Question 9.120: A body is traveling at 2.0 m/s along the positive direction ......

One approach is to choose a moving coordinate system that travels the center of mass of the body, and another is to do a little extra algebra analyzing it in the original coordinate system (in which the speed of the m = 8.0 kg mass is v_0 = 2 m/s, as given). Our solution is in terms of the latter approach since we are assuming that this is the approach most students would take. Conservation of linear momentum (along the direction of motion) requires

m v_0=m_1 v_1+m_2 v_2 \Rightarrow(8.0)(2.0)=(4.0) v_1+(4.0) v_2

which leads to v_2 = 4 -v_1 in SI units (m/s). We require

\Delta K=\left(\frac{1}{2}m_1 v_1^2+\frac{1}{2}m_2 v_2^2\right)-\frac{1}{2}m v_0^2 \Rightarrow 16=\left(\frac{1}{2}(4.0) v_1^2+\frac{1}{2}(4.0) v_2^2\right)-\frac{1}{2}(8.0)(2.0)^2

which simplifies to v_2^2=16-v_1^2 in SI units. If we substitute for v_2 from above, we find

\left(4-v_1\right)^2=16-v_1^2

which simplifies to 2 v_1^2-8 v_1=0, and yields either v_1 = 0 or v_1 = 4 m/s. If v_1 = 0 then v_2 =4 – v_1 = 4 m/s, and if v_1 = 4 m/s then v_2 = 0.

(a) Since the forward part continues to move in the original direction of motion, the speed of the rear part must be zero.

(b) The forward part has a velocity of 4.0 m/s along the original direction of motion.