Here, $m = 0.04 \, \text{kg}$. Position time graph shows that the particle movesfrom $x = 0$ at O to $x = 2$cm at A in 2 s. As x-t graph is a straight line, the motion is with a constant velocity $u = \frac{(2-0) \, \text{cm}}{(2-0) \, \text{s}}$ $= 1 \, \text{cm} \, \text{s}^{-1} = 10^{-2} \, \text{m} \, \text{s}^{-1}$ From $x = 2$cm at A, particle goes to $x = 0$ at B in 2 s. As AB is a straight line, motion is with constant velocity $v = \frac{(0-2) \, \text{cm}}{(4-2) \, \text{s}}$ $= -1 \, \text{cm} \, \text{s}^{-1} = -10^{-2} \, \text{cm} \, \text{s}^{-1}$ Negative sign indicates the reversal of direction of motion. This is being repeated. We can visualise a ball moving between two walls located at x = 0 and x = 2 cm, getting rebounded repeatedly on striking against each wall. On every collision with a wall, linear momentum of the ball changes. Therefore, the ball receives impulse after every two seconds.Magnitude of impulse = total change in linear momentum $= m u - m v = m(u - v)$ $= 0.04 [10^{-2} - (-10^{-2})]$$= 0.04 (10^{-2} + 10^{-2})$$= 0.08 \times 10^{-2} = 8 \times 10^{-4} \, \text{kg} \, \text{m} \, \text{s}^{-1}$