Question

A bullet of mass 3.7 g strikes a ballistic pendulum of mass 4.6 kg. The center of mass of the pendulum rises a vertical distance of 19 cm. Assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed.

Answer 1

Answer:

v = 2401.09 m/s

Explanation:

It is given that,

Mass of the bullet, $m_b = 3.7\ g=0.0037\ kg$

Mass of a ballistic pendulum, $m_p =4.6\ kg$

Vertical distance, h = 19 cm = 0.19 m

Here, the conservation of energy follows as the kinetic energy of bullet is converted to the potential energy as :

$\dfrac{1}{2}(m_b+m_p)V^2=(m_b+m_p)gh$..........(1)

$V=\sqrt{2gh}$

V is the speed of the bullet and the pendulum at the moment of collision.

Using the conservation of linear momentum as :

$m_bv=(m_p+m_b)V$

v is the speed of the bullet before collision.

$v=V\dfrac{m_b+m_p}{m_b}$

$v=\dfrac{m_b+m_p}{m_b}\times \sqrt{2gh}$

$v=\dfrac{0.0037+4.6}{0.0037}\times \sqrt{2\times 9.8\times 0.19}$

v = 2401.09 m/s

So, the bullet's initial speed is 2401.09 m/s. Hence, this is the required solution.