Question

suppose a clay model of a koala bear has a mass of 0.205 kg and slides on ice at a speed of 0.745 m/s. it runs into another clay model, which is initially motionless and has a mass of 0.345 kg. both being soft clay, they naturally stick together. what is their final velocity (in m/s)? (enter the magnitude )

Explanation:

Step1: Apply the law of conservation of momentum

The law of conservation of momentum states that \(m_1v_1 + m_2v_2=(m_1 + m_2)v_f\). Here, \(m_1 = 0.205\space kg\), \(v_1=0.745\space m/s\), \(m_2 = 0.345\space kg\), and \(v_2 = 0\space m/s\) (since the second clay - model is initially motionless).
Substitute the values into the formula: \((0.205\times0.745)+(0.345\times0)=(0.205 + 0.345)v_f\)

Step2: Solve for \(v_f\)

First, calculate the left - hand side: \(0.205\times0.745=0.152725\)
The right - hand side is \(0.55v_f\) (because \(0.205 + 0.345=0.55\))
So, \(v_f=\frac{0.152725}{0.55}\)

Answer:

\(v_f = 0.278\space m/s\)