2016: Problem 10

Topic: Dynamics
Concepts: Cutting strings vs. springs, Newton’s laws, Statics


Solution:

Initially when the blocks are at rest, the tension T1T_1 in the top string holds the weight of both blocks while the tension T2T_2 in the bottom string holds the weight of m2m_2 so

T1=(m1+m2)gT_1=(m_1+m_2)g
T2=m2gT_2=m_2g

If the top string is cut at the connection point to m1m_1, then at that instant m1m_1 no longer feels an upward tension force. On the other hand, the tension T2T_2 does not change instantaneously since it is sustained in an elastic string (effectively a spring). It takes time for a spring to change its length and we know the tension is determined by its elongation. Thus, we still have T2=m2gT_2=m_2g at that moment.

Applying Newton’s 2nd law to m1m_1,

m1a1=m1g+T2=m1g+m2gm_1a_1=m_1g+T_2=m_1g+m_2g
a1=(1+m2m1)g=(1+4kg2kg)(10m/s2)=30m/s2a_1=\left(1+\frac{m_2}{m_1}\right)g=\left(1+\frac{4\,\mathrm{kg}}{2\,\mathrm{kg}}\right)(10\,\mathrm{m/s^2})=30\,\mathrm{m/s^2}

Applying Newton’s 2nd law to m2m_2,

m2a2=m2gT2=0m_2a_2=m_2g-T_2=0
a2=0a_2=0

Thus, the answer is D.

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