Part 1/2
a neutron in a reactor makes an elastic headon collision with the nucleus of an atom initially at rest.
assume: the mass of the atomic nucleus is
about 14.1 the mass of the neutron.
what fraction of the neutron’s kinetic energy is transferred to the atomic nucleus?

part 2/2
if the initial kinetic energy of the neutron is
6.98 × 10−13 j, find its final kinetic energy.
answer in units of j.

0.247

5.25×10⁻¹³ J

Explanation:

Part 1/2

Elastic collision means both momentum and energy are conserved.

Momentum before = momentum after

m v = m v₁ + 14.1m v₂

v = v₁ + 14.1 v₂

Energy before = energy after

½ m v² = ½ m v₁² + ½ (14.1m) v₂²

v² = v₁² + 14.1 v₂²

We want to find the fraction of the neutron's kinetic energy is transferred to the atomic nucleus.

KE/KE = (½ (14.1m) v₂²) / (½ m v²)

KE/KE = 14.1 v₂² / v²

KE/KE = 14.1 (v₂ / v)²

We need to find the ratio v₂ / v.  Solve for v₁ in the momentum equation and substitute into the energy equation.

v₁ = v − 14.1 v₂

v² = (v − 14.1 v₂)² + 14.1 v₂²

v² = v² − 28.2 v v₂ + 198.81 v₂² + 14.1 v₂²

0 = -28.2 v v₂ + 212.91 v₂²

0 = -28.2 v + 212.91 v₂

28.2 v = 212.91 v₂

v₂ / v = 28.2 / 212.91

v₂ / v = 0.132

Therefore, the fraction of the kinetic energy transferred is:

KE/KE = 14.1 (0.132)²

KE/KE = 0.247

Part 2/2

If a fraction of 0.247 of the initial kinetic energy is transferred to the atomic nucleus, the remaining 0.753 fraction must be in the neutron.

Therefore, the final kinetic energy is:

KE = 0.753 (6.98×10⁻¹³ J)

KE = 5.25×10⁻¹³ J

you go to your settings preference and in the end youll see sth that says "i want to delete my acc" click on it and u'll acc will be deleted but u have to wait.

another example of kinetic energy would be,

kinetic energy   >     moving energy

answer   >   letter choice (c),   a hammer falling off a table.

hope that !                                         : )