Question

As shown in the diagram, a bullet of mass 7.0 g strikes a block of wood of mass 2.1 kg. The block of wood is suspended by a string of length 2.0 m, forming a pendulum. The bullet lodges in the wood, and together they swing upward a distance of 0.40 m from the original position of the wood block. What was the speed of the bullet just before it struck the wooden block?

Answer 1

bullet mass + wood mass = 2.107 kg

potential energy if up 0.4 meter:

I assume this is along the circumference
so
.4 = theta * 2 =
theta = .2 radians = 11.5 degrees

distance up from bottom = 2 (1-cos theta)
= .0803 meters

so energy at top = m g h =2.107*9.81*.0803 = 1.66 Joules

that is its kinetic energy at the bottom
(1/2) m v^2 = 1.66
.5*2.107*v^2 = 1.66
v = 1.26 m/s

momentum of block with bullet
= 2.107 * 1.26 = 2.64 kg m/s

that was the bullet momentum before the crash
.007 v = 2.64
v = 378 m/s

Answer 2

no