A neutron with mass 1.67e-27 kg moving with speed 5.0 × 101 km/s makes a head-on collision with a boron nucleus, originally at rest, with mass 1.66e-26 kg.
A) If the collision is completely inelastic, so the particles stick together, what is the final kinetic energy of the system, expressed as a fraction of the original kinetic energy? Express it as a decimal.
B) If the collision were perfectly elastic, what fraction of the original kinetic energy is transferred to the boron nucleus? Express it as a decimal.
a use conservation of momentum to find final velocity of the combined particle. Then, knowing final and initial velocities, tabulate the KE ratio
b. On this, you have to use conservation of energy and momentum, compute both final velocities.
To answer these questions, we need to apply the principles of conservation of momentum and conservation of kinetic energy.
A) In a completely inelastic collision, the particles stick together and move as one system after the collision. In this case, the initial kinetic energy of the system is given by the neutron, since the boron nucleus is initially at rest.
First, let's convert the speed of the neutron from kilometers per second to meters per second:
Given: speed = 5.0 × 101 km/s
To convert km/s to m/s, we need to multiply by 1000 (since 1 km = 1000 m).
So, speed = 5.0 × 101 km/s = 5.0 × 101 × 1000 m/s = 5.0 × 104 m/s
The initial kinetic energy of the system is given by:
K_initial = (1/2) * mass_neutron * speed^2
= (1/2) * 1.67e-27 kg * (5.0 × 104 m/s)^2
Now, after the collision, the neutron and the boron nucleus stick together. Therefore, the final kinetic energy of the system is zero, since the system comes to rest.
To find the fraction of the original kinetic energy, we divide the final kinetic energy by the initial kinetic energy:
Fraction = final kinetic energy / initial kinetic energy
= 0 / ((1/2) * 1.67e-27 kg * (5.0 × 104 m/s)^2)
= 0.
Therefore, the final kinetic energy of the system, expressed as a fraction of the original kinetic energy, is 0 (or 0%).
B) In a perfectly elastic collision, both momentum and kinetic energy are conserved. In this scenario, the particles rebound off each other without sticking together.
To find the fraction of the original kinetic energy transferred to the boron nucleus, we need to calculate the final kinetic energy of the boron nucleus when it rebounds.
Since the collision is head-on, the initial momentum of the system is the sum of the momenta of the neutron and the boron nucleus:
Initial momentum = (mass_neutron * velocity_neutron) + (mass_boron * velocity_boron)
= (1.67e-27 kg * 5.0 × 104 m/s) + (1.66e-26 kg * 0 m/s)
The final momentum of the system is given by:
Final momentum = (mass_neutron + mass_boron) * velocity_final
Since momentum is conserved, the initial momentum equals the final momentum:
(1.67e-27 kg * 5.0 × 104 m/s) + (1.66e-26 kg * 0 m/s) = (mass_neutron + mass_boron) * velocity_final
We can rearrange the equation to solve for the final velocity of the system:
velocity_final = [ (1.67e-27 kg * 5.0 × 104 m/s) ] / (mass_neutron + mass_boron)
Now, to find the fraction of the original kinetic energy transferred to the boron nucleus, we use the formula:
Fraction = (change in kinetic energy of the boron nucleus) / (initial kinetic energy of the system)
The change in kinetic energy of the boron nucleus is given by:
Change in KE_boron = (1/2) * mass_boron * (velocity_final)^2 - (1/2) * mass_boron * (0 m/s)^2
Finally, we can calculate the fraction of the original kinetic energy transferred to the boron nucleus:
Fraction = (1/2) * mass_boron * (velocity_final)^2 / ((1/2) * mass_neutron * speed^2).
To get the final answer, you can substitute the values for the respective masses, velocities, and speeds into the equations and perform the calculations.