Posted by **having trouble** on Sunday, November 21, 2010 at 9:06pm.

An 8.00 g bullet is fired into a 250 g block that is initially at rest at the edge of a table of 1.00 m height (Fig. P6.26). The bullet remains in the block, and after the impact the block lands 2.00 m from the bottom of the table. Determine the initial speed of the bullet.

- Physics -
**MathMate**, Sunday, November 21, 2010 at 9:18pm
So the block fell through a height of 1 m in time t, where

1.0=(1/2)gt²

or

t=√(2/9.81)

=0.452 s

During this time, the block has travelled horizontally 2m, or

horiz. velocity

=2/0.452

=4.43 m/s

This velocity, v, is the velocity of the block after impact, given by the equation of momentum before and after impact, namely

m1u1+m2u2=(m1+m2)v

or

v=(m1u1+m2u2)/(m1+m2)

where

m1=8g

m2=250g

u1= to be determined

u2=0 (block)

Solve for u1 (velocity of bullet)

- Physics -
**Damon**, Sunday, November 21, 2010 at 9:19pm
do the end first

mass m (does not matter what m is) lands 2 meters from the table

how long did it fall?

h = 0 at end

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

2 = 9.8 t^2

t = .45 seconds to fall to floor off table = time in air

distance = speed*time

2 = speed* .45

speed = 4.4 m/s horizontal

.008 v = .258 (4.4)

v = 141.9 m/s

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