Question

The 49-g arrow is launched so that it hits and embeds in a 1.45 kg block. The block hangs from strings. After the arrow joins the block, they swing up so that they are 0.44 m higher than the block's starting point.
Required:
How fast was the arrow moving before it joined the block?

Answer 1

Answer:

The arrow was moving at 16.2 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:

$\dfrac{1}{2}m_av^2 = (m_b+m_a)\Delta Hg$

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 v, we get:

$v = \sqrt{\frac{2(m_b+m_a)\Delta Hg}{m_a} }$

$m_a = 0.049kg,m_b = 1.45kg,\Delta H = 0.44m,g= 9.8m/s^2$

$v = \sqrt{\frac{2(0.049 + 1.45)0.44 * 9.8}{0.049} } \\= 16.2m/s$

The arrow was moving at 16.2 m/s