Question

A bullet of mass $0.050kg$ travelling at $300m{s}^{-1}$ penetrates inside a wooden block of $3kg$. The block now embedded with the bullet start moving together. What will be the combined velocity of the block and the bullet?

• A. $28.2m{s}^{-1}$
• B. $8.2m{s}^{-1}$
• C. $18.2m{s}^{-1}$
• D. $38.4m{s}^{-1}$

Answer 1

The correct option is D $38.4m{s}^{-1}$

Given:
mass of the bullet, $m_b=0.050kg$,
velocity of the bullet, $v_b=300m{s}^{-1}$ and
mass of the block, $m_w=3kg$.
Let the combined velocity of both bullet and the block be $v$.
According to the law of conservation of energy,
initial total energy of the system = final total energy of the system.
Here, system includes both bullet and the block.
$\therefore$ $\frac{1}{2}{m}_b{v}_b^2=\frac{1}{2}(m_b+m_w)v^2$
$\Rightarrow \frac{1}{2}\left(0.05\right)(300{)}^2=\frac{1}{2}(3+0.05)v^2$
$\Rightarrow 4500=\left(3.05\right)v^2$
$\Rightarrow v^2=1475.4$
$\Rightarrow v=38.4m{s}^{-1}$.