Question

A bullet is fired horizontally into an initially stationary block of wood suspended by a string and remains embedded in the block. The bullet’s mass is m = 0.0075 kg, while that of the block is M = 0.95 kg. After the collision the block/bullet system swings and reaches a maximum height of h = 1.1 m above its initial height. Neglect air resistance.
Calculate the ratio, expressed as a percent, of the kinetic energy of the block/bullet system immediately after the collision to the initial kinetic energy of the bullet.

KEf/KEi(%) =

Answer 1

The ratio of the kinetic energy of the block/bullet system immediately after the collision to the initial kinetic energy of the bullet is 0.78 %.

Final velocity of the block/bullet system

Apply the principle of conservation of energy to determine the final velocity of the block/bullet system.

K.E = P.E

¹/₂mv² = mgh

¹/₂v² = gh

v² = 2gh

v = √2gh

where;

• h is the maximum height reached by the system
• v is the initial velocity of the system

v = √(2 x 9.8 x 1.1)

v = 4.64 m/s

Initial velocity of the bullet

Apply the principle of conservation of linear momentum.

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

• u₁ is the initial velocity of the bullet
• u₂ is the initial velocity of the block
• v is the final velocity after collision
• m₁ is mass bullet
• m₂ is mass of block

(0.0075)u₁ + (0.95)(0) = 4.64(0.0075 + 0.95)

0.0075u₁ = 4.4428

u₁ = 4.4428/0.0075

u₁ = 592.37 m/s

Initial kinetic energy of the bullet

K.Ei = ¹/₂m₁u₁²

K.Ei = ¹/₂(0.0075)(592.37)²

K.Ei = 1,315.88 J

Final kinetic energy of the block/bullet system

K.Ef = ¹/₂(m₁ + m₂)v²

K.Ef = ¹/₂(0.0075 + 0.95)(4.64)²

K.Ef = 10.31 J

Ratio of final kinetic energy to initial kinetic energy

= K.Ef/K.Ei x 100%

= (10.31 / 1,315.88) x 100%

= 0.78 %

Learn more about kinetic energy here: brainly.com/question/25959744

#SPJ1