Physics clicker questions: bullet into block sequence

Physics 5A, Fall Semester 2007
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 7.44

Students were asked the following clicker questions (Classroom Performance System, einstruction.com) near the end of their learning cycle:

Object 1 = 0.020 kg bullet
Object 2 = 2.00 kg block
i = bullet before hitting block, block still stationary
f = just after bullet embedded into block, but before any sliding actually takes place

[0.6 participation points.] What is conserved for the process i→f?
(A) Momentum.
(B) Kinetic energy.
(C) Energy, with non-conservative work being done.
(D) (More than one of the above choices apply.)
(E) (Nothing is conserved.)
(F) (I'm lost, and don't know how to answer this.)

Sections 0906, 0907
(A) : 25 students
(B) : 3 students
(C) : 1 students
(D) : 7 students
(E) : 0 students
(F) : 0 students

Correct answer: (A)

Since there are no external horizontal impulses, momentum is conserved, such that v_12,f = (m_1*v_1,i)/(m_1 + m2). However, since there is deformation with no rebound and separation in this perfectly inelastic collision, kinetic energy is not conserved.

f = just after bullet embedded into block, but before any sliding actually takes place
ff = after block (with bullet inside) comes to a rest because of kinetic friction

[0.6 participation points.] What is conserved for the process f→ff?
(A) Momentum.
(B) Kinetic energy.
(C) Energy, with non-conservative work being done.
(D) (More than one of the above choices apply.)
(E) (Nothing is conserved.)
(F) (I'm lost, and don't know how to answer this.)

Sections 0906, 0907
(A) : 4 students
(B) : 9 students
(C) : 8 students
(D) : 8 students
(E) : 5 students
(F) : 1 students

Correct answer: (C)

Friction causes a horizontal impulse on the block (with embedded bullet), and thus momentum is not conserved. Friction also does work on the block/bullet, thus kinetic energy is not conserved, but energy is conserved if the non-conservative work done by friction is accounted for.

W_nc = delta(K_12),
-mu_k*(m_1 + m_2)*g*delta_x = (1/2)*(m_1 + m_2)*((v_12,f)^2 - 0).