Question

an 8 gram bullet fired into a 250 gram block that is initially at rest at the edge of a table of height 1 meter.The bullet remains in the block, and after the impact the block lands 2 meter from the bottom of the table. determinethe initial speed of the bullet.

Answer 1

Answer:

141.9 m/s

Explanation:

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)

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