Question

an object of mass initially hangs from a rod of length and negligible mass that can pivot freely in the vertical plane, as shown in the figure above. a projectile of mass is moving to the right at speed when it strikes the hanging object and then drops straight down. the hanging object then swings upward and comes to rest at the top of a half circle as shown. what is the initial speed of the projectile ?

Answer 1

The initial speed of the projectile C)
`(2M_(2)√(gL) )/(M_(2) )` .Therefore, C)
`(2M_(2)√(gL) )/(M_(2) )` is correct.

The initial energy of the projectile is 1/2 m2v2 .

After the collision, the energy is converted into the kinetic energy of the hanging object and the potential energy of the hanging object at the top of the half circle.

The kinetic energy of the hanging object is

1/2 m(lω) 2 , where ω is the angular velocity of the hanging object.

The potential energy of the hanging object at the top of the half circle is mgl.

Therefore, we have the following equation:

$\frac{1}{2} m_2 v^2 = \frac{1}{2} m (l \omega)^2 + m g l$

We can solve this equation for v to get:

$v = \sqrt{\frac{2mgl}{m_2}}$

it strikes the hanging object and then drops straight down. the hanging object then swings upward and comes to rest at the top of a half circle as

`(2M_(2)√(gL) )/(M_(2) )`.