Question

an 8.00 g bullet is fired into a 250 g block that is initially at rest at the edge of a 1 m high table. 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

Answer 1

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)

Answer 2

`v_o = 142.85 m/s`

Step-by-step explanation:

Time required by the block + bullet system to hit the ground is given as

`y = (1)/(2)gt^2`

now we have

`1 = (1)/(2)(9.81)t^2`

`t = 0.45 s`

now the speed of the bullet + block system is given as

`v = (d)/(t)`

`v = (2)/(0.45)`

`v = 4.43 m/s`

now by momentum conservation

`mv_o = (M + m)v`

`(8)v_o = (250 + 8)4.43`

`v_o = 142.85 m/s`