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Question

A vehicle of mass m is moving on a rough horizontal road with momentum P. If the coefficient of friction between the tyres and the road be μ, then the stopping distance is

A
p2μmg
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B
p22μmg
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C
p2μm2g
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D
p22μm2g
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Solution

The correct option is D p22μm2g

Initial velocity u=pm
Final velocity v=0 (as the vehicle must stop)
Force of friction =μmg
(where g is acceleration due to gravity)
Acceleration due to friction =μmgm=μg
(ve sign shows that it is retardation )
Using the kinematic expression
v2=u2=2as
and inserting various values we get stopping distance s
(0)2p2m2=2(μg)s
s=p22μ2μg

1177688_943701_ans_2ab26cd79f9e4d28840a19148aafc14e.png

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