a pool ball weighing 2 kg is traveling at 30 degree at 0.8 meter hits another ball moving at 0.5 m/s at 180 degree. if the second ball leaves the collision at 0 degree and the first moves away at 150 degree, find the final velocity vectors of the balls.
To find the final velocity vectors of the two balls after the collision, we can use the principles of conservation of momentum and conservation of kinetic energy.
Step 1: Determine the initial velocities of the balls.
The first ball is traveling at an angle of 30 degrees with a speed of 0.8 m/s. We can break down this velocity into horizontal and vertical components using trigonometry.
The x-component: vx1 = v1 * cos(30°) = 0.8 m/s * cos(30°) = 0.8 m/s * √3/2 = 0.6928 m/s.
The y-component: vy1 = v1 * sin(30°) = 0.8 m/s * sin(30°) = 0.8 m/s * 0.5 = 0.4 m/s.
The second ball is moving in the opposite direction (180 degrees) with a speed of 0.5 m/s. We need to consider the sign of the velocity later during calculations.
Step 2: Apply the conservation of momentum.
The total momentum before the collision equals the total momentum after the collision.
Momentum (p) = mass (m) * velocity (v).
Initial momentum: p_initial = (mass1 * v1) + (mass2 * v2)
= (2 kg * (0.8 m/s * √3/2)) + (-2 kg * 0.5 m/s)
= (2 kg * 0.5496 m/s) + (-1 kg * 0.5 m/s)
= 1.0992 kg·m/s - 0.5 kg·m/s
= 0.5992 kg·m/s.
Final momentum: p_final = (mass1 * v1') + (mass2 * v2')
= (2 kg * v1') + (-2 kg * v2').
Let v1' and v2' be the final velocities of the first and second balls, respectively.
Step 3: Apply the conservation of kinetic energy.
The total kinetic energy before the collision equals the total kinetic energy after the collision.
Kinetic Energy (KE) = (1/2) * mass * (velocity)^2.
Initial kinetic energy: KE_initial = (1/2) * mass1 * (v1)^2 + (1/2) * mass2 * (v2)^2
= (1/2) * 2 kg * (0.8 m/s)^2 + (1/2) * 2 kg * (0.5 m/s)^2
= 0.8 J + 0.5 J
= 1.3 J.
Final kinetic energy: KE_final = (1/2) * mass1 * (v1')^2 + (1/2) * mass2 * (v2')^2.
Step 4: Solve for the final velocities using the two equations.
We have two unknowns, v1' and v2', and two equations: p_final = 0.5992 kg·m/s and KE_final = 1.3 J.
Simplifying the momentum equation, we get:
2 kg * v1' - 2 kg * v2' = 0.5992 kg·m/s (equation 1).
Simplifying the kinetic energy equation, we get:
(1/2) * 2 kg * (v1')^2 + (1/2) * 2 kg * (v2')^2 = 1.3 J
(v1')^2 + (v2')^2 = 1.3 J / 2 kg
(v1')^2 + (v2')^2 = 0.65 m^2/s^2 (equation 2).
Now we have two equations (equation 1 and equation 2) with two unknowns (v1' and v2'). Solving these equations will give us the final velocities of the balls after the collision.