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

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2