Answer to Question #96803 in Mechanics | Relativity for srividya

Question

a proton undergo a head on elastic collision with a particle of unknown mass which is initially at rest and rebounds with 4/9 of its initial kinetic energy. Calculate the ratio of the unknown mass with respect to the mass of proton.

Answer 1

From the conservation of momentum:

$mv=-mv'+Mu$

We have:

$0.5mv'^2=\frac{4}{9}(0.5mv^2)\to v'=\frac{2}{3}v$

So,

$mv=-m\frac{2}{3}v+Mu\to u=v\frac{5}{3}\frac{m}{M}$

From the conservation of energy:

$0.5Mu^2=0.5mv^2-0.5mv'^2=\frac{5}{9}(0.5mv^2)$

$M\left(v\frac{5}{3}\frac{m}{M}\right)^2=\frac{5}{9}(mv^2)$

$\left(\frac{5}{3}\right)^2=\frac{5}{9}\frac{M}{m}$

$\frac{M}{m}=5$