When a force 1N acts on 1kg mass at rest for 1s, its final momentum is P. When 1N force acts on 1kg mass at rest through a distance of 1m, its final momentum is P'. The ratio of P to P is: 

In the first scenario, the body is initially at rest

F = 1 N,  t= 1 s

According to Newton's second law of motion,$\mathrm{F} = \frac{∆\mathrm{P}}{\mathrm{t}}$
$\Rightarrow ∆\mathrm{P} = \mathrm{F} \mathrm{x} \mathrm{t}$
P_i = 0 (As initial velocity is zero)
$\therefore$P_f = P = 1 x 1 = 1 kg m/s

In the second scenario,
F = 1 N, m= 1kg, s= 1 m

We know F= ma
$\Rightarrow \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}$= $\frac{1}{1}= 1 \mathrm{m}/{\mathrm{s}}^2$

We know that ${\mathrm{v}}^2 - {\mathrm{u}}^2= 2\mathrm{as}$
Initial velocity, u= 0
$\Rightarrow v^2= 2 x\left(1\right)x \left(1\right) = 2$
$v= \sqrt{2} m/s$
Initial momentum = 0 ($\because$Body is at rest)
Final momentum, P' = mv = $1 x \sqrt{2}$

Ratio P to P' :
$\frac{\mathrm{P}}{{\mathrm{P}}^{\text{'}}}=\frac{1}{\sqrt{2}}$