Answer:
The magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).
Explanation:
We can find the velocity of the two balls after the collision by conservation of linear momentum and energy:
[tex] P_{1} = P_{2} [/tex]
[tex] m_{1}v_{1_{i}} + m_{2}v_{2_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}} [/tex]
Where:
m₁: is the mass of the ball 1 = 100 g = 0.1 kg
m₂: is the mass of the ball 2 = 300 g = 0.3 kg
[tex]v_{1_{i}}[/tex]: is the initial velocity of the ball 1 = 6.20 m/s
[tex]v_{2_{i}}[/tex]: is the initial velocity of the ball 2 = 0 (it is at rest)
[tex]v_{1_{f}}[/tex]: is the final velocity of the ball 1 =?
[tex]v_{2_{f}}[/tex]: is the initial velocity of the ball 2 =?
[tex] m_{1}v_{1_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}} [/tex]
[tex] v_{1_{f}} = v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}} [/tex] (1)
Now, by conservation of kinetic energy (since they collide elastically):
[tex] \frac{1}{2}m_{1}v_{1_{i}}^{2} = \frac{1}{2}m_{1}v_{1_{f}}^{2} + \frac{1}{2}m_{2}v_{2_{f}}^{2} [/tex]
[tex] m_{1}v_{1_{i}}^{2} = m_{1}v_{1_{f}}^{2} + m_{2}v_{2_{f}}^{2} [/tex] (2)
By entering equation (1) into (2) we have:
[tex] m_{1}v_{1_{i}}^{2} = m_{1}(v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}})^{2} + m_{2}v_{2_{f}}^{2} [/tex]
[tex] 0.1 kg*(6.20 m/s)^{2} = 0.1 kg*(6.2 m/s - \frac{0.3 kg*v_{2_{f}}}{0.1 kg})^{2} + 0.3 kg(v_{2_{f}})^{2} [/tex]
By solving the above equation for [tex]v_{2_{f}}[/tex]:
[tex]v_{2_{f}} = 3.1 m/s [/tex]
Now, [tex]v_{1_{f}}[/tex] can be calculated with equation (1):
[tex] v_{1_{f}} = 6.20 m/s - \frac{0.3 kg*3.1 m/s}{0.1 kg} = -3.1 m/s [/tex]
The minus sign of [tex] v_{1_{f}}[/tex] means that the ball 1 (100g) is moving in the negative x-direction after the collision.
Therefore, the magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).
I hope it helps you!
