# Work -- KE related Question

1. Jan 25, 2017

### Arman777

1. The problem statement, all variables and given/known data
A father racing his son has half the kinetic energy of the son,who has half the mass of the father.The father speeds up by $1.0\frac {m} {s}$ and then has the same kinetic energy as the son.What are the original speeds of (a)the father and (b) the son

2. Relevant equations
$E_k=\frac 1 2 mv^2$

3. The attempt at a solution
$m_f=2m_s$
$2E_f=E_s$

İnital situation;
$2E_f= m_f(v_{f_0})^2=E_s=\frac 1 2 m_s(v_{s_0})^2$

so
$m_f=2m_s$

then we will obtain $4(v_{f_0})^2=(v_{s_0})^2$

Final situation

$\frac 1 2 m_f(v_{f_0}+1)^2=\frac 1 2 m_s(v_{s_0})^2$

then I get $v_{f_0}=1\frac {m} {s}$ which book says its $2.4\frac {m} {s}$

Where I am going wrong ? (My native language is not english so If made a mistake in understanding the question so sorry )

Thanks

Last edited: Jan 25, 2017
2. Jan 25, 2017

### Staff: Mentor

Everything looks good here. Show how you got your answer.

3. Jan 25, 2017

### Arman777

$\frac 1 2 m_f(v_{f_0}+1)^2=\frac 1 2 m_s(v_{s_0})^2$

we know that

$m_f=2m_s$
so
$\frac 1 2 2m_s(v_{f_0}+1)^2=\frac 1 2 m_s(v_{s_0})^2$

then we get ;

$2m_s(v_{f_0}+1)^2=m_s(v_{s_0})^2$

$2(v_{f_0}+1)^2=(v_{s_0})^2$ (1)

$\frac {(v_{s_0})^2} 2=2(v_{f_0})^2$ (2) From the $4(v_{f_0})^2=(v_{s_0})^2$

lets put this $Eq_2$ in the $Eq_1$

$(v_{f_0}+1)^2=2(v_{f_0})^2$

$(v_{f_0})^2+2(v_{f_0})+1=2(v_{f_0})^2$

$(v_{f_0})^2-2(v_{f_0})-1=0$

$((v_{f_0})-1)^2=0$

$(v_{f_0})=1\frac {m} {s}$

4. Jan 25, 2017

### Staff: Mentor

This is where you're messing up.

5. Jan 25, 2017

### Arman777

Oh I see.Thanks.unbelievable....