Question

A bullet with mass $200\text{gm}$ and velocity $v_b\text{m/s}$hits a ballistic pendulum bob of mass $800\text{gm}$ and rises along with pendulum to a height $h=0.2\text{m}$. Determine the value of $v_b.$ Take $g=10{\text{m/s}}^2$.

• A. $100\text{m/s}$
• B. $10\text{m/s}$
• C. $8\text{m/s}$
• D. $6\text{m/s}$

Answer 1

The correct option is B $10\text{m/s}$


Since, net external force on the system of (pendulum bob+ bullet) is zero in horizontal direction.
$v_p\to$initial velocity of pendulum bob
$v_b\to$initial velocity of bullet bfore collision
$v_s$ is the velocity of system of (bullet +pendulum bob) just after collision
Applying conservation of linear momentum in horizontal direction:
$P_i=P_f$
$m_b{v}_b+m_p{v}_p=(m_b+m_p)v_s$
$\Rightarrow v_s=\frac{m_b{v}_b+m_p{v}_p}{m_b+m_p}$
$\therefore v_s=\frac{0.2\times v_b+0}{1}=0.2v_b....\left(i\right)$



Applying mechanical energy conservation from position, just after collision till system reaches maximum height i.e $(v_s=0)$:

$\text{Loss in K.E of system}=\text{Gain in P.E of system}$
$\frac{1}{2}(m_b+m_p)v_s^2=(m_b+m_p)gh...\left(ii\right)$
$\Rightarrow v_s^2=2gh$
$\Rightarrow (0.2v_b{)}^2=2\times 10\times 0.2$
$\therefore v_b=\sqrt{\frac{2\times 10\times 0.2}{0.04}}=10\text{m/s}$