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

v = 2401.09 m/s

Step-by-step 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 :

`(1)/(2)(m_b+m_p)V^2=(m_b+m_p)gh`..........(1)

`V=√(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(m_b+m_p)/(m_b)`

`v=(m_b+m_p)/(m_b)* √(2gh)`

`v=(0.0037+4.6)/(0.0037)* √(2* 9.8* 0.19)`

v = 2401.09 m/s

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