A bullet of mass $30 \mathrm{~g}$ moving with $700 \mathrm{~ms}^{-1}$ collides with a block of mass $4…
A bullet of mass $30 \mathrm{~g}$ moving with $700 \mathrm{~ms}^{-1}$ collides with a block of mass $4 \mathrm{~kg}$ hanging by a string of length $0.4 \mathrm{~m}$. After collision, the block rises to a height of $0.2 \mathrm{~m}$. Then, find the velocity of the bullet when it comes out of the block.
$200 \mathrm{~ms}^{-1}$
$433 \mathrm{~ms}^{-1}$
$400 \mathrm{~ms}^{-1}$
$332 \mathrm{~ms}^{-1}$
Solution
Given, mass of bullet, $m_b=30 \mathrm{~g}=0.03 \mathrm{~kg}$
Velocity of bullet, $v_b=700 \mathrm{~ms}^{-1}$
Mass of block, $m_B=4 \mathrm{~kg}$
Height upto which block rises, h = 0. 2m
Before collision, momentum of bullet $=m_b \times v_b$
$
\begin{aligned}
& =0.03 \times 700 \\
& =21 \mathrm{~kg}-\mathrm{ms}^{-1}
\end{aligned}
$
Let $v_1, v_2$ be the velocities of bullet and block after the collision.
Using conservation of momentum,
$\Rightarrow \quad 21=0.03 \times v_1+4 v_2$...(i)
Using conservation of energy for the block,
Change in KE = Work done
$\begin{aligned} & \frac{1}{2} m_B v_2^2=m_B g h \\ \Rightarrow \quad & v_2^2=2 g h \\ \Rightarrow \quad & v_2=\sqrt{2 \times 9.8 \times 0.2} \\ & =1.979 \mathrm{~ms}^{-1}\end{aligned}$
Putting the value of $v_2$ in Eq. (i), we get
$
\begin{array}{rlrl}
& & 21=0.03 \times v_1+4 \times 1.979 \\
\Rightarrow & 21=0.03 \times v_1+7.919 \\
\Rightarrow & 21-7.919=0.03 \times v_1 \\
\Rightarrow \quad & v_1=\frac{13.084}{0.03} \mathrm{~ms}^{-1} \\
& =436.13 \mathrm{~ms}^{-1} \approx 433 \mathrm{~ms}^{-1}
\end{array}
$