Kevin S. Huang

2012: Problem 5

Topic: Collisions
Concepts: 1D inelastic collision, Kinetic energy

Solution:

The initial kinetic energy is given by

K_i=\frac{1}{2}Mv_1^2+\frac{1}{2}mv_2^2=\frac{1}{2}(12\,\mathrm{kg})(4\,\mathrm{m/s})^2+\frac{1}{2}(6\,\mathrm{kg})(2\,\mathrm{m/s})^2=108\,\mathrm{J}

Conserving linear momentum, we have

Mv_1-mv_2=(M+m)v_f

v_f=\frac{Mv_1-mv_2}{M+m}=\frac{(12\,\mathrm{kg})(4\,\mathrm{m/s})-(6\,\mathrm{kg})(2\,\mathrm{m/s})}{12\,\mathrm{kg}+6\,\mathrm{kg}}=2\,\mathrm{m/s}

where $v_f$ is the final velocity of both masses. The final kinetic energy is given by

K_f=\frac{1}{2}(M+m)v_f^2=\frac{1}{2}(18\,\mathrm{kg})(2\,\mathrm{m/s})^2=36\,\mathrm{J}

The loss in kinetic energy is

|\Delta K|=K_i-K_f=72\,\mathrm{J}

so the answer is D.