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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

$KE = \frac{P^{2}}{2 m} \Rightarrow \frac{P_{1}}{P_{2}} = \sqrt{\frac{m_{1}}{m_{2}}} \sqrt{\frac{{KE}_{1}}{{KE}_{2}}} = (2) \sqrt{\frac{3}{2}} = \sqrt{6}$

$F = \frac{dp}{dt}$

$\frac{t_{1}}{t_{2}} = (\frac{P_{1}}{P_{2}}) (\frac{F_{2}}{F_{1}}) = (\sqrt{6}) (\frac{2}{1}) = 2 \sqrt{6} : 1$

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