Conservation of Momentum EX 1

(a) The kinetic energy is related to the momentum as shown below: $K={\frac {1}{2}}mv^{2}={\frac {1}{2}}{\frac {m}{m}}mv^{2}={\frac {1}{2}}{\frac {m^{2}v^{2}}{m}}={\frac {1}{2}}{\frac {(mv)^{2}}{m}}={\frac {1}{2}}{\frac {p^{2}}{m}}$

Given that $p_{B}=p_{A}$, we can compute the ratio of their kinetic energies using the above expression as

${\frac {K_{B}}{K_{A}}}={{\frac {1}{2}}{\frac {{p_{B}}^{2}}{m_{B}}} \over {\frac {1}{2}}{\frac {{p_{A}}^{2}}{m_{A}}}}={{\frac {1}{m_{B}}} \over {\frac {1}{m_{A}}}}={\frac {m_{A}}{m_{B}}}={\frac {60}{0.020}}=3000$

(b) We can compute the momentum from the kinetic energy

$K={\frac {1}{2}}{\frac {p^{2}}{m}}$

$2mK=p^{2}$

$p={\sqrt {2mK}}$

Given that $K_{B}=K_{A}$, we can compute the ratio

${\frac {p_{B}}{p_{A}}}={{\sqrt {2m_{B}K_{B}}} \over {\sqrt {2m_{A}K_{A}}}}={\sqrt {\frac {m_{B}}{m_{A}}}}={\sqrt {\frac {0.020}{60}}}=0.018$