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 :
[tex]\dfrac{1}{2}(m_1+m_2)V^2=(m_1+m_2)gh[/tex]
[tex]V=\sqrt{2gh}[/tex]....................(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 :
[tex]m_1v=(m_1+m_2)V[/tex]
[tex]V=(\dfrac{m_1+m_2}{m_1})\sqrt{2gh}[/tex]
[tex]V=(\dfrac{0.012\ kg+2.2\ kg}{0.012 kg})\sqrt{2\times 9.8\ m.s^2\times 0.1\ m}[/tex]
V = 258.06 m/s
So, the initial speed of the bullet is 258.06 m/s. Hence, this is the required solution.
