An 7.47g bullet is fired into a 365g block that is initially at rest at the edge

Question

An 7.47g bullet is fired into a 365g block that is initially at rest at the edge of a frictionless table of height h = 1.12 m (see the figure below).

The bullet remains in the block, and after the impact the block lands d = 1.89 m from the bottom of the table. How much time does it take the block to reach the ground once it flies off the edge of the table?

What is the initial horizontal velocity of the block as it flies off the table? (assume this to be in the positive direction)

Determine the initial speed of the bullet.

Explanation / Answer

along vertical

initial velocity voy = 0

acceleration ay = -g

displacement y = -h

from equation of motion

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

-1.12 = 0 - (1/2)*9.8*t^2

t = 0.48 s

-------------------------

along horizontal

initial velocity vx = v

acceleration ax = 0

displacement x = d = 1.89 m

x = vx*t + (1/2)*ax*t^2

1.89 = vx*0.48

vx = 3.94 m/s

------------------

from momentum conservation

momentum before collision == momentm after collision

mu = (M+m)*v

7.47*10^-3*u = (0.365 + (7.47*10^-3))*3.94

u = 196.45 m/s <<<<-------answer

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