Question

A bullet weighing 2.5 g is shot at 350 m/s into a 5 kg wooden block suspended from a 10 m long string. The bullet strikes and is embedded in the block. The block is initially at rest. How high above its initial position does the block swing?

Answer 1

h = 1.56 mm

Step-by-step explanation:

given,

mass of bullet (m)= 2.5 g

speed of bullet (u)= 350 m/s

mass of wooden plank (M)= 5 Kg

length of string = 10 m

initial velocity of block = 0 m/s

height of block above original position = ?

using conservation of momentum

m u + M u = (m+ M)v

0.0025 x 350 + 0 = (5 +0.0025)v

v = 0.175 m/s

from the conservation of energy

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

`(1)/(2)v^2 =gh`

`h=(v^2)/(2g)`

`h=(0.175^2)/(2* 9.8)`

h = 1.56 mm