Question

A bullet of mass $20g$ hits a block of $1.98\text{kg}$ suspended from massless string of length $100\text{cm}$ and sticks to it. The bullet flies down at an angle of ${30}^{\circ }$ to the horizontal with a velocity of $200\text{m/s}$. Through what height will the block rise ?

• A. $0.15\text{m}$
• B. $0.30\text{m}$
• C. $0.45\text{m}$
• D. $0.75\text{m}$

Answer 1

The correct option is A $0.15\text{m}$

The horizontal component of the momentum of the bullet is equal to the momentum of the block with the bullet $mu\cos \alpha =(M+m)V.....\left(1\right)$
Where $V$ is the velocity of the block plus bullet embedded in it.
As the block can move as a pendiulum, the block rises till its kinetic energy is converted into potential energy , So, if the block rises upto a height h,
$\frac{1}{2}(M+m)V^2=(M+m)gh....\left(2\right)$
From (1) & (2)
$h={\left(\frac{m}{M+m}\right)}^2.\frac{u^2{\cos }^2\alpha }{2g}$
$={\left(\frac{20\times {10}^{-3}}{2}\right)}^2.(200{)}^2.\frac{{\cos }^2{30}^{\circ }}{\left(2\right)\left(10\right)}=0.15\text{m}$