1. Two blocks of mass 0.1 kg and 0.2 kg approach each other on a frictionless
surface at velocities of 0.4 and 1 m/s respectively. If the blocks collide and remain
together, calculate their joint velocity after the collision.
2. A cyclotron produces a 100-μA beam of 15-MeV deuterons. If the cyclotron
were 100% efficient in converting electrical energy into kinetic energy of the
deuterons, what is the minimum required power input, (in kilowatts)?
3. In a certain 25-W mercury-vapor ultraviolet lamp, 0.1% of the electric energy input
appears as UV radiation of wavelength 2537 AËš . What is the UV photon emission
rate per second from this lamp?
12
1. To solve this problem, we can use the principle of conservation of momentum.
The equation for conservation of momentum is:
(m1 * v1) + (m2 * v2) = (m1 + m2) * Vf
Where:
m1 and m2 are the masses of the two blocks,
v1 and v2 are the initial velocities of the two blocks, and
Vf is the final velocity of the combined blocks after the collision.
Given:
m1 = 0.1 kg (mass of the first block)
m2 = 0.2 kg (mass of the second block)
v1 = 0.4 m/s (velocity of the first block)
v2 = 1 m/s (velocity of the second block)
Using these values, we can substitute them into the equation and solve for Vf:
(0.1 kg * 0.4 m/s) + (0.2 kg * 1 m/s) = (0.1 kg + 0.2 kg) * Vf
(0.04 kg*m/s) + (0.2 kg*m/s) = (0.3 kg) * Vf
0.24 kg*m/s = 0.3 kg * Vf
Now we can solve for Vf:
Vf = 0.24 kg*m/s / 0.3 kg
Vf = 0.8 m/s
Therefore, the joint velocity of the blocks after the collision is 0.8 m/s.
2. To calculate the minimum required power input for the cyclotron, we can use the equation:
Power = (current * voltage) / efficiency
Given:
Current = 100 μA (microamperes) = 0.0001 A
Voltage = 15 MeV (million electron volts) = 15 * 10^6 eV
Efficiency = 100% = 1 (since the cyclotron is 100% efficient)
First, we need to convert the units of current and voltage:
1 A = 1 coulomb/second
1 eV = 1.6 * 10^-19 J (joules)
Converting current to amperes:
0.0001 A
Converting voltage to joules:
15 * 10^6 eV * (1.6 * 10^-19 J/eV) = 24 * 10^-13 J
Now we can calculate the power:
Power = (0.0001 A * 24 * 10^-13 J) / 1
Power = 24 * 10^-16 W = 2.4 * 10^-15 W
Finally, we can convert the power to kilowatts:
1 W = 0.001 kW
Power = (2.4 * 10^-15 W) * (0.001 kW / 1 W) = 2.4 * 10^-18 kW
Therefore, the minimum required power input for the cyclotron is 2.4 * 10^-18 kW.
3. To calculate the UV photon emission rate per second from the lamp, we can use the formula:
Emission rate = (electric energy input * efficiency) / (energy per photon * wavelength)
Given:
Electric energy input = 25 W
Efficiency = 0.1% = 0.001
Energy per photon = energy of a photon with wavelength 2537 A = E = hc / λ,
where h is the Planck's constant (6.63 * 10^-34 J·s) and c is the speed of light (3 * 10^8 m/s)
wavelength, λ = 2537 A = 2537 * 10^-10 m
Using these values, we can calculate the UV photon emission rate:
Emission rate = (25 W * 0.001) / ((6.63 * 10^-34 J·s * 3 * 10^8 m/s) / (2537 * 10^-10 m))
Simplifying:
Emission rate = (25 W * 0.001) / ((6.63 * 10^-34 J·s * 3 * 10^8 m/s) / (2537 * 10^-10 m))
Emission rate = 2.5 * 10^-3 / (6.63 * 3 * 10^8 / 2537) * 10^(-34-8+10)...
Emission rate = 2.5 * 10^-3 / (6.63 * 3 * 2537) * 10^-32
Emission rate = 3.94 * 10^(-3-1-32)
Emission rate = 3.94 * 10^-30 photons/s
Therefore, the UV photon emission rate per second from this lamp is 3.94 * 10^-30 photons/s.