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

To find the initial speed of the arrow, we use the conservation of energy principle, equating the potential energy at its highest point to the kinetic energy it had before embedding in the block and solving for the arrow's speed.

Step-by-step explanation:

The student's question is asking to find out the speed of an arrow before it embeds itself in a block and swings upward. To solve this physics problem, we can use the principle of conservation of energy. Since external forces like air resistance and friction are neglected, mechanical energy is conserved.

The potential energy gained by the block-arrow system at the highest point of its swing (when it is 0.44 m higher) is equal to the kinetic energy of the arrow just before impact. The potential energy (PE) at the height can be calculated using the formula PE = m*g*h, where m is the combined mass of the arrow and block, g is the acceleration due to gravity (9.8 m/s2), and h is the height (0.44 m). The kinetic energy (KE) just before the impact can be calculated with KE = 0.5*m*v2, where v is the speed we want to find.

Setting the two energies equal to each other (since KE_initial = PE_final) and solving for v, we can find the initial speed of the arrow before the collision.

Answer 2

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

Step-by-step 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;

`P_(initial) = P_(final)\\\\mv_(arrow) + mv_(block) = (m_1 + m_2)V\\\\0.049(v) + 1.45(0) = (0.049 + 1.45)2.937\\\\0.049v = 4.4026\\\\v = (4.4026)/(0.049) \\\\v = 89.85 \ m/s`

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