Question

A bullet of mass $20\text{g}$ strikes a pendulum of mass $5\text{kg}$ which is initially at rest. The center of mass of the pendulum rises a vertical distance of $10\text{cm}$. If the bullet gets embedded into the pendulum, calculate the initial speed of the bullet.

• A. $354.97\text{m/s}$
• B. $340.97\text{m/s}$
• C. $250.41\text{m/s}$
• D. $251\text{m/s}$

Answer 1

The correct option is A $354.97\text{m/s}$

Given: Mass of bullet, $m_1=0.02\text{kg}$
Mass of pendulum, $m_2=5\text{kg}$
Center of mass of pendulum rises to a height $h=0.1\text{m}$

Let speed of bullet $=u_1$
Initially pendulum is at rest, so $u_2=0$
Let the common velocity of bullet and pendulum just after the collision $=v$
$v=\frac{m_1{u}_1+m_2{u}_2}{m_1+m_2}=\frac{0.02\times u_1+5\times 0}{5+0.02}$
i.e $v=\frac{u_1}{251}$ ... (1)

After bullet gets embedded, applying conservation of energy
$(m_1+m_2)gh=\frac{1}{2}(m_1+m_2)v^2$
$v=\sqrt{2gh}=\sqrt{2\times 10\times 0.1}$
$v=\sqrt{2}\text{m/s}$
From eq.(1)
$\sqrt{2}=\frac{u_1}{251}$
or $u_1=354.97\text{m/s}$