Question

A bullet of mass $50\text{gm}$ is fired from below onto a bob of mass $450\text{gm}$ of a long simple pendulum as shown in the figure. The bullet remains inside the bob and the bob rises through a height of $1.8\text{m}$ vertically upwards. Find the speed of the bullet.

• A. $60\text{m/s}$
• B. $80\text{m/s}$
• C. $40\text{m/s}$
• D. $20\text{m/s}$

Answer 1

The correct option is A $60\text{m/s}$

Let the speed of the bullet be $v$ just before collision with the bob and velocity of system after collision$=v^{\prime }$
Given $m=50\text{gm}=$mass of bullet
and $M=450\text{gm}=$ mass of bob

$T=0$ in the string after collision, Considering (bullet+bob) as a system and ignoring the external force of gravity during the very short time interval of collision.


Hence, we can apply conservation of momentum in vertical direction for colliding systems:
$\Rightarrow P_i=P_f$
$mv+(M\times 0)=(M+m)v^{\prime }$
$\therefore v^{\prime }=\frac{mv}{M+m}=\frac{50\times v}{450+50}=0.1v...\left(i\right)$

Now, taking vertically upwards as $+vey$ axis,
$S=+1.8\text{m},v_y=0,u_y=v^{\prime }{a}_y=-g{\text{m/s}}^2$
Applying kinematic equation from initial position to final position of (bullet+bob) system:
$v_y^2=u_y^2+2a_{y}S$
$\Rightarrow 0^2=v^{\prime 2}-2g\times 1.8$
$\Rightarrow (0.1v{)}^2=36$
$\therefore v=\sqrt{3600}=60\text{m/s}$