A block of mass $1.9 \mathrm{~kg}$ is at rest at the edge of a table, of height $1 \mathrm{~m}$. A bullet of mass $0.1 \mathrm{~kg}$ collides with the block and sticks to it. If the velocity of the bullet is $20 \mathrm{~m} / \mathrm{s}$ in the horizontal direction just before the collision then the kinetic energy just before the combined system strikes the floor, is [Take $g=10 \mathrm{~m} / \mathrm{s}^{2}$. Assume there is no rotational motion and losss of energy after the collision is negligiable.
Correct Option: , 2
(2) Given,
Mass of block, $m_{1}=1.9 \mathrm{~kg}$
Mass of bullet, $m_{2}=0.1 \mathrm{~kg}$
Velocity of bullet, $v_{2}=20 \mathrm{~m} / \mathrm{s}$
Let $v$ be the velocity of the combined system. It is an
inelastic collision.
Using conservation of linear momentum
$m_{1} \times 0+m_{2} \times v_{2}=\left(m_{1}+m_{2}\right) v$
$\Rightarrow 0.1 \times 20=(0.1+1.9) \times v$
$\Rightarrow v=1 \mathrm{~m} / \mathrm{s}$
Using work energy theorem
Work done $=$ Change in Kinetic energy
Let $K$ be the Kinetic energy of combined system.
$\left(m_{1}+m_{2}\right) g h$
$=\mathrm{K}-\frac{1}{2}\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right) \mathrm{V}^{2}$
$\Rightarrow 2 \times g \times 1=K-\frac{1}{2} \times 2 \times 1^{2} \Rightarrow K=21 \mathrm{~J}$