Physics quiz question: bullet embedding in metal sphere

Physics 205A Quiz 4, fall semester 2015
Cuesta College, San Luis Obispo, CA

Alternative Ballistics 3D Product Intro - The Alternative In Lethal Force"
Alternative Ballistics
youtu.be/5TFuZ_ol7Hw

An Alternative Ballistics attachment[*] for a handgun holds a stationary hollow metal sphere at the end of the barrel, such that it catches the bullet to become a non-lethal projectile. The initial speed of the bullet[**] (mass 1.1×10^–2 kg) is 350 m/s, and after it embeds in the sphere, the combined bullet-sphere final speed is 70 m/s. Ignore friction, drag, and external forces during the brief amount of time that the bullet takes to embed inside the sphere. The mass of the hollow metal sphere is:
(A) 2.2×10^–3 kg.
(B) 4.4×10^–2 kg.
(C) 5.5×10^–2 kg.
(D) 0.77 kg.

[*] alternativeballistics.com.
[**] wki.pe/.40_S%26W.

Correct answer (highlight to unhide): (B)

Since there are no significant external impulses on the bullet-sphere system during this brief process, total momentum is conserved, so:

0 = m₁·(v_f1 – v₀₁) + m₂·(v_f2 – v₀₂),

where for the bullet, m₁ = 1.1×10^–2 kg, v₀₁ = +350 m/s (in the forward direction), v_f1 = +70 m/s (also in the forward direction); and for the sphere, m₂ = ?, v₀₂ = 0, v_f2 = +70 m/s (same as the final velocity of the bullet, as they travel together after this completely inelastic collision). So solving for the mass m₂ of the sphere:

0 = m₁·(v_f1 – v₀₁) + m₂·(v_f2 – 0),

–m₂·v_f2 = m₁·(v_f1 – v₀₁),

m₂ = –m₁·(v_f1 – v₀₁)/v_f2,

m₂ = –(1.1×10^–2 kg)·((+70 m/s) – (+350 m/s))/(+70 m/s) = 0.044 kg,

or to two significant figures, the mass of the sphere is 4.4×10^–2 kg.

(Response (A) is m₁·(v_f1/v₀₁); response (C) is m₁·(v₀₁/v_f1); response (D) is m₁·v₀₁.)

Student responses
Sections 70854, 70855, 73320
Exam code: quiz04w04K
(A) : 5 students
(B) : 44 student
(C) : 18 students
(D) : 5 students

Success level: 61%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.36