Question
There block $$A, B$$ and $$C$$ are lying on a smooth horizontal surface, as shown in the figure. $$A$$ and $$B$$ have equal masses, $$m$$ while $$C$$ has mass $$M.$$ Block $$A$$ is given an initial speed $$v$$ towards $$B$$ due to which it collides with $$B$$ perfectly inelastically. The combined mass collides with $$C,$$ also perfectly inelastically $$\frac{5}{6}th$$ of the initial kinetic energy is lost in whole process. What is value of $$M/m$$ ?
There block $$A, B$$ and $$C$$ are lying on a smooth horizontal surface, as shown in the figure. $$A$$ and $$B$$ have equal masses, $$m$$ while $$C$$ has mass $$M.$$ Block $$A$$ is given an initial speed $$v$$ towards $$B$$ due to which it collides with $$B$$ perfectly inelastically. The combined mass collides with $$C,$$ also perfectly inelastically $$\frac{5}{6}th$$ of the initial kinetic energy is lost in whole process. What is value of $$M/m$$ ?
A.
5
B.
2
C.
4
D.
3
Answer :
4
Solution :
Kinetic energy of block $$A$$
$${k_1} = \frac{1}{2}mv_0^2$$
$$\therefore $$ From principle of linear momentum conservation
$$\eqalign{ & m{v_0} = \left( {2m + M} \right){v_f} \Rightarrow {v_f}\frac{{m{v_0}}}{{2m + M}} \cr & K.{E_f} = \frac{1}{6}K.{E_i}\,\,\,\,\left( {{\text{given}}} \right) \cr & \frac{1}{2}\left( {2m + M} \right)v_f^2 = \frac{1}{6} \times \frac{1}{2}mv_0^2 \cr & 6\left( {2m + M} \right)\frac{{{m^2}v_0^2}}{{{{\left( {2m + M} \right)}^2}}} = mv_0^2 \cr & \Rightarrow 6m = 2m + M \cr & \Rightarrow 4m = M \cr & \therefore \frac{M}{m} = \frac{4}{1} \cr} $$
Kinetic energy of block $$A$$
$${k_1} = \frac{1}{2}mv_0^2$$
$$\therefore $$ From principle of linear momentum conservation
$$\eqalign{ & m{v_0} = \left( {2m + M} \right){v_f} \Rightarrow {v_f}\frac{{m{v_0}}}{{2m + M}} \cr & K.{E_f} = \frac{1}{6}K.{E_i}\,\,\,\,\left( {{\text{given}}} \right) \cr & \frac{1}{2}\left( {2m + M} \right)v_f^2 = \frac{1}{6} \times \frac{1}{2}mv_0^2 \cr & 6\left( {2m + M} \right)\frac{{{m^2}v_0^2}}{{{{\left( {2m + M} \right)}^2}}} = mv_0^2 \cr & \Rightarrow 6m = 2m + M \cr & \Rightarrow 4m = M \cr & \therefore \frac{M}{m} = \frac{4}{1} \cr} $$
