Question

A bullet of mass $50\text{g}$ is fired from below into the bob of mass $450\text{g}$ of a long simple pendulum, as shown in the figure. The bullet remains inside the bob and bob rises through a height of $1.8\text{m}$. Find the speed of the bullet.
[Take $g=10{\text{ms}}^{-2}$]

• A. $60{\text{ms}}^{-1}$
• B. $\sqrt{180}{\text{ms}}^{-1}$
• C. $36.6{\text{ms}}^{-1}$
• D. $60{\text{ms}}^{-1}$

Answer 1

The correct option is D $60{\text{ms}}^{-1}$

Given, $m_{bob}=0.45\text{kg};m_{bullet}=0.05\text{kg}$
$h=1.8\text{m}$


In collision, there is no external force involved. Thus, linear momentum of the system will be conserved.

Let initial velocity of the bullet and final velocity of (bob+bullet ) be $v$ and $V$ respectively.

$\therefore P_i=P_f$

$(0.05\times v)+(0.45\times 0)=(0.05+0.450)\times V$

$0.05v=0.5V$

$\therefore V=\frac{v}{10}$

After collision, the system of (bob+pullet) rises to a height of $1.8\text{m}$.

Using kinematic equations,

$v^2=u^2+2as$

Here,
$v=0;u=V;a=-g;h=1.8\text{m}$

$(0{)}^2=(V{)}^2+\left[2\right(-10)\times (1.8\left)\right]$

$\Rightarrow V^2=2\times 10\times 1.8$

$\Rightarrow {\left(\frac{v}{10}\right)}^2=36$

$\therefore v=60{\text{ms}}^{-1}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{D}\right)$ is the correct answer.