A bullet is fired at a plank of wood with a speed of $200 m{s}^{-1}$.  After passing through the plank, its speed reduces to $180 m{s}^{-1}$.  Another bullet, of the same mass and size but moving with a speed of $100 m{s}^{-1}$is fired at the same plank. What would be the speed of this bullet after passing through the plank? Assume that the resistance offered by the plank is the same for both the bullets

Let m be the mass of each bullet. Since the resistance offered by the plank is the same for the two bullets, the amount of work done by the plank is the same for the two bullets. From work- energy principle, the decrease in the kinetic energy is the same for the two bullets.

Decrease in KE of first bullet$=\frac{1}{2} m{u_1}^2 -\frac{1}{2}m{v_1}^2$

$= \frac{1}{2}m{\left(200\right)}^2-\frac{1}{2}m{\left(180\right)}^2 \left(i\right)$

If v₂ is the speed of the second bullet after passing through the plank, then 

Decrease in KE of second bullet $=\frac{1}{2} m{u_2}^2 -\frac{1}{2}m{v_2}^2$

$= \frac{1}{2}m{\left(100\right)}^2-\frac{1}{2}m{v_2}^2 \left(ii\right)$

Equating (i) and (ii) we have,

$\frac{1}{2}m{\left(200\right)}^2-\frac{1}{2}m{\left(180\right)}^2 = \frac{1}{2}m{\left(100\right)}^2-\frac{1}{2}m{v_2}^2$

which gives ${v_2}^2 = 2400 or v = 49 m{s}^{-1}$