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shtirl [24]
3 years ago
6

A projectile is fired in such a way that its horizontal range is equal to 14.5 times its maximum height. what is the angle of pr

ojection?
Physics
1 answer:
3241004551 [841]3 years ago
5 0

The maximum height is: hmax  = v0² sin² α / 2 g<span>
The horizontal range is: x = v0² sin 2 α / g</span>

So we are given that:

x = 14.5 hmax

v0² sin 2 α / g = 14.5 v0² sin² α / 2 g<span>
<span>7.25 sin² α = 2 sin α cos α    
<span>7.25 sin α = 2 cos α  
7.25 tan α = 2
<span>tan α = 0.27586
<span>α = tan^(-1) 0.27586
α = 15.42°<span>
<span>The angle of projection is 15.42</span></span><span>°</span></span></span></span></span></span>

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On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

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B)

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C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

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T=\frac{2\pi r}{v} (2)

From eq.(1), we can rewrite the velocity of the particle as

v=\frac{qBr}{m}

Substituting into(2), we can rewrite the period of revolution of the particle as:

T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

\omega=\frac{2\pi}{T} (3)

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The period has been found in part C:

T=\frac{2\pi m}{qB}

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}

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E)

For this part, we use again the relationship found in part B:

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which can be rewritten as

r=\frac{mv}{qB} (4)

The kinetic energy of the particle is written as

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So, from this we can find another expression for the velocity:

v=\sqrt{\frac{2K}{m}}

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r=\frac{\sqrt{2mK}}{qB}

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

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