Question

A 1.05-kg block of wood sits at the edge of a table, 0.770m above the floor. A 1.05

Answer 1

Conservation of momentum requires that the total momentum of both the block and bullet be the same both before and after the collision. Since the block was stationary before the collision it had no momentum so:

mbullet*vbullet0 = (mbullet+mblock)*vafter, so

vafter = mbullet/(mbullet+mblock)*vbullet

= (0.0105kg)/(1.05kg+.0105kg)*(740m/s) =7.327 m/s.

Since the block starts 0.77m above the floor, you can calculate the time it takes to hit the floor by:

d=1/2*a*t^2 => t=(2*d/a)^0.5

= (2*(0.77m)/(9.8m/s^2))^0.5 =0.3964 seconds

The horizontal distance the block can travel in that time is given by:

d= v*t = (7.327 m/s)*(0.378sec) = 2.904428 m