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Two trolleys of mass m and 3m are connected by a spring. They are compressed and released, they moves off in opposite direction and comes to rest after covering distances $s_{1} and s_{2}$ respectively. If the ratio of distances $s_{1} : s_{2} is 3 : 1$ then coefficient of the friction will be in the ratio of ........

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3 : 1

1 : 3

9 : 1

1 : 9

answer is C.

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Detailed Solution

Law of conservation of energy

$m_{1} = M$

$m_{2} = 3 M$

According L. C. M $m_{1} v_{1} = m_{2} v_{2}$

$\frac{m_{1}}{m_{2}} = \frac{v_{2}}{v_{1}}$

Now work – energy principle

$\frac{1}{2} m_{1} v_{1}^{2} = F {.S}_{1} = μ m_{1} {gs}_{1}$

$\frac{1}{2} m_{2} v_{2}^{2} = F {.S}_{2} = μ m_{2} {gs}_{2}$

$\frac{s_{1}}{s_{2}} = {(\frac{v_{1}}{v_{2}})}^{2} = {(\frac{m_{2}}{m_{1}})}^{2}$

$\frac{s_{1}}{s_{2}} = {(\frac{3m}{m})}^{2} = \frac{9}{1}$

$\frac{s_{1}}{s_{2}} = \frac{3}{1}$

$\frac{s_{1}}{s_{2}} = {(\frac{μ_{1}}{μ_{2}})}^{2} \times \frac{m_{1}}{m_{2}}$

$\frac{3}{1} = {(\frac{μ_{1}}{μ_{2}})}^{2} \times \frac{m}{3m}$

${(\frac{μ_{1}}{μ_{2}})}^{2} = \frac{9}{1}$

$\Rightarrow \frac{μ_{1}}{μ_{2}} = \frac{3}{1}$

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