Question: How fast was the arrow moving before it joined the block?
Answer:
The arrow was moving at 15.9 m/s.
Explanation:
The law of conservation of energy says that the kinetic energy of the arrow must be converted into the potential energy of the block and arrow after it they join:

where
is the mass of the arrow,
is the mass of the block,
of the change in height of the block after the collision, and
is the velocity of the arrow before it hit the block.
Solving for the velocity
, we get:

and we put in the numerical values
,



and simplify to get:

The arrow was moving at 15.9 m/s
Answer:
-0.7 m/sec
Explanation:
Mass of first block = m1 =3.0 kg
Mass of second block = m2= 5.0 kg
Velocity of first block = V1= 1.2 m/s
Velocity of second block = V2 = ?
Momentum of Center of mass MVcom is sum of both blocks momentum and is given by
MVcom= m1v1+m2v2
Where
M= mass of center of mass
Vcom= Velocity of center of mass=0 m/s (because center of mass is at rest , so Vcom = 0 m.sec)
Putting values, we get;
0= 3×1.2+5v2
==> v2= 3.6/5= - 0.7 m/s
-ve sign indicates that block 2 is moving in opposite direction of block 1