Answer: The velocity of the first ball is 0.875 m/s if the second ball travels at 1.5 m/s after collision.
Explanation:
Given: [tex]m_{1}[/tex] = 0.2 kg, [tex]m_{2}[/tex] = 0.15 kg
[tex]v_{1}[/tex] = 2 m/s, [tex]v_{2}[/tex] = 0 m/s, [tex]v'_{1}[/tex] = ?, [tex]v'_{2}[/tex] = 1.5 m/s
Formula used is as follows.
[tex]m_{1}v_{1} + m_{2}v_{2} = m_{1}v'_{1} + m_{2}v'_{2}[/tex]
where,
v = velocity before collision
v' = velocity after collision
Substitute the values into above formula as follows.
[tex]m_{1}v_{1} + m_{2}v_{2} = m_{1}v'_{1} + m_{2}v'_{2}\\0.2 kg \times 2 m/s + 0.15 kg \times 0 m/s = 0.2 kg \times v'_{1} + 0.15 kg \times 1.5 m/s\\0.4 kg m/s + 0 = 0.2v'_{1} + 0.225 kg m/s\\0.2v'_{1} kg = 0.175 kg m/s\\v'_{1} = \frac{0.175 kg m/s}{0.2 kg}\\= 0.875 m/s[/tex]
Thus, we can conclude that the velocity of the first ball is 0.875 m/s if the second ball travels at 1.5 m/s after collision.
