Question

A bullet of mass m = 8.00 g is fired into a block of mass M = 250 g that is initially at rest at the edge of a table of height h = 1.00 m. The bullet remains in the block, and after the impact the block lands d = 2.00 m from the bottom of the table. Determine the initial speed of the bullet.

Answer 1

after collsion

for the bullet + block system

along vertical

initial velocity voy = 0

acceleration ay = -9.8 m/s^2

displacement y = -1 m

y = voy*t + (1/2)*ay*t^2

-1 = - (1/2)*9.8*t^2

t = 0.452 s

along horizontal

x = v*t

v = x/t = 2/0.452 = 4.42 m/s

speed of bullet + block , v = 4.42 m/s

from momentum conservation

momentum before collsion = momentum after collision

m*vo = (M + m)*v

vo = (M + m)*v/m

vo = (0.25 + 0.008)*4.42/0.008

vo = 142.545 m/s <<<<------answer