A block of mass 1.9 kg is at rest at the edge of a table,


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.

  1. (1) $20 \mathrm{~J}$

  2. (2) $21 \mathrm{~J}$

  3. (3) $19 \mathrm{~J}$

  4. (4) $23 \mathrm{~J}$

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.

$\left(m_{1}+m_{2}\right) g h$

$=\mathrm{K}-\frac{1}{2}\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right) \mathrm{V}^{2}$

$\Rightarrow 2 \times g \times 1=K-\frac{1}{2} \times 2 \times 1^{2} \Rightarrow K=21 \mathrm{~J}$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now