Question

A shell of mass m at rest explodes in two parts with mass ratio 1:2. The surface has friction coefficient $\mu$. What is the ratio of stopping distance

• A. 4/1
• B. 3/2
• C. 3/1
• D. 2/1

Answer 1

The correct option is A

4/1

According to conservation of momentum ,

$\frac{m}{3}$ (V) + $\frac{2m}{3}$($V^1$) = 0 $\Rightarrow$ V = - 2 $V^1$

Friction, f = $\mu$ mg develop retardation for the masses i.e ., a = $\mu$g

$\therefore$ For lighter mass, let stopping distance be $s_1$

- = ${{V_1}^1}^2$ - 2 $\mu$g $S_2$ $\Rightarrow$ = $\frac{V^2}{8\mu g}$

$\therefore$ $\frac{S_1}{S_2}$ = $\frac{4}{1}$