COMEDK 2023 Shift 1 Paper

When a neutron undergoes a head-on elastic collision with a lead nucleus, we can analyze the process using the principles of conservation of momentum and kinetic energy, which apply since it's an elastic collision. The question involves calculating the fractional change in the kinetic energy of the neutron after the collision.
Let's denote:
mn as the mass of the neutron, mp as the mass of the lead nucleus, where given it is 206 times the mass of the neutron, so mp=206mn, vi as the initial velocity of the neutron before the collision, vf as the final velocity of the neutron after the collision, ui=0 as the initial velocity of the lead nucleus (since it's considered stationary before the collision), uf as the final velocity of the lead nucleus after the collision.
Since momentum is conserved in an elastic collision, the total initial momentum equals the total final momentum, which can be expressed as:
mnvi+mpui=mnvf+mpuf
Given ui=0, we can simplify this to:
mnvi=mnvf+mpuf
Because the collision is elastic, kinetic energy is also conserved. The change in kinetic energy we are interested in involves calculating the kinetic energy before and after the collision, specifically for the neutron, and then finding the fractional change.
To find the velocities after the collision, we use the formula derived for velocities in a one-dimensional elastic collision involving two masses:
The velocity of the neutron (vf) after the collision is given by:
‌vf=‌

(mn−mp)vi

mn+mp

‌‌ Substituting ‌mp=206mn,‌ we get: ‌
‌vf=‌

(mn−206mn)vi

mn+206mn

‌vf=‌

−205mn

207mn

vi
‌vf=−‌

205

207

vi
The fractional change in kinetic energy of the neutron is given by the change in kinetic energy relative to the initial kinetic energy. The kinetic energy (KE) formula is KE=‌

1

2

mv2, so the fractional change in kinetic energy for the neutron is:
∆KE=‌

‌ 1 2 mnvf2− 1 2 mnvi2

1 2 mnvi2

Substituting vf=−‌

205

207

vi, we have:
‌∆KE=‌

‌ 1 2 mn(− 205 207 vi)2− 1 2 mnvi2

1 2 mnvi2

‌=‌

‌ 1 2 mn( 205 207 )2vi2− 1 2 mnvi2

1 2 mnvi2

‌=‌

( 205 207 )2−1

1

Calculating this gives us a fractional change in the kinetic energy:
‌=(‌

2052−2072

2072

)
‌=(‌

42025−42849

42849

)
‌=‌

824

42849

This fraction simplifies approximately to a fractional decrease, and when calculated, it yields a value around -0.01923 , which is about a 2% decrease in kinetic energy. Hence, the correct answer is:
Option B: 2% decrease.