Answer:
v₂ = - 0.666[m/s]
Explanation:
To solve this problem we must use the principle of conservation of linear momentum, that is, the momentum is conserved before and after the firing of the cannonball.
P = m*v
where:
P = linear momentum. [kg*m/s]
m = mass [kg]
v = velocity [m/s]
[tex]P_{before} = P_{after}[/tex]
Before shooting the ball, the cannon and the ball are together therefore their mass is added, also share the same velocity which is zero, since this group is at rest.
After the shot, the ball comes out at a speed, while the cannon moves backwards due to the reaction (this same effect can be seen in real cannons, as well as when someone fires a gun or rifle)
(m₁ + m₂)*v₁ = (m₁*v₂) + (m₂*v₃)
where:
m₁ = mass of the cannon = 1500 [kg]
m₂ = mass of the ball = 10 {kg]
v₃ = velocity of the ball after shooting = 100 [m/s]
v₂ = velocity of the cannon after shooting [m/s]
(1500 + 10)*0 = (1500*v₂) + (10*100)
0 = 1500*v₂ + 1000
- 1000 = 1500*v₂
v₂ = - 0.666[m/s]
The negative sign indicates that the cannon effectively moves it in the opposite direction to the ball at the time of the shot.
