Question

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.

Short answer

All velocities have the same kinetic energy.

Step-by-step solution

The given data

The given velocities of the particle are

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

Understanding the concept of velocity and kinetic energy

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)

Calculation of the velocities to determine the rank of the velocities

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:

localid="1657177437323" $$\begin{gathered} v=\sqrt{{\left(-3\right)}^2+{\left(-4\right)}^2} \\ =5\mathrm{m}/\mathrm{s} \end{gathered}$$

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.