Question

1. an ice skater at rest on ice catches a dance partner moving 2.5 m/s during a performance. the ice skater has a mass of 68 kg and the dance partner has a mass of 54 kg. what is the speed of the ice skater and dance partner after the collision? (1 point) 1.4 m/s 2.3 m/s 2.0 m/s 1.1 m/s

Explanation:

Step1: Apply the law of conservation of momentum

The law of conservation of momentum states that \(m_1u_1 + m_2u_2=(m_1 + m_2)v\). Here, \(m_1 = 68\space kg\), \(u_1=0\space m/s\) (since the ice - skater is at rest), \(m_2 = 54\space kg\), \(u_2 = 2.5\space m/s\).
Substitute the values into the formula: \((68\times0)+(54\times2.5)=(68 + 54)v\)

Step2: Simplify the equation

First, calculate the left - hand side: \(0+135 = 122v\) (because \(54\times2.5=135\) and \(m_1 + m_2=68 + 54 = 122\space kg\))
Then, solve for \(v\): \(v=\frac{135}{122}\approx1.1\space m/s\)

Answer:

\(1.1\space m/s\)