A spring cannon is fastened to a glider at rest on a horizontal frictionless air track. The combined mass of the cannon and the glider is 0.560kg. The cannon fires a ball with a mass of 50g horizontally at a speed of 14 m/s. How fast does the cannon-glider assembly move backward after the cannon is fired?
Use SI units in your answer.
v_{cannon-glider} =
To solve this problem, we can use the principle of conservation of momentum.
According to this principle, the total momentum before an event is equal to the total momentum after the event, assuming no external forces act on the system.
Let's denote the initial velocity of the cannon-glider assembly as V_initial, and the final velocity after the cannon is fired as V_final.
Before the cannon is fired, the cannon and the glider are at rest, so the initial velocity of the system is zero:
V_initial = 0 m/s.
After the cannon is fired, the mass of the cannon and the glider remains the same, but the velocity of the ball changes.
Given:
Mass of the cannon and glider (m) = 0.560 kg.
Mass of the ball (m_ball) = 50 g = 0.05 kg.
Velocity of the ball (v_ball) = 14 m/s.
To calculate the final velocity of the cannon-glider assembly, we can use the conservation of momentum equation:
(m + m_ball) * V_initial = (m + m_ball) * V_final.
Substituting the given values:
(0.560 kg + 0.05 kg) * 0 m/s = (0.560 kg + 0.05 kg) * V_final.
Simplifying the equation:
0 kg m/s = 0.61 kg * V_final.
Now, let's solve for V_final:
V_final = 0 kg m/s / 0.61 kg.
V_final = 0 m/s.
Therefore, the cannon-glider assembly does not move backward after the cannon is fired.