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:

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

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

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

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