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?

the initial speed of the arrow before joining the block is 89.85 m/s

Explanation:

Given;

mass of the arrow, m₁ = 49 g = 0.049 kg

mass of block, m₂ = 1.45 kg

height reached by the arrow and the block, h = 0.44 m

The gravitational potential energy of the block and arrow system;

P.E = mgh

P.E = (1.45 + 0.049) x 9.8 x 0.44

P.E = 6.464 J

The final velocity of the system after collision is calculated as;

K.E = ¹/₂mv²

6.464 = ¹/₂(1.45 + 0.049)v²

6.464 = 0.7495v²

v² = 6.464 / 0.7495

v² = 8.6244

v = √8.6244

v = 2.937 m/s

Apply principle of conservation of linear momentum to determine the initial speed of the arrow;

Therefore, the initial speed of the arrow before joining the block is 89.85 m/s

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