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Ksenya-84 [330]

3 years ago

15

A stuntwoman is going to attempt a jump across a canyon that is 77 m wide. The ramp on the far side of the canyon is 25 m lower

than the ramp from which she will leave. The takeoff ramp is built with a 15º angle from horizontal. If the stuntwoman leaves the ramp with a velocity of 28 m/s, will she make the jump? Why?

1 answer:

liq [111]3 years ago

5 0

initial speed of the stuntman is given as

$v = 28 m/s$

angle of inclination is given as

$\theta = 15 degree$

now the components of the velocity is given as

$v_x = 28 cos15 = 27.04 m/s$

$v_y = 28 sin15 = 7.25 m/s$

here it is given that the ramp on the far side of the canyon is 25 m lower than the ramp from which she will leave.

So the displacement in vertical direction is given as

$\delta y = -25 m$

$\delta y = v_y * t + \frac{1}{2} at^2$

$-25 = 7.25 * t - \frac{1}{2}*9.8* t^2$

by solving above equation we have

$t = 3.12 s$

Now in the above interval of time the horizontal distance moved by it is given by

$d_x = v_x * t$

$d_x = 27.04 * 3.12 = 84.4 m$

since the canyon width is 77 m which is less than the horizontal distance covered by the stuntman so here we can say that stuntman will cross the canyon.

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A) $v=\sqrt{\frac{2qV}{m}}$

B) $r=\frac{mv}{qB}$

C) $T=\frac{2\pi m}{qB}$

D) $\omega=\frac{qB}{m}$

E) $r=\frac{\sqrt{2mK}}{qB}$

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

$\Delta U = qV$

where

q is the charge of the particle (positive)

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

$\Delta K=\frac{1}{2}mv^2$

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

$qV=\frac{1}{2}mv^2$

And solving for v, we find the speed v at which the particle enters the cyclotron:

$v=\sqrt{\frac{2qV}{m}}$

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

$F=qvB$

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This force acts as centripetal force, so we can write:

$F=m\frac{v^2}{r}$

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

$qvB=m\frac{v^2}{r}$

And solving for r, we find the radius of the orbit:

$r=\frac{mv}{qB}$ (1)

C)

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

It can be calculated as the ratio between the length of the circumference ( $2\pi r$ ) and the velocity of the particle (v):

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

where T is the period.

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:

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

And we see that also the angular frequency does not depend on the velocity.

E)

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

$v=\frac{qBr}{m}$

which can be rewritten as

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

The kinetic energy of the particle is written as

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

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

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