Answer to Question #90867 in Chemistry for Shanieka

Question

A father racing his son has 1/4 the kinetic energy of the son, who has 1/3 the mass of the father. The father speeds up by 1.6 m/s and then has the same kinetic energy as the son. What are the original speeds of (a) the father and (b) the son?

Answer 1

Kinetic energy formula:

$$E=1/2*mU^2$$

let's mark the speed of father U_F, the speed of son - U_S.

E_F=m_F*U²_F/2

E_S=m_S*U²_S/2

E_F=1/4*E_S

m_S=1/3m_F

E_S=m_F*(U_F+1.6)²/2

System of equations:

m_S*U²_S/2=m_F(U_F+1.6)²/2 (1)

m_F=3*m_S (2)

m_F*U²_F/2=1/4*(m_S*U²_S)/2 (3)

(2) --> (1)

(2) --> (3)

(1) : m_S*U²_S/2=3*m_S*(U_F+1.6)²/2 --> U²_S=3*(U_F+1.6)²

(3) : 3*m_S*U²_F/2=1/4*m_S*U²_S/2 --> 3U²_F=U²_S/2 --> U²_S=6*U²_S

6*U²_F=3*(U_F+1.6)²

2*U²_F=U²_F+3.2*U_F+1.6²

U_F=1/2*(3.2+/-4.5)

U_F=5.45 m/s

U_S=U_F*6^1/2=13.35 m/s

Answer: U_F=5.45 m/s, U_S=13.35 m/s