Question

A bullet of mass 50 g moving with velocity 200 m/s penetrates a wooden block of mass 5 kg suspended through a string from rigid support and comes to rest inside it. If length of the string is 10 m find the angular deflection of the string.

Answer 1

Initially only the bullet is moving with velocity $200m/s$ ,later the block and velocity move with some velocity v which can be found from conservation law of momentum:

$0.05\times 200+5\times 0=\left(5.05\right)\times v$

$10=5.05v$

$v=\frac{10}{5.05}=1.98m/s$

Due to this velocity the block has gained kinetic energy ,therefore because of energy conservation it will reach till the point its potential energy equals this kinetic energy.

Therefore,bevause of energy conservation ,

Initial energy=Final energy

$0=KE+PE$

$KE=PE$

$\frac{1}{2}m{v}^2=mgy$

$y=\frac{v^2}{2g}=0.2m$

Also, $y=l(1-\cos \theta )$

$\cos \theta =1-\frac{y}{l}=\frac{l-y}{l}$

$=\frac{10-0.2}{10}=0.98$

$\theta ={\cos }^{-1}\left(0.98\right)={11.48}^{\circ }$