Question

A proton with mass 1.67×10⁻²⁷kg is propelled at an initial speed of 3.00×10⁵m/s directly toward a uranium nucleus 5.00 m away. The proton is repelled by the uranium nucleus with a force of magnitude F=α/x², where is the separation between the two objects and α= 2.12×10⁻²⁶ N•m². Assume that the uranium nucleus remains at rest. What is the speed of the proton when it is 8.00×10⁻¹⁰m from the the uranium nucleus?

Answer 1

Answer:

$2.4\times 10^5 m/s$

Explanation:

We are given that

Mass of proton=$m_p=1.67\times 10^{-27}kg$

Initial speed of proton=$v_i=3.00\times 10^5 m/s$

$x_i=-5m$

$\alpha=2.12\times 10^{-26}Nm^2$

$F=-\frac{\alpha}{x^2}$

Where x is the separation between the two objects.

We have to find the speed of the proton when it is $8.00\times 10^{-10} m$ from the uranium nucleus.

It means we have to find $v_f \;if x_f=-8.00\times 10^{-10} m$

By work energy theorem

W=$\Delta K.E=\int_{x_i}^{x_f} Fdx$

Substitute the values

$W=\Delta K.E=\int_{x_i}^{x_f} -\frac{\alpha}{x^2}dx$

$W=-\int_{-5}^{-8\times 10^{-10}} \alpha x^{-2}dx$

$W=\alpha[x^{-1}]^{-8\times 10^{-10}}_{-5}$

$W=2.12\times 10^{-26}\times(-\frac{1}{8\times 10^{-10}}+\frac{1}{5})$

$W=-2.65\times 10^{-17}J$

$\frac{1}{2}m(v^2_f-v^2_i)=-2.65\times 10^{-17}$

$\frac{1}{2}\cdot 1.67\times 10^{-27}}(v^2_f-(3\times 10^5)^2)=-2.65\times 10^{-17}$

$v^2_f-9\times 10^{10}=\frac{2\cdot (-2.65)\times 10^{-17}}{1.67\times 10^{-27}}$

$V^2_f-9\times 10^{10}=-3.17\times 10^{10}$

$v^2_f=-3.17\times 10^{10}+9\times 10^{10}$

$v^2_f=5.83\times 10^{10}$

$v_f=\sqrt{5.83\times 10^{10}}=2.4\times 10^5 m/s$

Hence, the speed of proton when it is $8\times 10^{-10} m$ from the uranium nucleus=$v_f=2.4\times 10^5 m/s$