Figure 1 shows a fairground ride - AQA - A-Level Physics - Question 1 - 2022 - Paper 6

From the equation for power output, we have: $P = T \cdot \omega$.\n\nDuring the first 2 seconds (from the graph), the maximum angular velocity $\omega$ is approximately 1.5 rad/s. The net torque can be found from the maximum torque exerted during this period, calculated from the graph. Assuming a net torque of 960 N m:\n\n$P = T \cdot \omega = 960 \, N \, m \cdot 1.5 \, rad/s = 1440 \, W$\n\nSo, the power output is 1440 W.

The maximum torque $T_{max}$ can be identified from the graph by examining the steepest gradient during the first phase.\n\nHere, we find that during the initial segment of the torque, the maximum torque occurs at the slope around 2.0 seconds. Given that the angular velocity is maximized at this point, the torque value is about 960 N m.\n\nThus, the maximum torque during this period is 960 N m.

The angular impulse $J$ can be computed from the integral of torque over the time interval.\n\nBetween $t = 2.0 \, s$ and $t = 7.0 \, s$, the torque remains relatively constant and will be taken as its average. Assuming an average torque (based on the graph) of around 480 N m over the 5-second period:\n\n$J = \int_{t_1}^{t_2} T \, dt \approx T_{avg} \cdot (t_2 - t_1) = 480 \, N \, m \cdot (7.0 - 2.0) = 2400 \, N \, m \, s$\n\nThis results in an angular impulse of 2400 N m s.

Based on the torque characteristics observed in the angular velocity graph, the graph representing the variation of torque is the first one. Therefore, the correct choice is to tick the appropriate box corresponding to the first graph.