Q72P - University Physics with Modern Physics

Question

A proton with mass  $\mathbf{1}\mathbf{.}\mathbf{67}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{27}}\mathbf{ }\mathbf{kg}$ is propelled at an initial speed of $\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{ }\mathbf{m}\mathbf{/}\mathbf{s}$ directly toward a uranium nucleus 5.00 away. The proton is repelled by the uranium nucleus with a force of magnitude $\mathbf{F}\mathbf{=}\mathbf{\alpha }\mathbf{/}{\mathbf{x}}^{\mathbf{2}}$ , where x is the separation between the two objects and $\mathbf{\alpha }\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{12}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{26}}\mathbf{ }\mathbf{N}\mathbf{·}{\mathbf{m}}^{\mathbf{2}}$ . Assume that the uranium nucleus remains at rest. (a) What is the speed of the proton when it is $\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{ }\mathbf{m}$ from the uranium nucleus? (b) As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, after which the proton moves away from the uranium nucleus. How close to the uranium nucleus does the proton get? (c) What is the speed of the proton when it is again  5.00 m away from the uranium nucleus?

Step-by-Step Solution

Answer

1.  The speed of the proton is $2.414\times {10}^5 \mathrm{m}/\mathrm{s}$  when it is $8.00\times {10}^{-10} \mathrm{m}$ from the uranium nucleus.
2.  The closest distance to the uranium nucleus is $2.82\times {10}^{-10} \mathrm{m}$ .
3.  The speed of the proton is $3.00\times {10}^2 \mathrm{m}/\mathrm{s}$  when it is again 5.0 m.

Step 1: Force between Proton and Uranium nucleus

The force between the proton and uranium nucleus is due to the Coulombic repulsion between these two positive charges.

Step 2: Given data

Mass of the proton is, $1.67\times {10}^{-27} \mathrm{kg}$

The initial speed of the proton is, ${\mathrm{v}}_0=3.00\times {10}^5 \mathrm{m}/\mathrm{s}$ 

Uranium nucleus is at 5.0 m  from the proton 

Repelling force of proton is,  $\mathrm{F}=\frac{\mathrm{\alpha }}{{\mathrm{x}}^2}$ 

Here, x is the separation between the two objects and  $\alpha =2.12\times {10}^{26} \mathrm{N}·{\mathrm{m}}^2$

Step 3: (a) Find the speed of the proton when it is 8 . 00 × 10 - 10   m

Work done by the force when the proton is $8.00\times {10}^{-10}$ from the Uranium nucleus,

 

$$\begin{gathered} \mathrm{W}={\int }_{{\mathrm{x}}_0}^{\mathrm{x}}\left(\frac{\mathrm{\alpha }}{{\mathrm{x}}^2}\right)\mathrm{dx} \\ ={\left[\frac{-\mathrm{\alpha }}{\mathrm{x}}\right]}_{{\mathrm{x}}_0}^{\mathrm{x}} \\ ={\left[\frac{-2.12\times {10}^{-26}}{\mathrm{x}}\right]}_{5.0 \mathrm{m}}^{8.0\times {10}^{-10} \mathrm{m}} \\ =-2.56\times {10}^{-17} \mathrm{J} \end{gathered}$$  

 

Now, using the Work-Energy theorem, we get

 

 $$\begin{gathered} \mathrm{W}={\mathrm{K}}_2-{\mathrm{K}}_1 \\ -2.65\times {10}^{-17} \mathrm{J}=\frac{1}{2}{\mathrm{mv}}^2-\frac{1}{2}{\mathrm{mv}}_0^2 \\ -2.65\times {10}^{-17} \mathrm{J}=\frac{1}{2}\left(1.67\times {10}^{-27} \mathrm{kg}\right)\left[{\mathrm{v}}^2-\left(3.00\times {10}^5 \mathrm{m}·{\mathrm{s}}^{-2}\right)\right] \\ \mathrm{v}=2.414\times {10}^5 \mathrm{m}/\mathrm{s} \end{gathered}$$

 

Hence, the speed of the proton is $2.414\times {10}^5 \mathrm{m}/\mathrm{s}$ . 

Step 4: (b) Find how close to the uranium nucleus does the proton get

Applying the Work-Energy theorem again (but with $\mathrm{v}=0 \mathrm{m}/\mathrm{s}$  this time because the proton is momentarily at its closest distance to the uranium nucleus), we can find how close the uranium nucleus gets.

 

 $$\begin{gathered} \mathrm{W}={\mathrm{K}}_2-{\mathrm{K}}_1 \\ \int \mathrm{Fdx}={\mathrm{K}}_2-{\mathrm{K}}_1 \\ \int \left(\frac{\mathrm{\alpha }}{{\mathrm{x}}^2}\right)\mathrm{dx}=\frac{1}{2}{\mathrm{mv}}^2-\frac{1}{2}{\mathrm{mv}}_0^2 \\ -\frac{\mathrm{\alpha }}{\mathrm{x}}=-\frac{1}{2}{\mathrm{mv}}_0^2 \\ {\mathrm{x}}_{\min }=\frac{2\left(2.12\times {10}^{-26} \mathrm{N}·{\mathrm{m}}^2\right)}{\left(1.67\times {10}^{-27} \mathrm{kg}\right)\left(3.00\times {10}^5 \mathrm{m}/\mathrm{s}\right)} \\ {\mathrm{x}}_{\min }=2.82\times {10}^{-10} \mathrm{m} \end{gathered}$$  

 

Hence, the closest distance to the uranium nucleus is $2.82\times {10}^{-10} \mathrm{m}$ . 

Step 5: (c) Find the speed of the proton when it is again 5.00 m

Using the Work-Energy theorem yet again with the initial velocity equal to zero (when the proton is momentarily at rest), we can find the final velocity of the proton when it is again 5.0 m  away from the nucleus.

 

 $$\begin{gathered} \mathrm{W}={\mathrm{K}}_2-{\mathrm{K}}_1 \\ \int \mathrm{Fdx}={\mathrm{K}}_2-{\mathrm{K}}_1 \\ {\int }_{2.82\times {10}^{-10} \mathrm{m}}^{5.0 \mathrm{m}}\left(\frac{\mathrm{\alpha }}{{\mathrm{x}}^2}\right)\mathrm{dx}=\frac{1}{2}{\mathrm{mv}}^2-\frac{1}{2}{\mathrm{mv}}_0^2 \\ {\left[-\frac{\mathrm{\alpha }}{\mathrm{x}}\right]}_{2.82\times {10}^{-10} \mathrm{m}}^{5.0 \mathrm{m}}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{f}}^2 \\ \left[\frac{\left(-2.12\times {10}^{-26} \mathrm{N}·{\mathrm{m}}^2\right)}{5.0 \mathrm{m}}-\frac{\left(-2.12\times {10}^{-26} \mathrm{N}·{\mathrm{m}}^2\right)}{2.82\times {10}^{-10} \mathrm{m}}\right]=\left(0.5\right)\times \left(1.67\times {10}^{-27} \mathrm{kg}\right){\mathrm{v}}_{\mathrm{f}}^2 \\ {\mathrm{v}}_{\mathrm{f}}=3.00\times {10}^5 \mathrm{m}/\mathrm{s} \end{gathered}$$ 

 

Hence, the speed of the proton is $3.00\times {10}^5 \mathrm{m}/\mathrm{s}$ when it is again 5.0 m .