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? Answer in units of J.
B. What is the kinetic energy of the second bullet? Answer in units of J.
C. What is the ratio K2/K1 of their kinetic energies?
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
A) 0.4563 J
B) 1.00035 J
The kinetic energy of the bullet is the product of half time its mass and square of its velocity = (1/2)m*v^2
Kinetic energy k1=(2.6/1000)kg*(39.0m/s)^2 =3.9546J for the 1st bullet.
Kinetic for the 2nd bullet, k2= (1/2)(5.7/1000 kg)(39 m/s)^2
K2/k1 = 8.6697/3.9546=2.1923