Rank the following velocities according to the kinetic energy a particle will have with each velocity, greatest first: (a) $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}\mathbf{=}\mathbf{4}\hat{\mathbf{i}}\mathbf{+}\mathbf{3}\hat{\mathbf{j}}\mathbf{}\mathbf{,}\left(b\right)\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}\mathbf{=}\mathbf{4}\hat{\mathbf{i}}\mathbf{+}\mathbf{3}\hat{\mathbf{j}}\mathbf{}\mathbf{,}\left(c\right)\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}\mathbf{=}\mathbf{3}\hat{\mathbf{i}}\mathbf{+}\mathbf{4}\hat{\mathbf{j}}\mathbf{}\mathbf{,}\left(d\right)\mathbf{}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}\mathbf{=}\mathbf{3}\hat{\mathbf{i}}\mathbf{+}\mathbf{4}\hat{\mathbf{j}}\mathbf{}\mathbf{,}\left(e\right)\mathbf{}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}\mathbf{=}\mathbf{5}\hat{\mathbf{i}}\mathbf{}\mathbf{,}\left(f\right)\mathbf{}\mathbf{v}\mathbf{=}\mathbf{5}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{}\mathit{a}\mathit{a}\mathit{t}\mathbf{}\mathbf{30}\mathbf{°}$to the horizontal.

The given velocities of the particle are

$$\begin{gathered} a)\stackrel{\rightharpoonup }{\mathrm{v}}=4\hat{\mathrm{i}}+3\hat{\mathrm{j}} \\ \mathrm{b})\stackrel{\rightharpoonup }{\mathrm{v}}=-4\hat{\mathrm{i}}+3\hat{\mathrm{j}} \\ \mathrm{c})\stackrel{\rightharpoonup }{\mathrm{v}}=-3\hat{\mathrm{i}}+4\hat{\mathrm{j}} \\ \mathrm{d})\stackrel{\rightharpoonup }{\mathrm{v}}=-3\hat{\mathrm{i}}-4\hat{\mathrm{j}} \\ \mathrm{e})\stackrel{\rightharpoonup }{\mathrm{v}}=5\hat{\mathrm{i}} \\ \mathrm{f})\stackrel{\rightharpoonup }{\mathrm{v}}=5\mathrm{at}30°,\stackrel{\rightharpoonup }{\mathrm{v}}=5\cos 30\hat{\mathrm{i}}+5\sin 30\hat{\mathrm{j}} \end{gathered}$$

To rank kinetic energy, we have to find first find the resultant velocity, and then use the formula for kinetic energy in which the mass is the same for all velocities. Therefore, kinetic energy is ranked only on the magnitude of velocity.

Formulae:

The resultant velocity according to vector concept,

$v=\sqrt{v_x^2+v_y^2}$ (1)

The kinetic energy of a body,

$\mathrm{KE}=0.5\times \mathrm{m}\times {\mathrm{v}}^2$ (2)

The magnitude of the resultant velocity can be given using equation (i) as:

$$\begin{gathered} v=\sqrt{4^2+3^2} \\ =5\mathrm{m}/\mathrm{s} \end{gathered}$$

For (b)localid="1657176480423" $\stackrel{\rightharpoonup }{\mathrm{v}}=-4\hat{\mathrm{i}}+3\hat{\mathrm{j}}$

The magnitude of the resultant velocity can be given using equation (i) as:

$$\begin{gathered} \mathrm{v}=\sqrt{{\left(-4\right)}^2+3^2} \\ =5\mathrm{m}/\mathrm{s} \end{gathered}$$

For (c)localid="1657177424944" $-3\hat{\mathrm{i}}+4\hat{\mathrm{j}}$

The magnitude of the resultant velocity can be given using equation (i) as:

$$\begin{gathered} v=\sqrt{{\left(-3\right)}^2+4^2} \\ =5\mathrm{m}/\mathrm{s} \end{gathered}$$

For (d) $\stackrel{\rightharpoonup }{v}=-3\hat{\mathrm{i}}-4\hat{\mathrm{j}}$

The magnitude of the resultant velocity can be given using equation (i) as:

For (e)localid="1657177453609" $=5\mathrm{m}/\mathrm{s}$

The magnitude of the resultant velocity can be given using equation (i) as:

$$\begin{gathered} v=\sqrt{5^2} \\ =5\mathrm{m}/\mathrm{s} \end{gathered}$$

For (f)$\left(\mathrm{f}\right)\stackrel{\rightharpoonup }{v}=\left(5\cos 30°\right)\hat{\mathrm{i}}+\left(5\cos 30°\right)\hat{\mathrm{j}}$

The magnitude of the resultant velocity can be given using equation (i) as:

$$\begin{gathered} v=\sqrt{5^2\left({\cos }^{2}30°+{\sin }^{2}30°\right)} \\ =5\mathrm{m}/\mathrm{s} \end{gathered}$$

Assume m=1 kg

So, using equation (2), we can get that the kinetic energy of a body is directly proportional to the square of the velocity.

As the resultant velocity v=5 m/s is the same for all cases, kinetic energy is also the same for all.