Question

8.3 elastic and inelastic collisions

1. if a man with mass 70 kg, standing still, throws an object with mass 5 kg at 50 m/s, what will be the recoil velocity of the man, assuming he is standing on a frictionless surface?

a. -3.6 m/s
b. 0 m/s
c. 3.6 m/s

Explanation:

Step1: Apply the law of conservation of momentum

The initial momentum of the system (man + object) is \(0\) (since both are at rest initially). According to the law of conservation of momentum \(m_1v_1 + m_2v_2=0\), where \(m_1 = 70\space kg\) (mass of the man), \(v_1\) is the recoil velocity of the man, \(m_2=5\space kg\) (mass of the object), and \(v_2 = 50\space m/s\) (velocity of the object).

Step2: Solve for \(v_1\)

Substitute the values into the equation \(70v_1+5\times50 = 0\). Then \(70v_1=- 250\). So \(v_1=\frac{-250}{70}\approx - 3.6\space m/s\)

Answer:

A. \(-3.6\space m/s\)