A loaded bus (mass $m_{1}$ ) and an unloaded bus (mass $m_{2}$ ) are moving with the same kinetic energy. Brakes are applied to both the buses so as to exert equal retarding force. If $S_{1}$ and $S_{2}$ are the distances covered by the two buses respectively before coming to rest, then the ratio ( $S_{1}$ / $S_{2}$ ) is

\sqrt{\frac{m_{1}}{m_{2}}}

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{(\frac{m_{1}}{m_{2}})}^{2}

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\frac{m_{1}}{m_{2}}

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Solution

The correct option is D 1
${K . E .}_{1} = {K . E .}_{2}$

$\frac{1}{2} m_{1} {u_{1}}_{1}^{2} = \frac{1}{2} m_{2} {u_{2}}_{2}^{2} \Rightarrow m_{1} {u_{1}}_{1}^{2} = m_{2} {u_{2}}_{2}^{2} - - - (1)$

Using third equation of motion, we get,

$v^{2} - u^{2} = 2 (- a) s$

The accelaration due to the force applied by the brakes is negative as it is against the direction of motion of the buses.

Also, $v = 0 m s^{- 1}$
So, the equation becomes, $u = \sqrt{2 a s}$

And equation 1 becomes,

$m_{1} (2 a_{1} S_{1}) = m_{2} (2 a_{2} S_{2}) - - - (2)$

But, the retarding force is same. So, $F_{1} = F_{2}$

$\Rightarrow \frac{a_{1}}{a_{2}} = \frac{m_{2}}{m_{1}}$

Thus, from equation 2,

$\frac{S_{1}}{S_{2}} = \frac{m_{2} a_{2}}{m_{1} a_{1}} = \frac{m_{2} \times m_{1}}{m_{1} \times m_{2}} = 1$

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A loaded bus (mass m1) and an unloaded bus (mass m2) are moving with the same kinetic energy. Brakes are applied to both the buses so as to exert equal retarding force. If S1 and S2 are the distances covered by the two buses respectively before coming to rest, then the ratio (S1/S2) is