Question

A bullet of mass 12 g strikes a ballistic pendulum of mass 2.2 kg. The center of mass of the pendulum rises a vertical distance of 10 cm. Assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed.

Answer 1

Answer:

The  bullet's initial speed 258.06 m/s.

Explanation:

It is given that,

It is given that,

Mass of the bullet, m₁ = 12 g = 0.012 kg

Mass of the pendulum, m₂ = 2.2 kg

The center of mass of the pendulum rises a vertical distance i.e. h = 10 cm = 0.1 m

We need to find the initial speed of the bullet. Here, the kinetic energy of the bullet gets converted to potential energy of the system as :

$\dfrac{1}{2}(m_1+m_2)V^2=(m_1+m_2)gh$

$V=\sqrt{2gh}$....................(1)

Where

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

Let v be the initial speed of the bullet. Using conservation of momentum as :

$m_1v=(m_1+m_2)V$

$V=(\dfrac{m_1+m_2}{m_1})\sqrt{2gh}$

$V=(\dfrac{0.012\ kg+2.2\ kg}{0.012 kg})\sqrt{2\times 9.8\ m.s^2\times 0.1\ m}$

V = 258.06 m/s

So, the initial speed of the bullet is 258.06 m/s. Hence, this is the required solution.