An oxygen molecule of mass 5.32 x 10^-26 kg, moving at 6.0 x 10^4 m/s strikes a wall at 90 degrees and rebounds without loss of speed. If the duration of the impact is 10^-9 seconds, calculate..?

(a) the change in momentum of the molecule (b) the force exerted by the wall on the molecule (c) the force exerted by the molecule on the wall

Oct 23, 2016

Given

$m \to \text{Mass of the molecule"=5.32xx19^"-26} k g$

$u \to \text{Velocity of the molecule before}$
$\text{ bombardment} = - 6.0 \times {10}^{4} \frac{m}{s}$

$v \to \text{Velocity of the molecule after}$
$\text{ bombardment} = + 6.0 \times {10}^{4} \frac{m}{s}$

$t \to \text{Duration of impact} = {10}^{-} 9 s$

(a) So the change in momentum
$m v - \text{mu"=2xx5.32xx10^"-26} \times 6 \times {10}^{4}$

$= 63.84 \times {10}^{\text{-22"" kgm/s}}$

(b) The force exerted by the wall on the molecule is
$= \text{change in momentum"/"time}$

$= \frac{63.84 \times {10}^{\text{-22"" kgm/s}}}{{10}^{-} 9 s}$
$= 63.84 \times {10}^{-} 13 N$

(c) The molecule will exerts same amount of reactionary force on the wall in oppsite direction.

So this force will be $= - 63.84 \times {10}^{-} 13 N$