Two bullets have masses of 2.6 g and 5.7 g, respectively. Each is fired with a speed of 39.0 m/s.A. What is the kinetic energy of the first bullet?...

Given:

Mass of first bullet = m1 = 2.6 grams  = 2.6/1000 kg = 0.0026 kg

Mass of second bullet 2 = m1 = 5.7 grams  = 5.7/1000 kg = 0.0057 kg

common speed of the two bullets = v =30.0 m/s

Kinetic energy (K) is given by the formula:

K = (m*v^2)/2

Where m = mass and v = speed

Submitting values of mass and speed of the two bullets the above equation their kinetic energy is calculated below.

Kinetic energy of first bullet = (m1*v^2)/2 = (0.0026*39^2)/2 = 0.4563 J

Kinetic energy of second bullet = (m1*v^2)/2 = (0.0057*39^2)/2 = 1.00035 J

Ratio of kinetic energy of first and second bullet = m1/m2 = 5.7/2.6 = 2.1923

Answer:

A) 0.4563 J

B) 1.00035 J

C) 2.1923