Kevin S. Huang

2015: Problem 6

Topic: Collisions
Concepts: Conservation of linear momentum, Dynamics of CM

Solution:

Since there is no net external force on the system, we have conservation of linear momentum $p_{\text{tot}}$ . Recall that

M_{\text{tot}}v_{\text{CM}}=p_{\text{tot}}

v_{\text{CM}}=\frac{p_{\text{tot}}}{M_{\text{tot}}}

Since linear momentum is conserved, we can compute it at the beginning,

p_{\text{tot}}=m_1v_1-m_2v_2

We have $M_{\text{tot}}=m_1+m_2+m_3$ so

v_{\text{CM}}=\frac{m_1v_1-m_2v_2}{m_1+m_2+m_3}=\frac{(1.9)(1.7)-(1.1)(2.5)}{1.9+1.1+1.3}\,\mathrm{m/s}=0.11\,\mathrm{m/s}

Thus, the answer is A.