Question 7.SP.33: A spring unstretched is 200 mm long and connected to the two......

Since the same spring force acts on the two masses, but in opposite directions, the total linear impulse on the system of both masses is zero. Hence, the momentum of the two masses is constant, i.e., zero, and 2 v_{A}+3 v_{B}=0 . From this equation, v_{A}=-(3 / 2) v_{B} at all times.

Another equation involving v_{A} and v_{B} is necessary. The work done by the spring in expanding to its original length (by virtue of its potential energy) is equal to the change in kinetic energy of both masses. As the spring expands a distance x from its compressed position, its compressive force =2000(0.08-x) \mathrm{N} .

Hence, the total work done is

U=\int_{0}^{0.08} 2000(0.08-x) d x=6.4 \mathrm{~N} \cdot \mathrm{m}

Equate this to the final kinetic energy (initial is zero):

6.4=\frac{1}{2}(2) v_{A}^{2}+\frac{1}{2}(3) v_{B}^{2}

Solve simultaneously with the previous equation to obtain v_{B}=\underline{1.31 \mathrm{~m} / \mathrm{s}} (to the right) and v_{A}=\underline{1.96 \mathrm{~m} / \mathrm{s}} (to the left).