Let the mass of the moving particle
=m ∴ The mass of the stationary particle
=‌×m=‌Also, let the speed of moving particle
=v Now, by the law of conservation of linear momentum
Situation after collision
∴‌‌mv+(‌×0)=mv′+‌v′ .
⇒‌‌mv=m(v′+‌) ⇒ v=v′+‌‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)Also, coefficient of restitution,
e=‌| −(v′′−v′) |
| (0−v′) |
Since, it is elastic collision,
∴‌‌e=1 ⇒‌‌1=‌⇒v′′=v+v′ ⇒‌‌v′=v′′−v‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii)From Eqs. (i) and (ii), we get
v=v′′+‌ v′=[‌]‌‌‌⋅⋅⋅⋅⋅⋅⋅(iii)Now, kinetic energy transferred to the stationary particle
=(‌ kinetic energy ‌)‌final ‌−(‌ kinetic energy ‌)‌initial ‌ ⇒‌‌Kf′−Ki′=‌×‌×v′′2‌‌‌⋅⋅⋅⋅⋅⋅⋅(iv)Also, initial kinetic energy of mass
m =‌mv2‌‌‌⋅⋅⋅⋅⋅⋅⋅(v)∴ The fraction of
KE that gets transferred to the stationary particle from the moving particle
‌=‌=‌=‌(‌)2 ‌=‌(‌)2 ‌=‌‌=‌=‌