Question

A wooden block with a mass of 1000 grams is suspended by a rope from a tree. You shoot a bullet with a mass of 10 grams and a velocity of 1000 m/s. The bullet imbeds in the block causing it to swing upwards. How high in the vertical direction will the block swing?

Answer 1

Answer: 495.05 m

Step-by-step explanation:

Given:

Initial velocity of the bullet, u = 1000 m/s, mass of the bullet, m = 10 g, mass of the wooden block, M = 1000 g,

From the swinging of the wooden block just after the collision with the bullet to its maximum height, using conservation law of mechanical energy we have that,

1/2(m+M)
`v^(2)` = (m+M)gh

Where h = the vertical height of the block in the direction will the block swing, and g = acceleration due to gravity = 10
`ms^(-2)`

=> v=
`√(2gh)`

Using the conservation law of linear momentum,

mu = (m+M)v

mu = (m+M)
`√(2gh)` (where v=
`√(2gh)`)

=>h =
`(v^(2) m)/(m+M)`*
`(1)/(2g)`

h=
`(1000^(2) *10)/(10+1000)*(1)/(2*10) =495.05 m`