First, calculate the total momentum before the collision:

\(0.4 \times 2.5 - 0.5 \times 1.5\)

\(= 1.0 - 0.75 = 0.25 \text{ kg m s}^{-1}\)

Let the speed of P after the collision be \(v\) and the speed of Q be \(2v\). Using conservation of momentum:

\(0.4 \times 2.5 - 0.5 \times 1.5 = 0.4v + 0.5 \times 2v\)

\(0.25 = 0.4v + v\)

\(0.25 = 1.4v\)

\(v = \frac{0.25}{1.4} \approx 0.179 \text{ m s}^{-1}\)

Alternatively, consider the opposite direction for P:

\(0.4 \times 2.5 - 0.5 \times 1.5 = -0.4v + 0.5 \times 2v\)

\(0.25 = -0.4v + v\)

\(0.25 = 0.6v\)

\(v = \frac{0.25}{0.6} \approx 0.417 \text{ m s}^{-1}\)

Thus, the two possible speeds of P after the collision are 0.179 m s^-1 or 0.417 m s^-1.