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

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=$\sqrt{2gh}$

Using the conservation law of linear momentum,

mu = (m+M)v

mu = (m+M)$\sqrt{2gh}$ (where v=$\sqrt{2gh}$)

=>h = $\frac{v^{2} m}{m+M}$*$\frac{1}{2g}$

h=$\frac{1000^{2} *10}{10+1000}*\frac{1}{2*10} =495.05 m$