Two bullets have masses of 3.1 g and 4.4 g, respectively. Each is fired with a speed of 42.0 m/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) 2.734 J

(b) 3.881 J

(c) 1.42

Explanation:

Kinetic Energy: This is the energy a body posses due to motion. The unit of kinetic energy is Joules (J). It can be expressed mathematically as

K = 1/2mv² ... Equation 1

Where K = kinetic energy, m = mass, v = velocity.

(a) Given: m₁ = 3.1 g, = 0.0031 kg, v = 42.0 m/s.

Substituting into equation one above,

K₁ = 1/2 (0.0031) (42) ²

K₁ = 2.734 J

Thus Kinetic Energy of the first bullet = 2.734 J

(b) Given: m₂ = 4.4 g = 0.0044 kg, v = 42 m/s.

Substituting into equation one above,

K₂ = 1/2 (0.0044) (42) ²

K₂ = 3.881 J

Thus Kinetic energy of the second bullet = 3.881 J

(c) K₂/K₁ = 3.881/2.734

K₂/K₁ = 1.42