Answer to Question #245089 in Molecular Physics | Thermodynamics for kelani

Question

Three ice skaters meet at the center of a rink and each stands at rest facing the center, within arm's reach of the other two. On a signal, each skater pushes himself away from the other two across the frictionless ice. After the push, skater A with mass m_A = 70.0 kg

 moves in the negative y-direction at 2.00 m/s and skater B with mass m_B = 85.0 kg

 moves in the negative x-direction at 3.50 m/s. Find the x- and y-components of the 75.0 kg skater C's velocity (in m/s) after the push.

v_Cx=__ m/s

v_Cy=___ m/s

Answer 1

Three ice skaters meet at the center of a rink and we can consider that they are at rest before they start to move thus $v_{\text{skater A}_i} = v_{\text{skater B}_i} = v_{\text{skater C}_i} = 0\,m/s$, in consequence $\vec{p}_{\text{skater C}_i}=\vec{p}_{\text{skater B}_i}=\vec{p}_{\text{skater A}_i}=\vec{p_{i}}=\vec{0}$ .

Since momentum is conserved, this means that $\vec{p_i}=\vec{p_f}=\vec{p}_{\text{skater A}_f}+\vec{p}_{\text{skater B}_f}+\vec{p}_{\text{skater C}_f}=\vec{0}$. Now, we need to define the final conditions for the three skaters considering that A and B are moving along the negative y and x-axis, respectively:

$\vec{v}_{\text{skater A}_f} =[0\frac{m}s,-2\frac{m}s]; m_{\text{A}} =70.0\,kg \\ \vec{v}_{\text{skater B}_f} =[-3.5\frac{m}s,0\frac{m}s]; m_{\text{B}} =85.0\,kg \\ \vec{v}_{\text{skater C}_f} =[v_{c_x},v_{c_y}]; m_{\text{C}} =75.0\,kg$

Considering this we substitute for $\vec{p_f}= \vec{0}$ with the provided information when they started to move:

$\\ m_{A}\vec{v}_{\text{skater A}_f} +m_{B}\vec{v}_{\text{skater B}_f} + m_{C}\vec{v}_{\text{skater C}_f} =\vec{0} \\ (70\,kg) [0\frac{m}s,-2\frac{m}s]+(85\,kg)[-3.5\frac{m}s,0\frac{m}s]+ (75\,kg)[v_{c_x},v_{c_y}] =\vec{0} \\ [0,-140](\frac{kgm}s)+[-297.5,0](\frac{kgm}s)+ (75\,kg)[v_{c_x},v_{c_y}] =\vec{0}$

Following we can find $[v_{c_x},v_{c_y}]$ as

$[v_{c_x},v_{c_y}]= -\frac{(\frac{kgm}s)}{(75\,kg)}( [0,-140]+[-297.5,0]) \\ [v_{c_x},v_{c_y}]= -\frac{1}{(75)} [-297.5,-140](\frac{m}s) \\ [v_{c_x},v_{c_y}]= [+\frac{119}{30},+\frac{28}{15}](\frac{m}s)\approxeq[+3.966,+1.866](\frac{m}s)$

In conclusion, v_Cx= 3.966 m/s and v_Cy= 1.866 m/s. The skater C traves through the xy quadrant at v_C= 4.384 m/s.

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.