a bullet of mass ..21kg is fired from a gun of a mass 5.6kg. if gun is recoiled at a speed of 2.5 m/s .What speed must the bullet have as it leaves the gun
Assuming the gun/bullet combo was initially at rest, conserve momentum to get
(.21)(v) + (5.6)(-2.5) = 0
I think that's a heavy, slow bullet!
To determine the speed of the bullet as it leaves the gun, we can use the law of conservation of momentum. According to this law, the total momentum before the bullet is fired must be equal to the total momentum after the bullet is fired.
The momentum of an object can be calculated by multiplying its mass by its velocity. Let's denote the mass of the bullet as m1 and the velocity as v1, and the mass of the gun as m2 and the velocity of the gun as v2. The total initial momentum is given by:
Initial momentum = (mass of bullet x velocity of bullet) + (mass of gun x velocity of gun)
Final momentum is given by:
Final momentum = (mass of bullet x velocity of bullet) + (mass of gun x velocity of gun (after bullet is fired))
According to the law of conservation of momentum, the initial momentum is equal to the final momentum. Therefore,
(mass of bullet x velocity of bullet) + (mass of gun x velocity of gun) = (mass of bullet x final velocity of bullet) + (mass of gun x final velocity of gun (after bullet is fired))
Now, let's substitute the given values into the equation:
(0.21 kg x v1) + (5.6 kg x 2.5 m/s) = (0.21 kg x v2) + (5.6 kg x 0)
Simplifying the equation:
0.21 kg x v1 + 14 kg m/s = 0.21 kg x v2
Rearranging the equation to solve for v2:
v2 = (0.21 kg x v1 + 14 kg m/s) / 0.21 kg
Now, substitute the given value of v1 into the equation:
v2 = (0.21 kg x 2.5 m/s + 14 kg m/s) / 0.21 kg
Calculating the right side of the equation:
v2 = (0.525 kg m/s + 14 kg m/s) / 0.21 kg
v2 = 14.525 kg m/s / 0.21 kg
v2 ≈ 69.17 m/s
Therefore, the bullet must have a speed of approximately 69.17 m/s as it leaves the gun.