Question

1. two ice skaters want to prove conservation of momentum. skater a has a mass of 72 kg, and skater b has a mass of 48 kg. they are originally at rest. they hold hands and face each other. they then push off each other and move in opposite directions. skater b moves with a velocity of 3 m/s. what must the velocity of skater a be in the opposite direction after the push in order to prove conservation of momentum? (1 point) 1.5 m/s 3.0 m/s 2.0 m/s 4.5 m/s

Explanation:

Step1: Write the law of conservation of momentum formula

The law of conservation of momentum states that \(m_1v_1 + m_2v_2=m_{1}u_{1}+m_{2}u_{2}\). Since they are initially at rest \(u_1 = u_2=0\), so \(m_1v_1=-m_2v_2\) (negative because they move in opposite directions)

Step2: Substitute the given values

Let \(m_A = 72\space kg\), \(m_B = 48\space kg\), \(v_B=3\space m/s\). We need to find \(v_A\). From \(m_Av_A=-m_Bv_B\), we can solve for \(v_A\): \(v_A=\frac{-m_Bv_B}{m_A}\)
Substitute \(m_A = 72\), \(m_B = 48\), \(v_B = 3\) into the formula: \(v_A=\frac{- 48\times3}{72}\)
Calculate \(\frac{48\times3}{72}=\frac{144}{72} = 2\)

Answer:

\(2.0\space m/s\)