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An alpha particle (α), which is the same as a helium-4 nucleus, is momentarily at rest in a region of space occupied by an electric field. The particle then begins to move. Find the speed of the alpha particle after it has moved through a potential difference of â3.45Ã10^â3 V .The charge and the mass of an alpha particle are qα = 3.20Ã10^â19 C and mα = 6.68Ã10â27 kg , respectively.

Mechanical energy is conserved in the presence of which of the following types of forces?

a. electrostatic
b. frictional
c. magnetic
d. gravitational


Sagot :

Answer:

Speed = 575 m/s

Mechanical energy is conserved in electrostatic, magnetic and gravitational forces.

Explanation:

Given :

Potential difference, U = [tex]$-3.45 \times 10^{-3} \ V$[/tex]

Mass of the alpha particle, [tex]$m_{\alpha} = 6.68 \times 10^{-27} \ kg$[/tex]

Charge of the alpha particle is, [tex]$q_{\alpha} = 3.20 \times 10^{-19} \ C$[/tex]

So the potential difference for the alpha particle when it is accelerated through the potential difference is

[tex]$U=\Delta Vq_{\alpha}$[/tex]

And the kinetic energy gained by the alpha particle is

[tex]$K.E. =\frac{1}{2}m_{\alpha}v_{\alpha}^2 $[/tex]

From the law of conservation of energy, we get

[tex]$K.E. = U$[/tex]

[tex]$\frac{1}{2}m_{\alpha}v_{\alpha}^2 = \Delta V q_{\alpha}$[/tex]

[tex]$v_{\alpha} = \sqrt{\frac{2 \Delta V q_{\alpha}}{m_{\alpha}}}$[/tex]

[tex]$v_{\alpha} = \sqrt{\frac{2(3.45 \times 10^{-3 })(3.2 \times 10^{-19})}{6.68 \times 10^{-27}}}$[/tex]

[tex]$v_{\alpha} \approx 575 \ m/s$[/tex]

The mechanical energy is conserved in the presence of the following conservative forces :

-- electrostatic forces

-- magnetic forces

-- gravitational forces

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