Respuesta :
Answer:
v = 2401.09 m/s
Explanation:
It is given that,
Mass of the bullet, [tex]m_b = 3.7\ g=0.0037\ kg[/tex]
Mass of a ballistic pendulum, [tex]m_p =4.6\ kg[/tex]
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 :
[tex]\dfrac{1}{2}(m_b+m_p)V^2=(m_b+m_p)gh[/tex]..........(1)
[tex]V=\sqrt{2gh}[/tex]
V is the speed of the bullet and the pendulum at the moment of collision.
Using the conservation of linear momentum as :
[tex]m_bv=(m_p+m_b)V[/tex]
v is the speed of the bullet before collision.
[tex]v=V\dfrac{m_b+m_p}{m_b}[/tex]
[tex]v=\dfrac{m_b+m_p}{m_b}\times \sqrt{2gh}[/tex]
[tex]v=\dfrac{0.0037+4.6}{0.0037}\times \sqrt{2\times 9.8\times 0.19}[/tex]
v = 2401.09 m/s
So, the bullet's initial speed is 2401.09 m/s. Hence, this is the required solution.
