A lorry and a car of mass ratio 4:1 are moving with KE in the ratio $3 : 2$ on a horizontal road. Now brakes are applied and braking forces produced are in the ratio $1 : 2$ . Then the ratio of stopping timings of lorry and car is

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1 : 1

1 : 3

2 \sqrt{6} : 1

6 \sqrt{2} : 1

answer is C.

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Detailed Solution

W = F .S = {KE}_{f} - {KE}_{i}

$\Rightarrow F_{1} (S_{1}) = 0 - {KE}_{i}$

$\Rightarrow \frac{F_{1}}{F_{2}} (\frac{a_{1} t_{1}^{2}}{a_{2} t_{2}^{2}}) = \frac{{KE}_{1}}{{KE}_{2}}$

$\Rightarrow {(\frac{F_{1}}{F_{2}})}^{2} \times \frac{m_{2}}{m_{1}} \times \frac{t_{1}^{2}}{t_{2}^{2}} = \frac{{KE}_{1}}{{KE}_{2}}$

$\Rightarrow {(\frac{1}{2})}^{2} \times \frac{1}{4} \times \frac{t_{1}^{2}}{t_{2}^{2}} = \frac{3}{2}$

$\Rightarrow \frac{t_{1}^{2}}{t_{2}^{2}} = \frac{24}{1}$

$\Rightarrow \frac{t_{1}}{t_{2}} = \frac{2 \sqrt{6}}{1}$

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