Answer:
The correct answer is option 'c': Smaller stone rebounds while as larger stone remains stationary.
Explanation:
Let the velocity and the mass of the smaller stone be 'm' and 'v' respectively
and the mass of big rock be 'M'
Initial momentum of the system equals
[tex]p_i=mv+0=mv[/tex]
Now let after the collision the small stone move with a velocity v' and the big roch move with a velocity V'
Thus the final momentum of the system is
[tex]p_f=mv'+MV'[/tex]
Equating initial and the final momenta we get
[tex]mv=mv'+MV'\\\\m(v-v')=MV'.....i[/tex]
Now since the surface is frictionless thus the energy is also conserved thus
[tex]E_i=\frac{1}{2}mv^2[/tex]
Similarly the final energy becomes
[tex]E_f=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2[/tex]\
Equating initial and final energies we get
[tex]\frac{1}{2}mv^2=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\\\\mv^2=mv'^2+MV'^2\\\\m(v^2-v'^2)=MV'^2\\\\m(v-v')(v+v')=MV'^2......(ii)[/tex]
Solving i and ii we get
[tex]v+v'=V'[/tex]
Using this in equation i we get
[tex]v'=\frac{v(m-M)}{(M-m)}=-v[/tex]
Thus putting v = -v' in equation i we get V' = 0
This implies Smaller stone rebounds while as larger stone remains stationary.
