Question

The masses of Blocks $\text{A}$ , $\text{B}$ and $\text{C}$ are in the ratio $1:1:4$ . Block $\text{A}$ is given an initial velocity $v_0$ towards $\text{B}$ due to which it collides with $\text{B}$ perfectly inelastically. The combined mass collide with $\text{C}$ , also perfectly inelastically . If the loss in kinetic energy is $n$ times the initial kinetic energy , then the value of $n$ is

• A. $\frac{3}{4}$
• B. $\frac{4}{3}$
• C. $\frac{5}{6}$
• D. $\frac{6}{5}$

Answer 1

The correct option is C $\frac{5}{6}$

Given that,

$m_A:m_B:m_C=1:1:4$

Let the mass of Block $A$ be $m$

$\therefore m_A=m;m_B=m;m_C=4m$

Initial kinetic energy of block $(K{)}_i=\frac{1}{2}m{v}^2=\frac{1}{2}m{v}_0^2$

Using conservation of linear momentum,

$P_i=P_f$

$\Rightarrow m{v}_0=(m+m+4m)v$

$\Rightarrow v=\frac{v_0}{6}....\left(1\right)$

$\therefore$ Final $KE=\frac{1}{2}(m_A+m_B+m_C)v^2$

Substituting $\left(1\right)$ in the above equation,

$K_f=\frac{1}{2}\times 6m\times \frac{v_0^2}{36}\Rightarrow K_f=\frac{m{v}_0^2}{12}$

$\therefore$ Loss in kinetic energy =$K_i-K_f=\frac{m{v}_0^2}{2}-\frac{m{v}_0^2}{12}=\frac{5m{v}_0^2}{12}=\frac{5}{6}(KE{)}_i$

From the data given in the question,

$K_i-K_f=n{K}_i$

$\Rightarrow n$ = $\frac{5}{6}$

Hence, option (c) is the correct answer.