boldsymbol{B} = B_x\boldsymbol{\hat{i}} + B_y\boldsymbol{\hat{j}} + B_z\boldsymbol{\hat{k}}\) is \[\boldsymbol{A} \times \boldsymbol{B} = \begin{vmatrix} \boldsymbol{\hat{i}} & \boldsymbol{\hat{j}} & \boldsymbol{\hat{k}}\\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix}\] Applying the formula for cross product to the position and momentum vectors, we get \[\boldsymbol{L} = \begin{vmatrix} \boldsymbol{\hat{i}} & \boldsymbol{\hat{j}} & \boldsymbol{\hat{k}}\\ 5.0 & 0 & 0 \\ 0 & 1.0 & 0 \end{vmatrix} = (0) \boldsymbol{\hat{i}} - (0) \boldsymbol{\hat{j}} + (5.0)(1.0) \boldsymbol{\hat{k}} = 5.0 \,\mathrm{kg} \cdot \mathrm{m}^2/\mathrm{s}\, \boldsymbol{\hat{k}}\]

The particle's angular momentum about the origin is \(\boldsymbol{L} = 5.0 \,\mathrm{kg} \cdot \mathrm{m}^2/\mathrm{s}\, \boldsymbol{\hat{k}}\).