91Ó°ÊÓ

Mass of bullet = 10 gm = 0.01kg

speed of bullet = 400 m/sec

mass of door = 10 kg.

width of door = 1 m

bullet hits at edge of the door

Use conservation of momentum.

Initially momentum of door =0

momentum of bullet is total momentum

Li= mv_br_{b ..................................................}.(1)

where m = mass of bullet, v_b = velocity of bullet, r_b =1 m

Substitute the value we get

Li= (0.01 kg) (400 m/s) (1 m) = 4kgm²/s

Assume bullet get stuck in the door and traveling with door.

Momentum after strike is

$$\begin{gathered} {\mathrm{L}}_f=\left(I_f\right)\mathrm{Ó\neg } \\ {\mathrm{L}}_f=\left(I_d+I_b\right)\mathrm{Ó\neg }................................\left(2\right) \end{gathered}$$

Now find the inertia

moment of inertia of the door $=I_d=\frac{1}{3}M{r}^2$
and moment of inertia of the bullet $=I_b=m{r}^2$

Substitute the values in equation(2), we get

$$\begin{gathered} L_f=\left(\frac{1}{3}M{r}^2+m{r}^2\right)\mathrm{Ó\neg } \\ {L}_f=\left(\frac{1}{3}×10\mathrm{kg}×(1\mathrm{m}{)}^2+(0.01\mathrm{kg}\left)\right(1\mathrm{m}{)}^2\right)\mathrm{Ó\neg } \\ =3.34kg{m}^2\mathrm{Ó\neg }………………\left(2a\right) \end{gathered}$$

Equate initial momentum and final momentum, we get

3.34 Ó¬ = 4kgm²/s

Ó¬ = 1.197 rad/s = 1.2 rad/s