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 mA = 70.0 kg moves in the negative y-direction at 2.00 m/s and skater B with mass m = 85.0 kg moves in the negative x-direction at 2.50 m/s. Find the x- and y-components of the 80.0 kg skater C's velocity (in m/s) after the push. HINTvcx= ___ m/svcy=____m/s

Answer 1

To solve the problem it is necessary to apply the conservation equations for the moment, specifically for collision. In addition to that, the concepts of vector velocity are necessary, in which the components are obtained and if the total magnitude is necessary.

Conservation of Momentum equation is given by,

$m_1v_1+m_2v_2+m_3v_3 = 0$

We need to find the speed 3, therefore by readjusting the equation we have,

$v_c = \frac{-(m_1v_1+m_2v_2)}{m_3}$

$v_c = \frac{-(-70*2\hat{j}-85*2.5\hat{j})}{80}$

$v_c = 2.656\hat{i}+ 1.75\hat{j}$

Therefore the components of the velocity would be,

$v_{cx} = 2.656m/s$

$v_{cy} = 1.75m/s$