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:
[tex]\dfrac{1}{2}m_av^2 = (m_b+m_a)\Delta Hg[/tex]
where [tex]m_a[/tex] is the mass of the arrow, [tex]m_b[/tex] is the mass of the block, [tex]\Delta H[/tex] of the change in height of the block after the collision, and [tex]v[/tex] is the velocity of the arrow before it hit the block.
Solving for the velocity [tex]v[/tex], we get:
[tex]$v = \sqrt{\frac{2(m_b+m_a)\Delta Hg}{m_a} } $[/tex]
and we put in the numerical values
[tex]m_a = 0.045kg[/tex],
[tex]m_b = 1.40kg,[/tex]
[tex]\Delta H = 0.4m,[/tex]
[tex]g= 9.8m/s^2[/tex]
and simplify to get:
[tex]\boxed{ v= 15.9m/s}[/tex]
The arrow was moving at 15.9 m/s
