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 $22.0cm$.
Calculate the initial velocity of the bullet from the information provided.

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 Momentum is conserved before and after collision therefore,

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

initial velocity of wooden block is zero therefore substitute all value in above equation

$0.005\times v_i=(0.005+1.25)v_b{v}_i=251v_{b}m/s$

$v_i=251\times 2.08v_i=522.08m/s$

Hence,

initial velocity of bullet is $522.08m/s$