A stationary but moveable 450-kg cannon is fired and the 12-kg cannonball moves off to the right with a velocity of 20.0 m/s. What is the connon's velocity (both magnitude and direction) due to the kickback from the firing?
Masscannon*veloctiycannon+massbullet*velocity bullet=0 Momentum is conserved
cannons velocity= (1/masscannon)(-massbullet*velocitybullet)
10 m/s
To determine the cannon's velocity due to the kickback from the firing, you can use the principle of conservation of momentum. According to this principle, the total momentum before firing must be equal to the total momentum after firing.
The momentum of an object is given by the product of its mass and velocity:
Momentum = mass × velocity
Before firing, the cannon and the cannonball are at rest, so their momenta are both zero. After firing, the cannonball moves with a velocity of 20.0 m/s to the right. Let's assume the cannon's velocity due to kickback is V to the left.
Using conservation of momentum:
Total momentum before firing = Total momentum after firing
(0 kg) × (0 m/s) + (450 kg) × (0 m/s) = (12 kg) × (20.0 m/s) + (450 kg) × (V)
Simplifying the equation:
0 + 0 = 240 kg·m/s + 450 kg × V
0 = 240 kg·m/s + 450 kg × V
Rearranging the equation to solve for V:
450 kg × V = -240 kg·m/s
V = (-240 kg·m/s) / (450 kg)
V ≈ -0.533 m/s
Therefore, the cannon's velocity due to the kickback is approximately 0.533 m/s to the left. Note that the negative sign indicates the opposite direction to the cannonball.