A 5g bullet is fired from a 5000g gun. If the bullet leaves the gun going +800m/s, with what speed does the gun recoil?
momentum before = momentum after
0 = 5 (800) + 5000(v)
v = - 0.8 m/s
To find the speed at which the gun recoils when the bullet is fired, we can use the law of conservation of momentum. According to this law, the total momentum before the bullet is fired is 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, the mass of the gun as m2, the initial velocity of the bullet as v1, and the recoil velocity of the gun as v2.
Before the bullet is fired, the total momentum is given by:
Total Momentum Before = (Mass of Bullet) x (Initial Velocity of Bullet) + (Mass of Gun) x 0 (since the gun is initially at rest)
After the bullet is fired, the total momentum is given by:
Total Momentum After = (Mass of Bullet) x (Final Velocity of Bullet) + (Mass of Gun) x (Recoil Velocity of Gun)
According to the law of conservation of momentum:
Total Momentum Before = Total Momentum After
So, we have:
(m1 x v1) + (m2 x 0) = (m1 x 0) + (m2 x v2)
Since the mass of the bullet and gun are given as 5g and 5000g respectively (note: 1 kilogram (kg) = 1000 grams (g)), we can convert them to kilograms:
(m1 x 0.005kg x 800m/s) + (5kg x 0) = (0.005kg x 0) + (5kg x v2)
Simplifying the equation, we have:
0.004kgm/s = 5kg x v2
Now, we can solve for the recoil velocity of the gun (v2):
v2 = 0.004kgm/s / 5kg
v2 = 0.0008m/s
Therefore, the gun recoils at a speed of 0.0008 m/s in the opposite direction of the bullet.