The velocity of the 0.16 kg second ball after the collision with the first ball of equal mass is 1.2 m/s. The first ball's velocity before and after the collision is 1.2 and 0.85 m/s, respectively, and the second ball's velocity before the collision is 0.85 m/s.
We can find the velocity of the second ball after the collision by conservation of linear momentum:
[tex] p_{i} = p_{f} [/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 and m₂ are the masses of the first and second ball, respectively = m = 0.16 kg
[tex]v_{1}_{i}[/tex]: is the initial velocity of the first ball = 1.2 m/s
[tex]v_{1}_{f}[/tex]: is the final velocity of the first ball = 0.85 m/s
[tex]v_{2}_{i}[/tex]: is the initial velocity of the second ball = 0.85 m/s
[tex]v_{2}_{f}[/tex]: is the final velocity of the second ball =?
Taking the direction to the right as the positive direction of motion, we have:
[tex] m(v_{1}_{i} - v_{2}_{i}) = m(-v_{1}_{f} + v_{2}_{f}) [/tex]
[tex] v_{2}_{f} = v_{1}_{i} - v_{2}_{i} + v_{1}_{f} [/tex]
[tex] v_{2}_{f} = (1.2 - 0.85 + 0.85) m/s = 1.2 m/s [/tex]
Hence, the velocity of the second ball after the collision is 1.2 m/s to the right.
We can verify the answer by calculating the total kinetic energy before and after the collision, as follows:
[tex]K_{i} = \frac{1}{2}m(v_{1}_{i})^{2} + \frac{1}{2}m(v_{2}_{i})^{2} = \frac{1}{2}0.16 kg*[(1.2 m/s)^{2} + (0.85 m/s)^{2})] = 0.173 J[/tex]
[tex]K_{f} = \frac{1}{2}m[(v_{1}_{f})^{2} + (v_{2}_{f})^{2}] = \frac{1}{2}0.16 kg*[(0.85 m/s)^{2} + (1.2 m/s)^{2})] = 0.173 J[/tex]
Since the total kinetic energy before and after the collision is the same, the final velocity of the second ball after the collision is correct.
Therefore, the velocity of the second ball after the collision is 1.2 m/s.
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