Question

A block of mass $1.9\text{kg}$ is at rest at the edge of a table, of height $1\text{m}$. A bullet of mass $0.1\text{kg}$ collides with the block and sticks to it. If the velocity of the bullet is $20{\text{ms}}^{-1}$ in the horizontal direction just before the collision then the kinetic energy, just before the combined system strikes the floor, is [Take $g=10{\text{ms}}^{-2}$]. Assume there is no rotational motion and losses of energy after the collision is negligible.

• A. $20\text{J}$
• B. $21\text{J}$
• C. $19\text{J}$
• D. $23\text{J}$

Answer 1

The correct option is B $21\text{J}$

Given,
Mass of block, $m_1=1.9\text{kg}$
Mass of bullet, $m_2=0.1\text{kg}$
Velocity of bullet, $v_2=20{\text{ms}}^{-1}$

It is an inelastic collision.

Let $V$ be the velocity of the combined system.

Using conservation of linear momentum
$m_1\times 0+m_2\times v_2=(m_1+m_2)V$

$\Rightarrow 0.1\times 20=(0.1+1.9)\times V$

$\Rightarrow V=1{\text{ms}}^{-1}$

Let $K$ be the Kinetic energy of the combined system jus before striking the ground,

Using work energy theorem
$\text{Work done = Change in Kinetic energy}$

$(m_1+m_2)gh=K-\frac{1}{2}(m_1+m_2)V^2$

$\Rightarrow 2\times 10\times 1=K-(\frac{1}{2}\times 2\times 1^2)$

$\Rightarrow K=21\text{J}$

Hence, option $\left(B\right)$ is correct.