Conservation of momentum in head-on elastic collision

Steps:

• In a head-on collision, momentum conservation dictates that the total momentum before equals total after.
• For elastic collision, both momentum and kinetic energy are conserved.
• Let m_α be α-particle mass, v_α initial speed; M_gold gold mass, initially at rest.
• After backscattering, α speed is v_α', gold recoil speed v_gold; solving yields v_gold ≈ (2 m_α / M_gold) v_α for M_gold >> m_α.

Why C is correct:

• Gold nucleus mass (197 u) much greater than α-particle (4 u), so recoil speed v_gold is small per momentum conservation formula v_gold = (2 m_α v_α) / (m_α + M_gold) ≈ (2 m_α / M_gold) v_α.

Why the others are wrong:

• A: α-particle mass (4 u) is less than gold (197 u), opposite of stated.
• B: Head-on collision causes full backscattering, not slight deflection.
• D: Momentum decrease isn't the cause; collision dynamics determine recoil via conservation laws.

Final answer: C