A 30-caliber rifle with a mass of 2.80 kg fires a 9.84- g bullet with a speed of 726 m/s with respect to the ground. What kinetic energy is released by the explosion of the gunpowder that fires the bullet?
by conservation of momentum
2.8 * speed of rifle back = .00984*726
solve for speed of rifle back, Vr
Ke = (1/2)(2.8)(Vr)^2 + (1/2)(.00984)(726)^2
To find the kinetic energy released by the explosion of the gunpowder, we can use the formula for kinetic energy:
Kinetic Energy = (1/2) * mass * velocity^2
In this case, the mass of the bullet is given as 9.84 grams, which we need to convert to kilograms by dividing by 1000:
mass = 9.84 g / 1000 = 0.00984 kg
The velocity of the bullet is given as 726 m/s.
Now we can plug these values into the formula:
Kinetic Energy = (1/2) * 0.00984 kg * (726 m/s)^2
≈ 0.00984 kg * 526476 m^2/s^2
≈ 5171.334944 J
Therefore, the kinetic energy released by the explosion of the gunpowder is approximately 5171.33 Joules.