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

Answer:

$v_o = 142.85 m/s$

Explanation:

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

$y = \frac{1}{2}gt^2$

now we have

$1 = \frac{1}{2}(9.81)t^2$

$t = 0.45 s$

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

$v = \frac{d}{t}$

$v = \frac{2}{0.45}$

$v = 4.43 m/s$

now by momentum conservation

$mv_o = (M + m)v$

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

$v_o = 142.85 m/s$