Question

A $1.25-kg$ wooden block rests on a table over a large hole as in Figure $P9.71$. A $5.00-g$ bullet with an initial velocity $v_i$ is fired upward into the bottom of the block and remains in the block after the collision. The block and bullet rise to a maximum height of $22.0cm$.
Describe how you would find the initial velocity of the bullet using ideas you have learned in this chapter.

Answer 1

Given,

Mass of wooden block is $1.25kg$

mass of bullet is $5g=0.0$

maximum height is $22cm=0.22m$

We know that final velocity is

$v_b=\sqrt{2gh}$

Substitute all value in above equation

$v_b=\sqrt{2\times 9.8\times 0.22}{v}_b=2.08m/s$

as this collision is inelastic collision therefore momentum will be conserved before and after collision.

$m_1{v}_i+m_2{v}_2=(m_1+m_2)v_b$

Substitute all value in above equation

$0.005\times v_i+1.25\times 0=(0.005+1.25)2.08v_i=\frac{2.61}{0.005}{v}_i=522.08m/s$

Hence,

Initial velocity of the bullet is $522.08m/s$