The change in momentum of a body is $2.5 k g m s^{- 1}$ due to an impulse. How much is the force acted on it if the duration of the force acting is $0.2 s$ ?

5 N

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1.25 N

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12.5 N

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25 N

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Solution

The correct option is C 12.5 N
Given that, change in momentum, $△ (m v) = 2.5 k g m s^{- 1}$
Duration of the force acting, $t = 0.2 s e c$
We know that:
Impulse =Change in momentum = $F \times t$
$△ (m v) = F \times t$
$F = \frac{△ (m v)}{t} = \frac{2.5}{0.2}$
$\Rightarrow$ $F = 12.5 N$

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The change in momentum of a body is 2.5 kg ms−1 due to an impulse. How much is the force acted on it if the duration of the force acting is 0.2 s ?

The correct option is C 12.5 NGiven that, change in momentum,△(mv)=2.5 kg ms−1 Duration of the force acting, t=0.2 sec We know that: Impulse =Change in momentum =F×t △(mv)=F×t F=△(mv)t=2.50.2 ⇒ F=12.5N

The change in momentum of a body is $2.5 k g m s^{- 1}$ due to an impulse. How much is the force acted on it if the duration of the force acting is $0.2 s$ ?

The correct option is C 12.5 N
Given that, change in momentum, $△ (m v) = 2.5 k g m s^{- 1}$
Duration of the force acting, $t = 0.2 s e c$
We know that:
Impulse =Change in momentum = $F \times t$
$△ (m v) = F \times t$
$F = \frac{△ (m v)}{t} = \frac{2.5}{0.2}$
$\Rightarrow$ $F = 12.5 N$