Energy-momentum Formula Questions:

1) A particle is moving with an energy of 400 kJ (kilojoules), it has a mass of 2*10^(-9)g, what is it momentum?

Answer:

From the formula of the energy we find the momentum:

P = √ (E² – (m₀ c²)²)/c

P = √ ((400*10³ J)² – (2*10^(-12) Kg (3*10⁸ m/s)²)²)/(3*10⁸ m/s)

P = √ (127600000000 J²)/(3*10⁸ m/s)

P = 357211.4 J / (3*10⁸ m/s) = 119070.4 *10⁸ Kg*m/s

2) The momentum of a mass is 10¹² Kg*m/s, if it rest mas is 10 ng, what is its energy?

Answer:

using the formula of the energy,

E = √ ((10¹² Kg*m/s 3*10⁸ m/s)² + (10*10^(-12) Kg (3*10⁸ m/s)²))

E = √ (9 *10⁴⁰ J²) = 3 *10²⁰ J

Photon Energy Formula Questions:

1) Calculate the energy of a photon which has a wavelength of 2.3 μm.

Answer:

We replaced the wavelength, Plank’s constant and the speed of light in the photon’s energy equation

E = (6.626 x 10^-34 J.s)(3.0 x 10⁸ m/s) / 2.3 x 10^-6 m

E = 8.64 x 10^-20 J

2) Calculate the wavelength which has a photon with an energy of 6.0 x 10 ^-20 J

Answer:

First we cleared the photon’s wavelength of the photon’s energy equation

λ = hc / E

Now we replaced the data

λ = (6.626 x 10^-34 J.s)(3.0 x 10⁸ m/s) / 6.0 x 10^-20 J

λ = 3.3 x 10 ^-6 m

 

 

α = angular acceleration, (radians/s²)

Δω = change in angular velocity (radians/s)

Δt = change in time (s)

ω₁ = initial angular velocity (radians/s)

ω₂= final angular velocity (radians/s)

t₁ = initial time (s)

t₂= final time (s)

Angular Acceleration Formula Questions:

1) A disc in a DVD player starts from rest, and then begins spinning when the user presses “Play”. After 4.00 s, the disc is spinning at 160 radians/s. What was the average angular acceleration of the disc?

Answer: At the initial time (t₁ = 0 s), the angular velocity was ω₁ = 0 radians/s. At the final time (t₂ = 4.00 s), the angular velocity of the disc was ω₁ = 160 radians/s. The average angular acceleration can be found using the formula:

 

 

 

 

 

Between the initial and final times, the average angular acceleration of the disc was 40.0 radians/s².

2) At certain high-speed roller coaster has a magnetic braking system that is meant so slow the roller coaster cars as they approach the station. The magnetic brakes slow the cars just enough for another braking system to take over. If the wheels of the cars are initially spinning at 400.0 revolutions per second, and the magnetic brakes apply a constant angular acceleration of -440.0 radians/s² for 5.00 s, what will be the angular velocity of the roller coaster cars after the magnetic brakes have been applied?

Answer: The angular velocity at the initial time (t₁ = 0 s) is given in terms of revolutions per second. To use this value in the angular acceleration formula, the value must be converted to radians per second. There are 2π radians per revolution, and so the initial angular velocity is:

ω₁ = 400.0 revolutions/s

 

 

The final angular velocity at time t₁ = 5.0 s can be found by rearranging the angular acceleration formula:

 

 

 

 

 

 

 

The angular velocity after the magnetic brakes have been applied is approximately 313 radians/s. This is equivalent to 49.8 revolutions/s.

 

 

 

H = maximum height (m)

v₀ = initial velocity (m/s)

g = acceleration due to gravity (9.80 m/s²)

θ = angle of the initial velocity from the horizontal plane (radians or degrees)

Maximum Height Formula Questions:

1) A firefighter aims a fire hose upward, toward a fire in a skyscraper. The water leaving the hose has a velocity of 32.0 m/s. If the firefighter holds the hose at an angle of 78.5°, what is the maximum height of the water stream?

Answer: The water droplets leaving the hose can be treated as projectiles, and so the maximum height can be found using the formula:

 

 

 

 

 

The maximum height of the water from the hose is 50.2 m.

2) An athlete in a high jump competition leaves the ground at a velocity of 5.80 m/s, and an angle of 87.4°. What is the maximum height of the athlete’s center of mass?

Answer: The center of mass of the athlete can be found by treating the person as a projectile, and using the maximum height formula:

 

 

 

 

 

The maximum height of the athlete’s center of mass is 1.71 m.

Note: High jumpers use a technique called the “Fosbury flop” to lower their centers of mass below their bodies. Using this technique means that they can jump over bars approximately 20 cm higher than their centers of mass.

Relativistic Energy Formula Questions:

1) What is the energy of a particle whit mass 4.2 x 10 ^-27 kg and velocity 270.0 x 10⁶m/s?

Answer:

We replace the data in the relativistic energy equation:

E = (4.2 x 10^-27 kg)(3.0 x 10⁸ m/s)² / sqrt [1 – (270.0 x 10⁶ m/s / 3.0 x 10⁸ m/s)²]

E = 7.03 x 10^-10 J

 

2) Find the velocity of a particle whose relativistic energy is 5.9 x 10^-8 J and has a mass of 3.8 x 10 ^-27 kg

Answer:

We cleared the velocity of the relativistic energy equation:

v = c sqrt (1 – (mc² / E)²

Then we replace the data:

v = 3.0 x 10⁸ m/s sqrt [1 – (3.8 x 10^-27 kg(3.0 x 10⁸ m/s)² / 5.9 x 10^-8 J)²]

v = 299.9 x 10⁶ m/s

Schrodinger Equation Formula Questions:

1) A particle is confined to a box, in one dimension, over the x axe. The wavefunction describing the particle is Ae^{(I n x)}, where n is a integer number and I is the imaginary number obeying I²=-1. What is the energy of the particle given in terms of n?

Answer: 

We replace the data in the equation above:

-ℏ²/2m ∂²/∂x² (Ae^{(I n x))} =

=-ℏ²/2m A ∂²/∂x² (e^{(I n x)}) =

=-ℏ²/2m A I² n² e^{(I n x)}

Then regrouping the terms conveniently, we find the form

=-(I²) ℏ²/2m n² (Ae^{(I n x)})

=-(-1) ℏ²/2m n² (Ae^{(I n x)})

And the expression in the parenthesis is exactly the wavefunction

= ℏ²/2m n² Ψ

Then, because this is only the left-hand side of the Schrodinger equation, the complete equation is

ℏ²/2m n² Ψ = E Ψ

So, the energy is

E= ℏ²/2m n²

 

2) Consider the same particle above, what is its velocity after those 5 seconds?

Answer:

The formula for the velocity is obtained deriving with respect to time the equation for the position.

V=d/dt (r₀ + v₀ t +1/2 a t²)=a*t

Then, the velocity is

V=2 m/s² * 5 s = 10 m/s

Uncertainty Principle Formula Questions:

1) Assume an electron is confined to a atom of size 0.4 nm, what is the energy average of the particle in the atom?

Answer:

From the equation above we find Δp, which is the average momentum of the particle in the atom

Δp ≤ ℏ/2(0.4 nm) = 1.66*10^(-24) Kg*m/s

The energy is given by Δp²/2m, where 9.10938356 × 10^-31 kg is the mass of the electron

ΔE ≤ 9.4 eV

2) Consider the same particle above, what is the time average of the electron in the atom?

Answer:

The formula for the time comes from the second equation of the uncertainty principle

Δt ≤ ℏ/ 2ΔE

Then Δt ≤ 4.13*10⁽¹⁵⁾ eV*s/2(1.66*10^(-24) Kg*m/s)

Δt ≤ 1.24*10⁽³⁹⁾ s ≈ 3.94*10⁽³¹⁾ years

Instantaneous Velocity Formula Questions:

1) A cat that is walking toward a house along the top of a fence is moving at a varying velocity. The cat’s position on the fence is . Position x is in meters, and time t is in seconds. What is the cat’s instantaneous velocity at time t = 10.0 s?

Answer: The cat’s velocity can be found using the formula:

The cat’s position has only one component since it is moving in a straight line along the fence. The positive x direction is chosen to be toward the house. The magnitude of the velocity in the x direction is:

This derivative can be solved using the scalar multiple rule and the power rule for derivatives:

The magnitude of the instantaneous velocity is:

The magnitude of the cat’s instantaneous velocity at t = 10.0 s is:

The cat’s instantaneous velocity at t=10.0 s is 0.05 m/s in the -x direction (away from the house).

2) A child kicks a ball horizontally, off the edge of a cliff. The horizontal position of the ball is given by the function x(t) = bt, where b = 6.0 m/s. The vertical position of the ball is given by the function y(t) = ct², where c = -4.90 m/s². At t = 4.0 s, what are the horizontal and vertical components of the instantaneous velocity?

Answer: The components of the instantaneous velocity can be found using the formula:

If the horizontal direction of the ball is defined as the positive x direction, and vertically upward is defined as the positive y direction, then the magnitudes of the x and y components of the instantaneous velocity are:

and

These can be solved using the scalar multiple rule and the power rule for derivatives:

The horizontal instantaneous velocity is:

The horizontal velocity of the ball is a constant value of 6.0 m/s in the +x direction.

The vertical instantaneous velocity is:

v_y = c(2t)

v_y = 2ct

At t = 4.0 s, the vertical instantaneous velocity is:

v_y = 2ct

v_y = 2(-4.90 m/s²)(4.0 s)

The vertical instantaneous velocity at t = 4.0 s is 39.2 m/s in the -y direction.

Wavelength to Frequency Formula Questions:

1) One of the violet lines of a Krypton laser is at 406.7 nm. What is its frequency?

Answer: The velocity, v, of light is 3 x 10⁸m/s.

The wavelength, λ = 406.7 nm = 406.7 x 10^-9 m.

λ = v/f

f = v/λ

f = 3 x 10⁸ / 406.7 x 10^-9

f = 1.22 x 10 ²⁰ Hz

2) If a wave is moving through a medium at 700 m/s, and 2500 waves pass point A in one minute, what is the wavelength?

Answer:

The velocity of the wave, v = 700 m/s. The frequency of the wave, f = 2500/min x 1 min/60 sec = 41.67 waves/sec.

λ = v/f

λ = (700 m/s) / (41.67 waves/sec)

λ = 16.8 m

3) A beam of light has a wavelength of 350 m. What is the frequency of the light?

Answer: The speed of light, v = 3 x 10⁸m/s and the wavelength, λ = 350 m.

λ = v/f

f = v/λ

f = (3 x 10⁸ m/sec)/ (350 m)

f = 857142.86 waves/sec = Hz