A ball has a kinetic energy of 100 J. What would be the kinetic energy of a ball with twice the mass and half the momentum?

May 12, 2018

$\text{KE"_2=12.5color(white)(l)"J}$

Explanation:

There exist a relationship between $\text{KE}$ and $p$, the kinetic energy and momentum of an object of mass $m$.

$\text{KE} = \frac{1}{2 m} \cdot {p}^{2}$

Proof for this formula

$\text{L.H.S."="KE}$
$\textcolor{w h i t e}{\text{L.H.S.}} = \frac{1}{2} m \cdot {v}^{2}$
$\textcolor{w h i t e}{\text{L.H.S.}} = \frac{1}{2} \cdot {\left(m \cdot v\right)}^{2} \cdot \frac{1}{m}$
$\textcolor{w h i t e}{\text{L.H.S.}} = \frac{1}{2 m} \cdot {p}^{2}$
color(white)("L.H.S.")="R.H.S."

The question states that

• ${p}_{2} = \frac{1}{2} {p}_{1}$ and
• ${m}_{2} = 2 {m}_{1}$

where ${p}_{1}$, ${p}_{2}$, ${m}_{1}$, and ${m}_{2}$ the mass and velocity of the first and second ball, respectively,

The kinetic energy of the second ball would be

${\text{KE}}_{2} = \frac{1}{2 \textcolor{\mathrm{da} r k b l u e}{{m}_{2}}} \cdot {\textcolor{p u r p \le}{{p}_{2}}}^{2}$
$\textcolor{w h i t e}{{\text{KE}}_{2}} = \frac{1}{2 \cdot \textcolor{\mathrm{da} r k b l u e}{2 \cdot {m}_{1}}} \cdot {\textcolor{p u r p \le}{\left(\frac{1}{2} {p}_{1}\right)}}^{2}$
$\textcolor{w h i t e}{{\text{KE}}_{2}} = \frac{1}{2 \textcolor{\mathrm{da} r k b l u e}{{m}_{1}}} \cdot {\textcolor{p u r p \le}{{p}_{1}}}^{2} \cdot \textcolor{\mathrm{da} r k b l u e}{\frac{1}{2}} \cdot {\left(\textcolor{p u r p \le}{\frac{1}{2}}\right)}^{2}$
$\textcolor{w h i t e}{{\text{KE}}_{2}} = \frac{1}{8} \cdot \frac{1}{2 \textcolor{\mathrm{da} r k b l u e}{{m}_{1}}} \cdot {\textcolor{p u r p \le}{{p}_{1}}}^{2}$
color(white)("KE"_2)=1/8*"KE"_1
color(white)("KE"_2)=1/8*100color(white)(l)"J"
color(white)("KE"_2)=12.5color(white)(l)"J"