Question

A proton's speed as it passes point A is 4.00×104 m/s . It follows the trajectory shown in the figure. What is the proton's speed at point B?

Answer 1

We know that:

$0.5m(vB^{2} -vA^{2} )=q*[VB-VA]\\== > vB^{2} =vA^{2} +(\frac{2q}{m})*[VB-VA]\\ =(4.30*10^{4} )^{2} +(\frac{2*1.6*10^{-19} }{2*1.67*10^{-27} }) [-10-30]$

$=18.49*10^{8} -76.64*10^{8}$

$=-58.15*10^{8} \\== > vB=-7.62*10^{4} m/s$

Answer 2

The proton's speed at point B is  -7.62 x ×10⁴ m/s.

What is proton?

The proton is a subatomic particle lies inside the nucleus of an atom of element. The number of proton gives an idea of the atomic number of the element.

The proton's speed as it passes point A is 4.00×10⁴ m/s . It follows the trajectory shown.

1/2m (vb² - va²) = q (Vb - Va)

vb² =  va² +2q/m (Vb - Va)

where v is the velocity and V is the potential at point A and B, q ia the charge on proton and m is the mass of proton.

Substitute the given values, we get

vb² = (4.00×10⁴ ) + (2x1.6 x 10⁻¹⁹/1.67 x 10⁻²⁷) (-10 - 30)

vb = -7.62 x ×10⁴ m/s

Thus, proton's speed at point B is  -7.62 x ×10⁴ m/s.

Learn more about  proton.

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