Question

In Fig. 4-32, particle${}^{\mathbf{A}}$moves along the linelocalid="1654157716964" ${}^{\mathit{y}\mathbf{=}\mathbf{30}\mathbf{}\mathit{m}}$with a constant velocitylocalid="1654157724548" $\stackrel{\mathbf{\rightharpoonup }}{\mathit{v}}$of magnitude localid="1654157733335" ${}^{\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}}$and parallel to thelocalid="1654157740474" ${}^{\mathbf{x}}$axis. At the instant particlelocalid="1654157754808" ${}^{\mathbf{A}}$passes thelocalid="1654157747127" ${}^{\mathbf{y}}$axis, particlelocalid="1654157760143" $\mathit{B}$leaves the origin with a zero initial speed and a constant accelerationlocalid="1654157772700" ${}^{\mathit{a}}$of magnitudelocalid="1654157766903" $\mathbf{0}\mathbf{.}\mathbf{40}\mathbf{}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$. What anglelocalid="1654157780936" ${}^{\mathbf{\theta }}$between $\overline{\mathbf{a}}$and the positive direction of thelocalid="1654157787877" ${}^{\mathbf{y}}$axis would result in a collision?

Short answer

The angle${}^{\theta }$between${}^{a\text{'}}$and the positive direction of the y axis that results in a collision is${}^{60°}$

Step-by-step solution

Given information

Particle A moves along the${}^{\mathrm{y}=30\mathrm{m}}$with constant velocity${}^{\mathrm{v}=3\mathrm{m}/s}$parallel to x axis.

Particle B moves with${}^{v=0\mathrm{m}/s}$and constant acceleration ${}^{\stackrel{\rightharpoonup }{\mathrm{a}}=0.40m/s^2}$

Determining the concept of kinematic equations

This problem is based on kinematic equations that describe the motion of an object with constant acceleration. Applying second kinematic equation for both the particles A and B along x and y axes, the angle between vector$\overline{\mathbf{a}}$ and${}^{\mathbf{+}\mathbf{y}}$axis can be found.

Formula:

The displacement in kinematic equation is given by,

$x=v_{0}t+\frac{1}{2}a{t}^2$ (i)

Where ${}^{v_0}$is the initial velocity

To derive the equation of motion of particle A and particle B along y axis

Equation of motion of particle A and particle B along y axis can be written using the second kinematic equation as,

$$\begin{gathered} \mathrm{y}={\mathrm{v}}_{\mathrm{y}}\mathrm{t}+\frac{1}{2}{\mathrm{a}}_{\mathrm{y}}{\mathrm{t}}^2\mathrm{s} \\ 30\mathrm{m}=\frac{1}{2}\left[\left(0.40\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)\mathrm{cos\theta }\right]{\mathrm{t}}^2 \end{gathered}$$ (ii)

To derive the equation of motion of particle A and particle B along x axis

Equation of motion of particle A and particle B along x axis can be written using the second kinematic equation as,

$$\begin{gathered} \mathrm{x}={\mathrm{v}}_{\mathrm{x}}\mathrm{t}+\frac{1}{2}{\mathrm{a}}_{\mathrm{x}}{\mathrm{t}}^2 \\ \left(3.0\mathrm{m}/\mathrm{s}\right)\mathrm{t}=\frac{1}{2}\left[\left(0.40\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)\mathrm{sin\theta }\right]{\mathrm{t}}^2 \end{gathered}$$

Solving above equation for t,

$t=\frac{\left(3.0{\displaystyle \frac{m}{s}}\right)2}{\left(0.40{\displaystyle \frac{m}{s^2}}\right)\sin \theta }$ (iii)

To find the angleθ betweenaand the positive direction of the y axis

Substituting equation (iii) in equation (ii),

$$\begin{gathered} 30m-\frac{1}{2}\left[\left(0.40\frac{m}{s^2}\right)\cos \theta \right]\left[\frac{2\left(3.0{\displaystyle \frac{m}{s}}\right)}{\left(0.40{\displaystyle \frac{m}{s^2}}\right)\sin \theta }\right] \\ As{\sin }^2\theta =1-{\cos }^2\theta \\ 1-{\cos }^2\theta -\frac{9.0}{\left(0.20\right)\left(30\right)}\cos \theta \\ {\cos }^2\theta -1.0\cos \theta -1-0 \\ \cos \theta -\frac{1}{2} \\ \theta ={\cos }^{-1}\left(\frac{1}{2}\right) \\ \theta =60° \end{gathered}$$