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

`2.4* 10^5 m/s`

Step-by-step explanation:

We are given that

Mass of proton=
`m_p=1.67* 10^(-27)kg`

Initial speed of proton=
`v_i=3.00* 10^5 m/s`

`x_i=-5m`

`\alpha=2.12* 10^(-26)Nm^2`

`F=-(\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* 10^(-10) m` from the uranium nucleus.

It means we have to find
`v_f \;if x_f=-8.00* 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) -(\alpha)/(x^2)dx`

`W=-\int_(-5)^{-8* 10^(-10)} \alpha x^(-2)dx`

`W=\alpha[x^(-1)]^{-8* 10^(-10)}_(-5)`

`W=2.12* 10^(-26)*(-(1)/(8* 10^(-10))+(1)/(5))`

`W=-2.65* 10^(-17)J`

`(1)/(2)m(v^2_f-v^2_i)=-2.65* 10^(-17)`

`(1)/(2)\cdot 1.67* 10^(-27)}(v^2_f-(3* 10^5)^2)=-2.65* 10^(-17)`

`v^2_f-9* 10^(10)=(2\cdot (-2.65)* 10^(-17))/(1.67* 10^(-27))`

`V^2_f-9* 10^(10)=-3.17* 10^(10)`

`v^2_f=-3.17* 10^(10)+9* 10^(10)`

`v^2_f=5.83* 10^(10)`

`v_f=\sqrt{5.83* 10^(10)}=2.4* 10^5 m/s`

Hence, the speed of proton when it is
`8* 10^(-10) m` from the uranium nucleus=
`v_f=2.4* 10^5 m/s`