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A body $A$ is at rest and body $B$ is moving at uniform velocity of $40 m s^{- 1}$ . Force $F_{1}$ acts on $A$ and $F_{2}$ acts on $B$ for $10$ seconds. The velocity of $A$ and $B$ now is $40 m s^{- 1}$ and $80 m s^{- 1}$ respectively. If $A$ and $B$ are of equal mass, then :

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$F_{1} = F_{2}$

$F_{1} = 2 F_{2}$

$F_{2} = 2 F_{1}$

$F_{1} = \frac{F_{2}}{4}$

answer is A.

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Detailed Solution

$F ∆ t = ∆ P F o r A : F_{1} \times 10 = m (40 - 0) F_{1} \times 10 = 40 m F o r B : F_{2} \times 10 = 40 m F_{1} = F_{2}$

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