Two bullets have masses of 2.5 g and 6.5 g, respectively. Each is fired with a speed of 42.0m/s. a) What is the kinetic energy of the first bullet? Answer in units of Jb) What is the kinetic energy of the second bullet? Answer in units of Jc) What is the ratio K2/K1 of their kinetic energies?

(a) K₁ = 2.205 J.

(b) K₂ = 5.733 J.

(c) K₂/K₁ = 2.6.

Explanation:

Kinetic Energy: This can be defined as the energy of a body in motion. The Unit of kinetic energy is J.

Mathematically kinetic energy can be represented as,

K = 1/2mv²

(a)

K₁ = 1/2m₁v₁² ... Equation 1

Where K₁ = kinetic energy of the smaller bullet, m₁ = mass of the smaller bullet, v₁ = velocity of the smaller bullet.

Given: m₁ = 2.5 g = 2.5/1000 = 0.0025 kg, v = 42 m/s.

Substituting these values into equation 1

K₁ = 1/2 (0.0025) (42) ²

K₁ = 4.41/2

K₁ = 2.205 J.

(b)

K₂ = 1/2m₂v₂² ... Equation 2

Where K₂ = kinetic energy of the heavier bullet, m₂ = mass of the heavier bullet, v₂ = velocity of the heavier bullet.

Given: m₂ = 6.5 g = 6.5/1000 = 0.0065 kg, v₂ = 42.0 m/s

Substituting into equation 2,

K₂ = 1/2 (0.0065) (42) ²

K₂ = 11.466/2

K₂ = 5.733 J.

(c)

The ratio: K₂/K₁ = 5.733/2.205

K₂/K₁ = 2.6.