Question

A 15 g bullet is fired horizontally into a block of wood

withmass 2.5 kg and embedded in the block. Initially the block
ofwood hangs vertically and the impact causes the block to swing
sothat its center of mass rises 15cm. Find the velocity of
thebullet just before the impact.

Answer 1

290.4 m/s

Step-by-step explanation:

Let g = 10m/s2

15g = 0.015 kg

After getting momentum from the bullet, the block-bullet system has a kinetic energy that raises itselft upward, this kinetic energy is converted to potential energy as the mass stops at 15cm = 0.15m high:

`E_p = E_k`

`mgh = mv^2/2`

where m is the system mass and h is the vertical distance traveled, v is the system initial velocity, which is what we are looking for. We can divide both sides of the equation by m:

`v^2 = 2gh = 2*10*0.15 = 3`

`v = √(3) = 1.732 m/s`

At the impact, according to the conservation of momentum:

`m_uv_u + m_bv_b = (m_u + m_b)v`

where
`m_u` = 0.015 kg is the mass of the bullet,
`v_u` is the velocity of the bullet before the impact,
`m_b` = 2.5 kg is the mass of the block,
`v_b` = 0 is the velocity of the block before the impact. v = 1.732 m/s is the velocity of the system after the impact.

`0.015v_u + 0 = (0.015 + 2.5)1.732`

`v_u = ((0.015 + 2.5)1.732)/(0.015) = 290.4 m/s`