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professor190 [17]

3 years ago

5

If a proton and an electron are released when they are 6.50×10⁻¹⁰ m apart (typical atomic distances), find the initial accelerat

ion of each of them.
a) Acceleration of electron
b) Acceleration of proton

1 answer:

Cerrena [4.2K]3 years ago

3 0

Answer:

(a) Acceleration of electron= 5.993×10²⁰ m/s²

(b) Acceleration of proton= 3.264×10¹⁷ m/s²

Explanation:

Given Data

distance r= 6.50×10⁻¹⁰ m

Mass of electron Me=9.109×10⁻³¹ kg

Mass of proton Mp=1.673×10⁻²⁷ kg

Charge of electron qe= -e = -1.602×10⁻¹⁹C

Charge of electron qp= e = 1.602×10⁻¹⁹C

To find

(a) Acceleration of electron

(b) Acceleration of proton

Solution

Since the charges are opposite the Coulomb Force is attractive

So

$F=\frac{1}{4(\pi)Eo }\frac{|qp*qe|}{r^{2} }\\ F=(8.988*10^{9}Nm^{2}/C^{2})*\frac{(1.602*10^{-19})^{2} }{(6.50*10^{-10} )^{2} } \\F=5.46*10^{-10}N$

From Newtons Second Law of motion

F=ma

a=F/m

For (a) Acceleration of electron

$a=F/Me\\a=(5.46*10^{-10} )/9.109*10^{-31}\\ a=5.993*10^{20}m/s^{2}$

For(b) Acceleration of proton

$a=F/Mp\\a=(5.46*10^{-10} )/1.673*10^{-27} \\a=3.264*10^{17}m/s^{2}$

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a) The time of flight is 3.78 s

b) The initial speed is 17.0 m/s

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Explanation:

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a)

The motion of the ball is a projectile motion, which consists of two independent motions:

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- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction

We start by considering the vertical motion, to find the time of flight of the ball. We do it by using the following suvat equation: for the y-displacement:

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where we have:

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$u_y=u sin \theta$ is the initial vertical velocity, with u being the initial velocity (unknown) and $\theta=-27.0^{\circ}$ the angle of projection

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Along the x-direction, the equation of motion is instead

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where $ucos \theta$ is the horizontal component of the velocity. Rewriting this equation as

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And solving for t, we find the time of flight:

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b)

We can now find the initial speed, u, by using the equation of motion along the x-direction

$x=u cos \theta t$

where we know that:

x = 59.0 m is the horizontal range

$\theta=-23^{\circ}$ is the angle of projection

$t=3.78 s$ is the time of flight

Solving for u, we find the initial speed:

$u=\frac{x}{cos \theta t}=\frac{59.0}{(cos (-23^{\circ}))(3.78)}=17.0 m/s$

c)

First of all, we notice that the horizontal component of the velocity remains constant during the motion, and it is equal to

$v_x = u cos \theta = (17.0)(cos (-23^{\circ})=15.6 m/s$

The vertical velocity instead changes according to the equation

$v_y = u sin \theta + gt$

Substituting all the values and t = 3.78 s, the time of flight, we find the vertical velocity at the time of impact:

$v_y = (17.0)(sin (-23^{\circ}))+(-9.8)(3.78)=-43.7 m/s$

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