Two 3.4g bullets are fired with speeds of 42.9m/s and 77.4m/s, respectively a) if the kinetic energy of the first bullet is 3.128697 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?
3 Answers | Add Yours
The formula for kinetic energy is
Ek = 1/2(mv^2)
In this equation, m is the mass in kilograms and v is the velocity in meters per second. This produces an answer in Joules.
Using this formula, the kinetic energy of the second bullet is 10.18429 J.
We get this result by using the formula .5*(.0034*77.4*77.4), using the weight and speed of the second bullet as given in your question.
The ratio of their kinetic energies is found by dividing the kinetic energy of the second bullet by that of the first: 10.18429/3.128697.
This yields a ratio of 3.255122:1.
Please not that the kinetic energy information provided in the question is not essential for answering either part a) or part b) of the question. Kinetic energy of first bullet can also be calculated the same way as that that of second bullet.
Mass of the bullets = m 3.4 g = 3.4/1000 kg = 0.0034 kg
Speed of first bullet = v1 = 42.9 m/s
Speed of second bullet = v2 = 77.4 m/s
Kinetic energy of first bullet = K1 = 3.12869 J
b) Kinetic energy of second Bullet
Kinetic energy (K) of a moving object is given by the formula:
K = (m*v^2)/2
Therefore kinetic energy of the second bullet (K2) is given by:
K2 = (m*v^2)/2 = (0.0034*77.4^2)/2 = 10.184292 J
c) Ratio of kinetic energies
Ratio of kinetic Energies = K2/K1 = 3.12869/10.184292 = 3.255122
Please note that it is not necessary to calculate calculate or know the kinetic energies to calculate the ratio of kinetic energies. It can also be worked out as:
K2/K1 = (v2^2)/(v1^2) = (77.4^2)/(42.9^2) = 3.255122
The kinetic energy of a moving body is the product of half the mass and square of its velocity. When the units of mass is expressed in kiolograms and velocity is in meters, the generated kinetic energy is expressed in joules.Therefore,
K= (1/2)mv^2, where , K is the kinetic energy in joules of the body of mass of m kilograms moving with a velocity of v meters/second.
1sr bullets mass m = 3.4 grams =3.4/1000 kg =0.0034 kg.
1st bullete's velocity = 42.9 m/s
So,the kinetic energy K1 of the 1st bullet: K1 = (1/2) (0.0034 kg)(42.9m)^2 =3.128697J
The 2nd bullet's mass = 3.4g = 0.0034 kg, velocity =77.4m/s
Therefore, the kinetic energy, K2 of the 2nd bullet: K2= (1/2)(0.0034kg)(77.4m)^2 =10.184292 J
The ratio of the kinetic energies K2/K1 = 10.184292/3.128697 = 3.2551225
We’ve answered 319,199 questions. We can answer yours, too.Ask a question