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Normal Reaction

A lighter bod...

Question

A lighter body and heavier body both are moving with same momentum and are being applied the same retarding force.Their stopping distances and stopping times are $S_{1}, S_{2} a n d t_{1}, t_{2}$ respectively.Then,

A

S₁ > S₂

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B

S₁ < S₂

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C

t₁ > t₂

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D

t₁ < t₂

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Solution

The correct option is B
S₁ < S₂

$G i v e n m_{1} v_{1} = m_{2} v_{2} . L e t m_{1} > m_{2} \Rightarrow v_{2} < v_{1}$
$a_{1} = \frac{F}{m_{1}} a n d a_{2} = \frac{F}{m_{2}},$ where F is the applied force.
We have,
$O = v_{1} - a_{1} t_{1} \Rightarrow t_{1} = \frac{v_{1}}{a_{1}} = \frac{m_{1} v_{1}}{f}$
$O = v_{2} - a_{2} t_{2} \Rightarrow t_{2} = \frac{v_{2}}{a_{2}} = \frac{m_{2} v_{2}}{f}$
$\Rightarrow t_{1} = t_{2}$
Also, $O = v_{1}^{2} - 2 a_{1} s_{1} \Rightarrow s_{1} = \frac{v_{1}^{2}}{2 a_{1}} = (\frac{m_{1} v_{1}}{2 F}) v_{1}$
$O = v_{2}^{2} - 2 a_{2} s_{2} \Rightarrow s_{2} = \frac{v_{2}^{2}}{2 a_{2}} = (\frac{m_{2} v_{2}}{2 F}) v_{2}$
$S_{2} > s_{1} s i n c e v_{2} > v_{1}$

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